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Solutions to

LE. Irodev's Problems in General Physics Volume I Mechanics • Heat • Electrodynamics

SECOND EDITION

ABHAY KUMAR SINGH Director Abhay's 1.1. T. Physics Teaching Centre Patna-6

CBS

CBS PUBLISHERS & DISTRIBUTORS 4596/1A, 11 DARYAGANJ, NEW DELHI -110 002 (INDIA)

Dedicated to my Teacher

Prof. (Dr.) J. Thakur (Department of Physics, Patna University, Patna-4)

ISBN : 81-239-0399-5

First Edition : 1995 Reprint : 1997 Second Edition : 1998 Reprint : 2000 Reprint : 2001 Reprint : 2002 Reprint : 2003 Reprint : 2004 Reprint : 2005 Copyright © Author & Publisher All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system without permission, in writing, from the publisher. Published by S.K. Jain for CBS Publishers & Distributors, 4596/1A, 11 Darya Ganj, New Delhi - 110 002 (India) Printed at India Binding House, Delhi - 110 032

FOREWORD

Science, in general, and physics, in particular, have evolved out of man's quest to know beyond unknowns. Matter, radiation and their mutual interactions are basically studied in physics. Essentially, this is an experimental science. By observing appropriate phenomena in nature one arrives at a set of rules which goes to establish some basic fundamental concepts. Entire physics rests on them. Mere knowledge of them is however not enough. Ability to apply them to real day-to-day problems is required. Prof. Irodov's book contains one such set of numerical exercises spread over a wide spectrum of physical disciplines. Some of the problems of the book long appeared to be notorious to pose serious challenges to students as well as to their teachers. This book by Prof. Singh on the solutions of problems of Irodov's book, at the outset, seems to remove the sense of awe which at one time prevailed. Traditionally a difficult exercise to solve continues to draw the attention of concerned persons over a sufficiently long time. Once a logical solution for it becomes available, the difficulties associated with its solutions are forgotten very soon. This statement is not only valid for the solutions of simple physical problems but also to various physical phenomena. Nevertheless, Prof. Singh's attempt to write a book of this magnitude deserves an all out praise. His ways of solving problems are elegant, straight forward, simple and direct. By writing this book he has definitely contributed to the cause of physics education. A word of advice to its users is however necessary. The solution to a particular problem as given in this book is never to be consulted unless an all out effort in solving it independently has been already made. Only by such judicious uses of this book one would be able to reap better benefits out of it. As a teacher who has taught physics and who has been in touch with physics curricula at I.I.T., Delhi for over thirty years, I earnestly feel that this book will certainly be of benefit to younger students in their formative years.

Dr. Dilip Kumar Roy Professor of Physics Indian Institute of Technology, Delhi New Delhi-110016.

FOREWORD

A. proper understanding of the physical laws and principles that govern nature require solutions of related problems which exemplify the principle in question and leads to a better grasp of the principles involved. It is only through experiments or through solutions of multifarious problem-oriented questions can a student master the intricacies and fall outs of a physical law. According to Ira M. Freeman, professor of physics of the state university of new Jersy at Rutgers and author of "physic—principles and Insights" -"In certain situations mathematical formulation actually promotes intuitive understanding Sometimes a mathematical formulation is not feasible, so that ordinary language must take the place of mathematics in both roles. However, Mathematics is far more rigorous and its concepts more precise than those of language. Any science that is able to make extensive use of mathematical symbolism and procedures is justly called an exact science". I.E. Irodov's problems in General Physics fulfills such a need. This book originally published in Russia contains about 1900 problems on mechanics, thermodynamics, molecular physics, electrodynamics, waves and oscillations, optics, atomic and nuclear physics. The book Jas survived the test of class room for many years as is evident from its number of reprint editions, which have appeared since the first English edition of 1981, including an Indian Edition at affordable price for Indian students. Abhay Kumar Singh's present book containing solutions to Dr. I.E. Irodov's Problems in General Physics is a welcome attempt to develop a student's problem solving skills. The book should be very useful for the students studying a general course in physics and also in developing their skills to answer questions normally encountered in national level entrance examinations conducted each year by various bodies for admissions to professional colleges in science and technology. B.P. PAL Professor of Physics I.I.T., Delhi

PREFACE TO THE SECOND EDITION Nothing succeeds like success, they say. Now, consequent upon the warm welcome on the part of students and the teaching fraternity this revised and enlarged edition of this volume is before you. In order to make it more up-to-date and viable, a large number of problems have been streamlined with special focus on the complicated and ticklish ones, to cater to the needs of the aspiring students. I extend my deep sense of gratitude to all those who have directly or indirectly engineered the cause of its existing status in the book world.

Patna June 1997

Abhay Kumar Singh

PREFACE TO THE FIRST EDITION

When you invisage to write a book of solutions to problems, one pertinent question crops up in the mind that—why solution! Is this to prove one's erudition? My only defence against this is that the solution is a challenge to save the scientific man hours by channelizing thoughts in a right direction. The book entitled "Problems in General Physics" authored by I.E. Irodov (a noted Russian physicist and mathematician) contains 1877 intriguing problems divided into six chapters. After the acceptance of my first book "Problems in Physics", published by Wiley Eastern Limited, I have got the courage to acknowledge the fact that good and honest ultimately win in the market place. This stimulation provided me insight to come up with my second attempt—"Solutions to I.E. Irodov's Problems in General Physics." This first volume encompasses solutions of first three chapters containing 1052 problems. Although a large number of problems can be solved by different methods, I have adopted standard methods and in many of the problems with helping hints for other methods. In the solutions of chapter three, the emf of a cell is represented by (xi) in contrast to the notation used in figures and in the problem book, due to some printing difficulty. I am thankful to my students Mr. Omprakash, Miss Neera and Miss Punam for their valuable co-operation even in my hard days while authoring the present book. I am also thankful to my younger sister Prof. Ranju Singh, my younger brother Mr. Ratan Kumar Singh, my junior friend Miss Anupama Bharti, other well wishers and friends for their emotional support. At last and above all I am grateful to my Ma and Pappaji for their blessings and encouragement. ABHAY KUMAR SINGH

CONTENTS Foreword Preface to the second edition Preface to the first edition

iii vi

PART ONE PHYSICAL FUNDAMENTALS OF MECHANICS 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Kinematics The Fundamental Equation of Dynamics Laws of Conservation of Energy, Momemtum, and Angular Momentum Universal Gravitation Dynamics of a Solid Body Elastic Deformations of a Solid Body Hydrodynamics Relativistic Mechanics

1-34 35-65

66-101 102-117 118-143 144-155 156-167 168-183

PART TWO THERMODYNAMICS AND MOLECULAR PHYSICS 2.1 2.2 2.3 2.4 2.5 2.6 2.7

Equation of the Gas State. Processes The first Law of Thermodynamics. Heat Capacity Kinetic theory of Gases. Boltzmann's Law and Maxwell's Distribution The Second Law of Thermodynamics. Entropy Liquids. Capillary Effects Phase Transformations Transport Phenomena

184-195 196-212 213-226 227-241 242-247 248-256 257-266

PART THREE ELECTRODYNAMICS 3.1 3.2 3.3 3.4 3.5 3.6 3.7

Constant Electric Field in Vacuum Conductors and Dielectrics in an Electric Field Electric Capacitance. Energy of an Electric Field Electric Current Constant Magnetic Field. Magnetics Electromagnetic Induction. Maxwell's Equations Motion of Charged Particles in Electric and Magnetic Fields

267-288 289-305 306-324 325-353 354-379 380-407 408-424

PART ONE

PHYSICAL FUNDAMENTALS OF MECHANICS

1.1

KINEMATICS

1.1 Let v

be the stream velocity and v' the velocity of motorboat with respect to water. The o motorboat reached point B while going downstream with velocity (vo + v') and then returned

with velocity (v' — vo) and passed the raft at point C. Let t be the time for the raft (which flows with stream with velocity vo) to move from point A to C, during which the motorboat moves from A to B and then from B to C. Therefore (vo+ v')-v — I j v 0

v

0)

A

On solving we get v_

•.s--------(Vat 171)1 - - - 1

2

1.2 Let s be the total distance traversed by the point and t1 the time taken to cover half the distance. Further let 2t be the time to cover the rest half of the distance. —= vo t1 2

Therefore

s —

and

2

or

(vi + v2) t or a

(1)

t = 1 2 vo

+

V2

(2)

Hence the sought average velocity + ) 2 v (v1 v2 mg v1 + v2 + 2 v 2t [s/2 vo ]+ [ s/(vi + v2).] o 13 As the car starts from rest and finally comes to a stop, and the rate of acceleration and deceleration are equal, the distances as well as the times taken are same in these phases of motion. s

s

Let At be the time for which the car moves uniformly. Then the acceleration / deceleration .ti—At i t me s each. So, 2

it =

or

A

Hence

— At )1 + 4 j

1 21 w

22 az Es

2

4 w

V 1 — 4

At =

At ) At 2

= 15s.

W

1.4

(a) Sought average velocity

s

t

1111111111111■

SRI

20 1111111111111111111111SIMIR

200 cm = 10 cm/s 20s

1111111111111111M1101111 1111111111111164 11111110

ds

(b) For the maximum velocity, — should be dt maximum. From the figure dis maximum for all points on the line -ac, thus the sought maximum velocity becomes average velocity for the line ac and is equal to :

be= ab

111111111115M111111111111111 1111111115111 11111111111111 ■11111111■ 111111MICUIP 11121111MIM EMU gite:211111111111111111111111■

cm 4s

25 cm/s

ds should pass through the (c) Time to should be such that corresponding to it the slope ds s — = To-. From figure the tangent at point d point 0 (origin), to satisfy the relationship dt passes through the origin and thus corresponding time t = to =

16 s.

1.5 ,Let the particles collide at the point A (Fig.), whose position vector is 73*(say). If t be the time taken by each particle to reach at point A, from triangle law of vector addition : 2t

3= 1 +u t— 2+ r1— r2

SO,

(v2 — v 1) t

(1) (2)

0

or,

--a. —0 ri r2 I

r2I

=

vz— v1 I v2 — v 11

, which is the sought relationship.

1.6 We -have V

= V—V0

From the vector diagram [of Eq. (1)] and using properties of triangle

(1)

3

v' — V v(2) + v2 + 2 vo v cos cc, and

v' sin (it — cp)

v

=

= 39.7 km/hr v sin cp

or, sin 0 —

sm 0

(2)

vi

0 . sin_i (v sin q)) v'

or

Using (2) and putting the values of v and d

i

0 = 19.1°

1.7 Let one of the swimmer (say 1) cross the river along AB, which is obviously the shortest path. Time taken to cross the river by the swimmer 1. t1=

d . 7.7..v.2.._ , (where AB = d is the width of the river)

(1)

o For the other swimmer (say 2), which follows the quickest -path, the time taken to cross the river. d t2 = —,v

A

(2)

B.i „ r . _ . . . _ •. . . . x ______.).0

d

1

1

VI

A

In the time t2, drifting of the swimmer 2, becomes x = vo t2 =

vo --i v

d, (using Eq. 2)

If t3 be the time for swimmer 2 to walk the distance x to come from C to B

°

(Fig.), then

v0 d

X

1..1 at

(3)

— -•U

(using Eq. 3)

V? U

According to the problem ti. = t2 + t3 d or, V7 . V; v — 2

d vo d 1 + v Vu

On solving we get vo u

= 3 km/hr.

_1 i V

12

- 1

(4)

4

1.8

Let

1

the

b e

t he

r i v et

d i s t a

Let

w a t e r .

nc e

vo

Th e r e f or e

c ov e r e d

b e

the

t i me

b y

strea

th e

boat

m

vel oc i t y

b y

the

t a k en

A

a

a l o ng

nd

the

v '

b oat

A

in

the

its

1

riv e r

as

vel oc i t y

w ell

o f

as

b y

e a c h

th e

b o a t

boa t

B

w ith

res

ac

n

pe c t

j ourne y

1

tA VI

a nd

for

th e

bo

Vo

-

V o

1

B

a t

4-

1

t

2/

+ B V

-

V2 0

V

v

VI2 -

o2

V

112

to wh e r e

Henc e , VI2 -

tB

On

1 . 9

sub s t i t u t i o n

Le t

v o

be

tA / t B

th e

s t r e a

m

ve l

V t)

V

th e

ve l

i

)

1

T12 -

F8

o c i t y

a nd

v '

oc i t y

o f

bo a t

wi t h

r e s pe c t

t o

wa t e r .

A

v o -7

Let

=

2

i 7v

m a ke

>

0,

s ome

a n

d r i f t i n g

a ngl e

0

of

with

t th i s

tim e

l f o w

(wh e r e V

In

boa t

i nt e r v a l ,

is

i nev i t a bl e .

d i r e c t i

d

i s

o f

t he

on

(Fig . ) ,

t h e w i d th

th e

t hen

of

th e

ri

t i me

ta

ken

to

c r os s

th e

ri ve

v e r )

0

s i n

dr i f t i n

h t e

g

bo a t

x= (v' cos 0 + vo) t (v' cos 0 + v„)

= v'

x• •

Fo r

d

ot

0

( c ot

0

+

cos e c

0

+

co s e c

0 )

0,

whi c h

y i e l d s

1

1 0

-

11

He nc e ,

Th e

0

so l

ut i on

e

o f

bo d y

d

(c

cos

1 . 1 0

0 )

0

d r i f t i ng )

(mi ni mu m

d

sin

t hi s

probl e

t hrown

up

ght

0

1 20°

be c o me s

m

s t r a i

=

2

si

be

1

mpl e

a nd

in

the

t he

o t h e r

fra

me

bo d y

be

with

a t t a c h ed

2,

th e

n

Let th f r a

me

of

2

So ,

f r o m

...►

-10

r12-

ro

T12 'a

Of

'

ra

But

1

1 7 :1 1 . .

j

=

kinema t i c

th e

from

(be c a

1

4( 1 2 ) 1

1

v+0 2 1 =

H e nc e ,

the

so

o f

2 Vo

VO(12) u gh t

for

on

nt

a c c e l e r a t i on

W12

-2

us e

c on s t a

42

1 -1'

v00 2 ) t ,

proper t i e s

qu a t i

(12) + VO (12)

i i c2

=

0

t

and

r

2)

=

0)

( 1 )

'1

So ,

e

o

t ria

ngle

2 Vo —

dista

nc e

1 7

: 2 1

=

2

v0

vo

1 /

vo

c os

2

(1

( 7 t /2

-

-

0 0 )

sin 0) t=

2

2

m

:

for

o ne

the

of

the

b od y

bod i e s .

1

in

th e

5 1 . 1 1 Let the velocities of the paricles (say ii17' and v2

) becomes mutually perpendicular after

time t. Then their velocitis become ' V1

' =

Vi+gt;

V2

it

gt

V2

(1)

N

.

V; Hence,

t

{171 72

(3)

g Now form the Eq. r

=r 12

;Tz —

+v 0(12)

t+ 0(12)

a12)j2) I t, (because W12

1

-0

2

W I12

2

= 0 and 02) = 0)

here Hence the sought distance v1 v2

+ v2

(as I v—0*02) = v1 + v2)

r12 I.— g 1.12 From the symmetry of the problem all the three points are always located at the vertices of equilateral triangles of varying side length and finally meet at the centriod of the initial equilateral triangle whose side length is a, in the sought time interval (say t ). --> 15

Let us consider an arbitrary equilateral triangle of edge length 1 (say). Then the rate by which 1 approaches 2, 2 approches 3, and 3 approches 1, becomes :

— dl

V - V COS (

2n --)

3

dt 0

On integrating :

f al = 3v— a

2

f dt 0

2a

a= — 2

3

vt so t= — 3v

6 1.13 Let us locate the points A and B at an arbitrary instant of time (Fig.). If A and B are separated by the distance s at this moment, then the points converge or — ds point A approaches B with velocity = v — u cos a where angle a varies with time. dt On intergating, ds = f (v — u cos a) dt,

f

(where T is the sought time.) or

1= f (v — u cos a) dt 0

As both A and B cover the same distance in x-direction during the sought time interval, so the other condition which is required, can be obtained by the equation ex=

So,

fx v

dt

uT = f v cos a dt

Solving (1) and (2), we get T

(2)

—u

One can see that if u = v, or u < v, point A cannot catch B. 1.14 In the reference frame fixed to the train, the distance between the two events is obviously equal to L Suppose the train starts moving at time t= 0 in the positive x direction and take the origin ( x = 0) at the head-light of the train at t = 0. Then the coordinate of first event in the earth's frame is 1 2, x1= —2 WI and similarly the coordinate of the second event is x,4

1 — w(t +

—1

The distance between the two events is obviously. xi —x2= 1— wt (t 4- t/2 ) = 0.242 km in the reference frame fixed on the earth.. For the two events to occur at the same point in the reference frame K, moving with constant velocity V relative to the earth, the distance travelled by the frame in the time interval T must be equal to the above distance. Thus So,

V'T mg 1 — w t + T/2) —/— w ( t + T/2 )

4.03 m/s

The frame K must clearly be moving in a direction opposite to the train so that if (for example) the origin of the frame coincides with the point x1 on the earth at time t it coincides with the point x2 at time t + T.

7 1.15 (a) One good way to solve the problem is to observer at its bottom (Fig.).

work in the elevator's frame having the

Let us denote the separation between floor and of celing by h = 2.7 m. and the acceleration the elevator by w = 1.2 m/s2 From the kinematical formula Y= Here and

yo + voy t +1 wy t

(1)

2

y = 0, yo = + h, vg, = 0 wy

watt (y) (- g)

Wele (y)

- (w) =

(g + w)

So,

1 0= h+ - {--(g+w)}t2 2

or,

t=

= 2 -7 m. e t1.0=1.2 in

=0.7 s.

g+w (b) At the moment the bolt loses contact with the elevator, it has already aquired the velocity equal to elevator1given by : vo = (1.2) (2) =

2.4 m/s

In the reference frame attached with the elevator shaft (ground) and pointing the y-axis upward, we have for the displacement of the bolt.,

r •

= voy t +1 wy t2 T

1 = v„t+-., (-g)t22 or,

Ay = (2.4) (0.7) +

a

4

(- 9-8) (0.7)2 =

f

- 0.7 m.

2 Hence the bolt comes down or displaces downward relative to the point, when it loses contact with the elevator by the amount 0.7 m (Fig.). Obviously the total distance covered by the bolt during its free fall time

s= I Ay

1.16 Let the particle 1 and

+ 2 (yg) = 0.7 m +

(2-4) 2 m= 1.3m. (9.8)

2 be at points B and A at t = 0 at the distances /1 and

12 from

intersection point 0. Let us fix the inertial frame with the particle 2. Now the particle 1 moves in relative to this reference frame with a relative velocity712 = v its trajectory is the straight line BP. Obviously, the minimum distance between the particles is equal to the length of the perpendicular AP dropped from point A on to the straight line BP (Fig.).

8

From Fig. (b),

v12

vi and tan 0 = V2

vi

(1)

The shortest distance

AP= AM sin 0 = (OA - OM) sin 0 = (12 - /1 cot 0) sin 0 v2 or

v

AP = (12- 11H _ 1

71

, -,

vi v vi + v-2

v1 I2 -

V2 11

,

(using 1)

7

v vi + V2

The sought time can be obtained directly from the condition -that (4 - v1 t)2 + (12 - v2 t)2 2 v2 11 v1 + 1 is minimum. This gives t = 2 2 • Vi + V2

1.17 Let the car turn off the highway at a distance x from the point D. So, CD = x, and if the speed of the car in the field is v, then the time taken by the car to cover the distance AC = AD - x on the highway

ti=

AD - x ri v

(1) C c

and the time taken to travel the distance CB in the field ‘112 4-X2 t2 -

(2)

V

So, the total time elapsed to move the car from point A to B

t ti+ t2

,, AD - x 1V

12 + V

For t to be minimum

dt dx Or

1 [

0

Or

12x2.

1

- - + I, 11 V 12 +x2 2 +x2 or 1

=0

X

9

1.18 To plot x (t), s (t) and wx (t) let us partion the given plot vx (t) into five segments

(for

detailed analysis) as shown in the figure. For the part oa: wx= 1 and vx = t = v t t2 Thus, 0x1 (t) = fvx dt = fdt = — = si (t) 2

0

Putting t = 1, we get, A xi = s =

a-

46a2

Putting

1

0

1 unit

2

3

4

e 5 6

-1

-2

d

For the part ab : wx = 0 and vx = v = constant =

Thus

b

VX

1

= f vx dt = f dt = (t -1) = s2 (t) t = 3, Ax2 = s2 = 2 unit

For the part b4 : wx =

1 and vx = 1 - (t - 3) = 4 - t) = v

Thus

Putting For the part 4d : So,

Thus Putting Similarly Putting

t2 A x4 (t) = f (1 - t) dt = 4 t 4 t= 6, Ax4= -1

-8

2 t s4 (t) = f I vx I dt = f (t - 4) dt = —2 - 4t + 8 4 t = 6, 54 = 2 unit

For the part d 7 :

wx= 2 and vx = - 2 + 2 (t - 6) = 2 (t - 7) v= I vx I

2 (7 - t)

for

t 4-- 7

6 Now,

Ax (t ) = f2 (t - 7) dt = t

Putting

- 14t + 48

t = 4, bas = -1 6

Putting Similarly

2

s5

= f 2 (7 - t) dt = 14 t - t 2 - 48 t = 7, s5 = 1

On the basis of these obtained expressions wx (t), x (t) and s (t) plots can be easily plotted as shown in the figure of answersheet.

10 1.19

(a) Mean velocity —

Total distance covered Time elapsed

s =

R =

50 cm/s

=

V

(1)

(b) Modulus of mean velocity vector -D. I A Fi = 2R II= — = 32 cm/s At (c) Let the point moves from i to at

(2) along the half circle (Fig.) and vo and v be the spe

f

the points respectively. dv ■■• 1111

We have

w

dt

or,

f So,

( v o

w,

+

t)

dt

vo + (vo + wit)

=

vo

2

f

+V

2

fit

So, from (1) and (3) +

n R

2 Now the modulus of the mean vector of total acceleration I < i;7> I

1.20

i(7, I

vo + v

I All= At

(see Fig.)

(a) we have

d

and

w

-

2

a

a

dt

(b) From the equation

51 (1 — a t), 1

71." 0, at t= 0 and also at t= At

a

1 So, the sought time At = a As

So,

- 2 a t)

v

1o=

a (1 — 2

a

t )

for s t

v=

a(2at-1)

fort>

1 2a 1 2a

(3

11

Hence, the sought distance 1/2 a

1/a

Simplifying, we get, s = 1.21

t 2c (a) As the particle leaves the origin at t = 0 So,

Ax = x = f vx dt

(1)

v = Ti;(1 — —,:),

As

where vo). is directed towards the +ve x-axis v

So,

x = v0 (1 — —t)

(2)

t

From (1) and (2), r (3)

Hence x coordinate of the particle at t =

Similarly at

6 s.

x= 10 x 6 (1 — — —) = 24 cm = 0.24m 15 2x t = 10 s 10

x= 10 x 10 (1 — 2 and at

)— 0 5

t= 20s

x= 10 x 20 (1 —

° = —200 cm = — 2 m 2x 5 (b) At the moments the particle is at a distance of 10 cm from the origin, x = ± 10 cm. Putting

x = + 10 in Eq. (3)

10= 10t 1- 01) or, t

2 — 10t + 10 = 0,

10± V 100 — 40 So,

t=t=

2 x = — 10in Eqn (3)

Now putting

— 10 = 10 (1 — 1 10 ' On solving,

t = 5 ± 075 s

As t cannot be negative, so, t = (5 +

) ss

= 5± NifY s

12

Hence the particle is at a distance of 10 cm from the origin at three moments of time : t = 5 ± -43- s, 5 + Z-5- s (c) We have

v '70 (I- — 1 x

So,

vo (1 —1 t

v= NI =

vo (I- — 1) t

forts x

for t > T

r s= f vo (11 — IT) dt for t s t = vo t (1 — fit) 0

So

Y

r

and

r>T

(A) 4

4

V0

S=f

0

And for t = 8 s ■

5

(1

8

s= f 10 (1 - i) dt + f 10 (5 - 1) d t 0

5

On integrating and simplifying, we get s= 34 cm. On the basis of Eqs. (3) and (4), x (t) and s (t) plots can be drawn as shown in the answer sheet. 1.22 As particle is in unidirectional motion it is directed along the x-axis all the time. As at

t= 0, x= 0 So, Therefore, or, (1)

As, r v ,,,2

On integrating,

f dv . f =

a2 Or,

v=

2

t

(2)

o

o

2

dt

13

(b) Let s be the time to cover first s m of the path. From the Eq. s= f v dt

r s=

a2

2

f

0 or

t.

a2 t2

,

a t = -- -2 2

(using 2)

2 _ f— vs a

(3)

The mean velocity of particle 2 iS-/a

a2

— f V (t)

=

1.23

dt

fdt

=

f0

2

t dt a 'is2

2 'IF/ a

According to the problem

-

v dv

ds

.. a y (as v decreases with time)

0 s - f Vv- dv = a f ds v0 o

or,

2 On integrating we get s = — vo

3a

3/2

Again according to the problem

- cc: = a Nr; or - 41-' = a dt v 0 Ndc v .

or,

a .1* dt vo 2 V— vo

Thus

1.24 (a) As So, and therefore

t=

--+

a

r = att-.. bt2 j

A

3

..-1.

x= at, y= -bt2 —bx2 a2 Y —

X

14

which is Eq. of a parabola, whose graph is shown in the Fig. v

(b) As

dr

--11. =

atr-bt2j*

-

= at-rill.2btr

d t

(c)

cos a =

17t11.

(art2btn.( -2k1 1

vw or,

(1)

a2 + 4 b2 t2) 2b 2bt

cos a-

Va2+4b2t2 so, or,

tan a a= tanLan

_

a 2bt

a 2bt

(d) The mean velocity vector

-+ Hence,

i7dt

f(ait 2 btndt

fdt

at-btj t

a2+ jI= V (-bt)2 =

V112 +b2 t2

1.25 (a) We have x= at and y= at(1-at)

(1)

Hence, y (x ) becomes, ax (i-ax) Y= a = a

2 a x (parabola)

(b) Differentiating Eq. (1) we get vx= a and vy= a(1-2at)

(2)

-

15

So,

v= Vv2 +v2 x

=aV1 +(1 - 2 at)2 y

Diff. Eq. (2) with respect to time wx = 0 and w = - 2 a cc z w= "vi 14/2+ w = 2aa

So,

(c) From Eqs. (2) and (3) -0 v- ai+a(1-2at)i

We have So,

7r cos

v•w

1

—

and w = 2aa -a(1 - 2 ato)2 act

—

,

4

aV

+(1-2ato)2 2aa

On simplifying.

1- 2 a to= *1

As,

to PI

1 0 ,

rf

=

a

t'

1.26

Differentiating motion law : x= a sin wt, y= a(1- cos w t ), with respect to time, = a co cos cot , vy = a co sin cot So,

cia.= a w cos cot

and

a w sin coti

(1)

v = aw= Coast.

(2)

Differentiating Eq. (1) with respect to time --10

dV,11' 2 • 2 = - a w sm cott+aw cos cot)

$4, =

(3)

—dt

(a) The distance s traversed by the point during the time t is given by

s =

( using 2 )

vdt- facodt- a cot

f

Thus, 1711 w, i.e., the angle between velocity vector and acceleration vector equals —71 2 1.27

Accordiing to the problem w= w(-j) dvx So,

= wx

dvv = 0 and w =

dt

Y

= -w dt

Differentiating Eq. of trajectory, y = a x - b x 2, with respect to time adx _ dt

dt

2bx dx dt

16

dyl dt

So,

x-0

. a dx dt

x..0

,Again differentiating with respect to time d2y _

2

dx

ad

2/3 (-cii)

d t2

d r2 d x

or,

-2bx

- w = a (0) - 2 b ( -c

2

Tt

d x cI t =

or,

d2X dt2

-2

.1177 2L

b x (0) (using 1)

(using 1)

(-

117, Using (3) in (2)

ifdlt I

I x- o

=

a

2b

Hence, the velocity of the particle at the origin

v=

v = Viib'-

Hence,

1.28

(1 + a2)

As the body is under gravity of constant accelration r, it's velocity vector and displacemen vectors are:

--a. --). v = vo + g

--0

t

(1:

—3.

and

Ar =

r=v

,--). 1 2 = t+— t r 0 2g k

_ u at t = 0)

So,

=

r

--3IP

vo + 2

Hence from Eq. (3), ---OP =

gt_

--I.

— = = At t

(3)

over the first t seconds

---)

VA + u

t

(4)

2

But we have v = v0 at t =

0 and

Also at t = t (Fig.) (also from energy conservation)

(2;

17 Hence using this propety in Eq. (5) vii = vo + 2 (1.70'•git + g2-c2 2 (1,0 • e As

t * 0,

SO, t-

2 g

Putting this value of t in Eq. (4), the average velocity over the time of flight , „(vo • gi = vo - g

2 g

t.29 The body thrown in air with velocity vo at an angle a from the horizontal lands at point P on the Earth's surface at same horizontal level (Fig.). The point of projection is taken as origin, so, Ax = x and Ay = y (a) From the Eq. Ay = v0 J.+ 2-

1

wy t 2 1

0= vo sin As

t * 0,

at-

gt2

2 2 vo sin a

so, time of motion t =

g (b) At the maximum height of ascent, vy =

0

so, from the Eq.

2 2 vo sin

Hence maximum height H =

a

2g

During the time of motion the net horizontal displacement or horizontal range, will be obtained by the equation Ax= v, t + -

1 - w t2 2 x

1 R= vo cos a t - - (0) t2 = vo cos a 2 T-

Or,

when

v2,, sin 2 a g

R= H 2

vo sin a g

2

vo sin

2

a or tan a= 4, so, a= tan-1 4

2g

(c) For the body, x (t) and y (t) are x = vo cos a t

(1)

18

y= vo sin a t - —

and

t2 2g Hence putting the value of t from (1) into (2) we get, 2 2 x 1 x gX y = vo s ) = x tan a a „ in (v cos a)co - 2 g (vo cos a 2 vo cos- a'

(2)

Which is the sought equation of trajectory i.e. y (x) (d) As the body thrown in air follows a curve, it has some normal acceleration at all the moments of time during it's motion in air. At the initial point (x = 0, y = 0), from the equation : Wn = R—R , (where R is the radius of curvature) 2 Vo g cos a = — (see Fig.) or Ro =

2 Vo

Ro

g cos a

At the peak point vy = 0, v = vx = vo cos a and the angential acceleration is zero. 2 Now from the Eq. wn = R 2 2 Vo cos a v20 cos2 a g= R , or R-

g Note : We may use the formula of curvature radius of a trajectory y (x), to solve part (d), 2 3 r [1 + (dy / dx) R= :_ld2Y/dr21 1.30 We have, vx = vo cos cc, vy = vo sm a - gt

Thus Now, v (vo sin cc - g t) . -g TY r

= - —g V t

.g iv I I wt I ___ _z_

Hence

v

2 =

Now w= 1/ IV - w n

g2 _ g2 2122.

t 11?

or

Vx Wn = g — (where vx = v

1./ 2 2 \ vt - vy )

19 A uring time of motion 7 I v„ d

As

v w = w = - g --L v t v On the basis of obtained expressions or facts the sought plots can be drawn as shown in the figure of answer sheet. (1) (Ox) at point 0 (origin) with velocity vo = Viiii 1.31 The ball strikes the inclined plane As the ball elastically rebounds, it recalls with same velocity vo, at the same angle a from the normal or y axis (Fig.). Let the ball strikes the incline second time at P, which is at a distance 1 (say) from the point 0, along the incline. From the equation 1 y= voy

t + -i wy t2

1 0= vo cos a lc — — g cos a t2 2 where t is the time of motion of ball in air while moving from 0 to P. 2 vo As T * 0, SO, T (2) g Now from the equation. x= v Ck

1 2 t+- w t 2 x

1= vo sin a t 1- -1 g sin a T2 2

1=

SO,

vo sin a (2 vo) + i1 _ sn a (2 vo) )

S

26

g

•

4 vo2 sm a

(using 2)

g Hence the sought distance, 1=

4 (2 gh) sin a =

8h sin a (Using Eq. 1)

g 1.32 Total time of motion 2 vo sin a t—

II_

9.8 t

2 vo

2 x 240

or sin a = g

(1)

and horizontal range R R . vo cos a t or cos a = vo T

=

5100 85 = 2401 4t

From Eqs. (1) and (2) ,T24. (85)2 (9.8)2 ( (480)2 4 t2)2 —

On simplifying

T4 - 2400 T2 4- 1083750

1

=

0

(2)

20

Solving for r2 we get : 2400 ± V 1425000

2400 ± 1194 I`

T2—

Thus

T=

2 42.39 s = 0.71 min and

T = 2455

2

s = 0.41 min depending on the angle a.

1.33 Let the shells collide at the point P (x, y). If the first shell takes t s to collide With second and At be the time interval between the firings, then x= vo cos 01 t = vo cos and y= vo sin 0

A0

(1)

g (t — At)2

(2)

02 (t —

t — — gt2 ' 2

= vo sin 02 (t — At) — 2

Y

Integrating

f

t

dy = vo f dt or y = vo t

(1)

o o dr — ay or dx = a y dt = a vot dt dt=

And also we have

x

(using 1)

t

f dx = a vo f t dt, or, x= la v 2 o 0 (b) According to the problem So,

1 a y2

t2 = — 0 2

(using 1)

o

vy = vo and vx = a y So, Therefore

v=

\rvx277. v; . V v20 +

dv a2 y dy )4, = = V q) + a y2 dt t dt

(2)

a2 y2

a2 y 11 1 + (ay I v 0)2

Diff. Eq. (2) with respect to time.

dv dtY = So,

d vx 14, Y=

0 and

w= 1%1= a vo

=w

dt

x=

dy a -di = a

Vo

21

Hence wn = V w2

- q - V a2 v2o —

a vo

a4

y2 1 + (ay I vo)2 .

1.35 (a) The velocity vector of the particle --a. -.-.

v= at+bxj

d= a : A = bx dt

And

dy = bx dt = bat dt

Integrating

1 vo)

2

--1.

So,

Y

V 1 + (ay

t

5 dy = ab f t dt or, y= —

o

o

1

2

ab t

2

From Eqs. (2) and (3), we get, y =--- x2 (b) The curvature radius of trajectory y (x) is : 3 [ 1 +

2 (dy Idx)2 1

R=

(5)

Id2y1dx21 2

Let us differentiate the path Eq. y = b

with respect to x,

d2

= b x and y b dx a dx2 a From Eqs. (5) and (6), the sought curvature radius : 3 2

R

12

7a- 1+ (—b x)

1.36 In accordance with the problem -410 =

OT,

So,

wt

= ds d

or v dv = w ds

v dv = criti)ds=a+-dr-*

v dv = ai•d r adx (because

--10.

a Is directed towards the x-axis) z

fvdv= a fdx 0

Hence

•"V

v v

But So,

a

0

v2 = 2 ax or, v = V2crx

(6)

22

1.37 The velocity of the particle v = at dv

So,

w = a r

dt

v2

a2t 2

R

R

wn = — in

And

(1) (using v = at)

(2)

s = fvdt r

From

1 2 —at

-2/ERTI= a fv d t = o

2

4 x i . t2 _ a R

So,

(3)

From Eqs. (2) and (3) wn = 41cari Hence w = = V a2 + (4 It a i)2 = a V 1 + 16 x2 ri2 = 0.8 m/s2 1.38 According to the problem lwrI'm Iwnl v2 .—

- dv

For v (t),

dr

R

Integrating this equation from vo s vs v and Os ts t

v , —f dv 1 -= i

vo

or, v =

(1 +vot)

o

0 v dv

dt

v

R

2

NJW for v (s), - —ds = —R , Integrating this equation from vo s vs v and Os ss s

v Iv o

s

dv

SO,

v

Hence

1

v

s

R 0

cis or, In TI = o

v= voe

R

- siA

(b) The normal acceleration of the point

2 v

V2 e-2s

/R

(2)

wn = R

R

And as accordance with the problem

(using 2)

and w I wr 1

SO,

w=

=

A A 1 t ut wn Un

1 %1

2 Vn = a -=-e- 2"

a

ig

w

R

1,2

= a LR

-▪

23

v=a

1.39 From the equation

dy

a ds

dt 2

25 dt

w

-

a2 —2 '

and

a2 S

v w =

2

a vi a -Cs =

=

R R As wr is a positive constant, the speed of the particle increases with time, and the tangential acceleration vector and velocity vector coincides in direction. A

Hence the angle between v aand

wis equal to between wt is an w,

by means of the formula : tan a

a 2 s/R = 2s

1%1

=

wrI 1.40 From the equation

and a can be found

a2/2

R

1= a sin ou t

So,

dl — = a co cos co t dt v dv 2 • wr —dt – a co sin w t , and v2 a2 CO2 COS2 CO t —It at 0, sin w t = 0 and cos w t = ± wM

(a) At the point 1= Hence

w=

W

1 so, wt = 0 , n etc.

a2 €02 R

Similarly at 1= ± a, sin cot = t 1 and cos cot = 0, so, rvn = 0 2 Hence w= I wt I = a0) 1.41 As wt = a and at t = 0, the point is at rest (1) Let R be the curvature radius, then w= v

2

2 2

at

R

2 as (using 1)

But according to the problem bt 4 So, bt4 =

alt 2

or, R =

2 a

2 a 2 bs

2=

bt

Therefore

w=1/7w-1- rt

=

Hence

w= a1/1 +(4bs2/a3)2

(using 1)

a2 + (2 as / R)2

(2)

a2 + ( 4 bs2 / a2)2 (using 2)

24

r)2

1.42 (a) Let us differentiate twice the path equation y (x) with respect to time.

41. 2 ax cr cPy . 2a[ dt

dt ' dt2

d2X

dt

+ x dt2

Since the particle moves uniformly, its acceleration at all points of the path is normal and at the point x = 0 it coincides with the direction of derivative d2 y/dt 2. Keeping in mind that at the point x = 0,

i

111

v,

d

1 c/_21 dt 2

We get

NE

2 a v2 = wn

x- 0

w= 2

2a So,

v2 =

1

V

—

In

or R= — R' 2a Note that we can also calculate it from the formula of problem (1.35 b) (b) Differentiating the equation of the trajectory with respect to time we see that IVn

b2x -d—r + a2y

dt

(1)

A( = 0 dt

which implies that the vector (b2x r+ a2yri is normal to the velocity vector --). dx „A. li/ •r• V = t + dt dt j which, of course, is along the tangent. Thus the former vactor is along the normal and the normal component of acceleration is clearly 2 x d2X

b Wn me

on using

dt2

2 d2y a dt2 dt2

( b4x 2 + a4y2 )112

--0 ---, wn = w •n / l i t l . A t x =

0,y=

±

bandsoatx=0

Adl ± dt2 ... 0 Differentiating (1)

=0 II= 0 at x = 0 dt

Also from (1)

So

±v

(since Thus

and

tangential velocity is constant = v )

( 4.-2Y- ) = ± b- v2 dt2

bV2 I Wn I m

a2

v2

2

This gives R = a2/b .

-R

25 1.43 Let us fix the co-ordinate system at the point

0 as shown in the figure, such that the

radius vector rof point A makes an angle 0 with x axis at the moment shown. Note that the radius vector of the particle A rotates clockwise and we here take line ox as reference line, so in this case obviously the ( d0 - angular velocity co= taking dt anticlockwise sense of angular displacement as positive. Also from the geometry of the triangle OAC R = r or, r= 2 R cos 0. sin 0 sin ( lt - 2 0 ) Let us write, -0. -o -0 2 -0 • r = r cos 0 i + r sm Oi = 2 R cos 0 i + R sm 2 07 Differentiating with respect to time. d 0 ---* — •+ 2R cos20 -4107* dt 1 dt -1

dcrit or i7':- 2R2cos0(-sin0)

-.--0 (-d 0 ) -0 [sm201-cos20j] or, v = 2R dt -4. --*2 or, v= 2Rw(sin20i-cas So,

74.

0j )

Ivi or v.

2wR=0.4m/s.

As w is constant, v is also constant and w

So,

w= wn -

t

dv — dt

0,

v2 _( 2 (.0 R )2 . 4 w2 R .. 0.32 m/s2 R R

Alternate : From the Fig. the angular velocity of the point A, with respect to centre of the circle C becomes d (d2t0 -

2.

2

(-d0 dt

= 2 w . constant

Thus we have the problem of finding the velocity and acceleration of a particle moving along a circle of radius R with constant angular velocity 2 w. Hence

v= 2wR and

v2 ( 2 w R )2 w= wn = ---R R 1.44 Differentiating q) ( t) with respect to time S

R

-

2 4w R

d cp 2at c dt '1' oz =

(1)

For fixed axis rotation, the speed of the point A: v=coR= 2atR or R.

v ta t

(2)

26

Differentiating with respect to time dv v — = 2 a R = -, (using 1) dt

w

2

But

w11/4

=RV

V2

v/2at

So,

wr

,

=2atv

(using 2)

w 2 n =NI (vIt)2+(2atv)2

=Y-111+4a2t4 t 1.45 The shell acquires a constant angular acceleration at the same time as it accelerates linearly.

The two are related by (assuming both are constant) w 1

2nn

Where w = linear acceleration and f3 =angular acceleration 2 Then, w= V 2.132 itn.V 2•—w ( 2 n n) 1

But

v 2= 2 w 1, hence finally 2 nnv co. 1

1.46 Let us take the rotation axis as z-axis whose positive direction is associated with the

positive direction of the cordinate p, the rotation angle, in accordance with the right-hand screw rule (Fig.) (a) Defferentiating tp ( t) with respect to time. dcl9= a -3bt2= coz dt d2c1)

do)

dt2 -

dt

= fiz

-6bt

(1) and

(2)

From (1) the solid comes to stop at .6, t = t =

N

r

— 3b

The angular velocity co = a- 3 bt2, for 0 s t s Nra73T; ri3b (a-3bt2)dt

fdt Similarly

13

fiz

Va

dt 0 6b t for all values of t.

27

ff3 dt

So,

3/;17,3b Jo 6br dr

= fdt

m V3711-,

i"-a/$3b

i dt 0

(b) From Eq. (2) pz = - 6b t So,

( pz ), =vri/rt; = -6b Nrli = -2 11(17,

Hence

i3 = I

) (~z t- irdig I= 2-iSiTE

1.47 Angle a is related with I w, I and w,, by means of the fomula : tan a =

Wn

l wri

where w,, = w 2 R and I w1 I ... 13R ,

(1)

where R is the radius of the circle which an arbitrary point of the body circumscribes. dco dco — = at (here t3 = dr ' as 13 is positive for all values of t) dr

From the given equation (3 i=

2

So,

w . r.,2

At

R. at

4

R . a2t R

-

lwrk 13-R= atR

and

2

2

4

Putting the values of I w, I and w„ in Eq. (1), we get, 2 tan a = a

3 R/ 4 = ar or, t= [ ( 4- ) tan a 4 atR

113

t4

a

1.48 In accordance with the problem, I3z < 0 dco

Thus - dt

Or,

k Ira7 , where k is proportionality constant

- 1 da) .- kidtor, 10-)= .1/.0 . v 0) ° 0

When co = 0, total time of rotation t - x = 2

kt

-42

(1) “1130

k

28 2 I(0 0 / k

f

1 CO0 4- k 24t 2

fwdt Average angular velocity < w > idt

0

-kt 1/7 (0) dt

2 V i T o ; / k 2VEnik Hence < co > = coo t + k{

3

1122

k'vrw— t21 2 °

-

/ 2 -legr)° = co0/3 k

o

1.49 We have co = coo - a q) =

dt Integratin this Eq. within its limit for (q)) t

d

= f d t or ,, In

wo -kg)

-10 (00 — icci) 0 Hence

cp =

too

--kt

( 1 - el")

(b) From the Eq., w = coc, k cp and Eq. (1) or by differentiating Eq. (1) -kt

(02.

w °6

1.50 Let us choose the positive direction of z-axis (stationary rotation axis) along the vector . In accordance with the equation o O r ,

O r ,

Hence

(1)

w

4,

f 2 Wz

2

1 3 0 f

coz

c o s c p t h p 0 = 130 sin cp tV2 Po sin fp The plot coz (q)) is shown in the Fig. It can be seen that as the angle cp grows, the vector iiifirst increases, coinciding with the direction of the vector

g ((,)>

0), reaches the maximum at cp q)/2,

then starts decreasing and finally turns into zero at q) = r. After that the body starts rotating in the opposite direction in a similar fashion (coz < 0). As a result, the body will oscillate about the position fp = cp/2 with an amplitude equal to n/2.

29 1.51 Rotating disc moves along the x-axis, in plane motion in x - y plane. Plane motion of a solid can be imagined to be in pure rotation about a point (say I) at a certain instant known as instantaneous centre of rotation. The instantaneous axis whose positive sense is directed along (Trof the solid and which passes through the point I, is known as instantaneous axis of rotation. Therefore the velocity vector of an arbitrary point (P) of the solid can be represented as : V

30

-IP -■ (0 X TEr

(1)

On the basis of Eq. (1) for the C. M. (C) of the disc -• vc x rd (2) 2it

A

CO

According to the problem v t t i and

---)4

ri.e. cTilx -y plane, so to satisy the

rcr

Eqn. (2) rc./ is directed along (-n. Hence point I is

>3C

IUNIO •

at a distance r 1= y, above the centre of the

C

0

disc along y - axis. Using all these facts in Eq. (2), we get Vc

coy or y

(3) CO

(a) From the angular kinematical equation (4)

woz + 13z t co 80 On the other hand x

t.

v t, (where x is the x coordinate of the C.M.) x

or,

t

(5)

From Eqs. (4) and (5), w =

Using this value of to in Eq. (3) we get y

—

V2

f3 x/v

px

( hyperbola )

(b) As centre C moves with constant acceleration w, with zero initial velocity So,

Therefore,

Hence

1 2

Wt X= —

vc

2

and v,= wt

v2 x w

▪

-

30

132 The plane motion of a solid can be imagined as the combination of translation of the C.M. and rotation about C.M. -4. So, we may write VA = vc + VA Vc

W

A

"

(1) and

+ CO X rA

--•

—

wc

wA c

2 W

+ CO

( — r A

rAC

C

)

+

(fix

(2)

)

rAC

s the position of vector of A with respect to C. i

In the problem vc = v - constant, and the rolling is without slipping i.e., vc = v. w R , So; we = 0 and 13=

0. Using these conditions in Eq. (2) 2 A

2

V

A

R A

Here, uA c is the unit vector directed along 2 V Hence WA = — and WA is directed along R wheel.

rA

A

(-u

or directed toward the centre of the A C

(b) Let the centre of the wheel move toward right (positive x-axis) then for pure tolling on the rigid horizontal surface, wheel will have to rotate in clockwise sense. If w be the vc angular velocity of the wheel then co= •

R

Let the point A touches the horizontal surface at t = at 0, further let us locate the point A t= t, When it makes 0 = w t at the centre of the wheel. -a. -a. -4 -4. From Eqn. (1) VA- V CO X rAc c

+ R sin 0 (-

Vi+03(—iC)X[Rcos0(—j)

or,

vA

=

v i

-

R

[ cos

co t (— i)

+ sin

I

cot j

7* au

So,

(V — COS W O

+ V S M

(as v= caR)

COO

(v sin cot)2-

VA= V (V — V COS (002

= v V 2 (I - cos cot) = 2 v sin (cot / 2) Hence distance covered by the point A during

T=

2n/co

2a/au

s= f

vA dt =

f 2 v sin(cot/2)

8 dt =

v

= 8

R

0

133 Let us fix the co-ordinate axis xyz as shown in the fig. As the ball rolls without slipping along the rigid surface so, on the basis of the solution of problem 1.52 : v0= V c

Thus ye

0) X roc

wR and (1-0* t (-

u —4.

k)

as

vc

7.} t

t

(1)

• ▪▪

•

31

03c + 134°X roc = 0

Vc1WR=VA

Vc wt and

(a)

Le t

us

attached

As

fix

wi t h

point 0

—

- -

R

R

the the

is

w an d

(using

1)

V8

R

co-ordinate

rigid

the

—

13 =

surface

system as

instantaneous

wi t h

s h o wn

in

centre

the

the

fram e

Fig.

of rotation

of the ball

mo me n t

at the

so, "I

Now,

0,

vA

vc

+

Vc

So,

A

-0 Si mi l a l r y

v

-0 -

-0

=

2 vc

2

x

+

wt i

rAc

CO (- k)

(using

x

R

)

(v

c+

1)

rBc

v c

+

(-

÷ (DR

(- j ).

v;

at

k ) x

R

(i )

vc. t + vc (- j ) -0

vB = Nr2- vc =

wt

and

Wnbc + (02 (—

+

c'c

angle

45°

b o t h

fro m

is

the

W0 gm

unit

( -

vector

t o wa r d s

WA =

u m,

)

(using

1)

along

roc

We

( u s i n g

2 )

a n d

w0

is

(02R

R the

centre

of the

ball

+ co2 (-

) rx rAc

= wr+o)2R(—.13+13(—r)xR7

w4 t 4

So,

W

4)2R B

R

Now

( Fi g . )

-10

uoc

directed

j

R

t2 SO,

- 0

i and

A

A w here

an

A

) = —

CO2 (""'

is

rx 770.c

‘,2 vc az

WA=

4

wh +

R2

V =

( wt 2 ) 2

2

1

+ 2R

14r8 im

S i mi l a r l y

se

i

--I.

vc. + co x

V c

So,

wR)

W i+0)2R

(I )

+ (.02 (-

(-

0

rx

+

t3 (—k)xR(i) (- ji

(using

1)

s h o wn

in

Fig.

32

2

—

w (-1) (using 2) w2t 2 2

So,

WB =

r

2

w—

+W

1.54 Let us draw the kinematical diagram of the rolling cylinder on the basis of the solutioi of problem 1.53.

we-07.

A VA:72Vc

0 As, an arbitrary point of the cylinder follows a curve, its normal accelerationand radius of curvature are related by the well known equation v2 Wn =

1F so, for point A,

2 vA WA (n)

RA

4 v2

or,

RA =

2 ac

4r (because v s (or, for pure rolling)

Similarly for point B, 2 VB

RB

RB los 2 v

Or,

2 =2 Dr

r

1.55 The angular velocity is a vector as infinitesimal rotation commute. Then the relative angular velocity of the body 1 with respect to the body 2 is clearly. c°2 as for relative linear velocity. The relative acceleration of 1 w.r.t 2 is (1 )12

22 (1 )1

(

) dt

s•

-

33

where S' is a frame corotating with the second body and S is a space fixed frame with origin coinciding with the point of intersection of the two axes,

( dirii ) but

(dcTi;) + (7). .. 7.

dt

2 x 4-1)1

dt

S s, • -0 Since S' rotates with angular velocity w2 . However

d4 — . 0 as the first body rotates dt s

with constant angular velocity in space, thus --1. --a. P12= CO 1 X CO2.

Note that for any vector bthe relation in space forced frame (k) and a frame (k') rotating with angular velocity (ins d El dt

—, + co x 6*

d dt K

1.56 We have So,

0..)

2

IC

74.

= at 1 + bt j

(1)

w .. V (at)2 + (t2)2 , thus, colt_ 10.=

7.81 rad/s

Differentiating Eq. (1) with respect to time p s. —

dt

So,

,, ,_ Z.

at+zotj

(2)

(3 = V a2 + (2 bt)2

Pit- 10. -

and

1.3 rad/s2

(b)

Putting the values of (a) and (b) land' taking t = a fe 17° 1.57 (a) Let the axis o1 the cone

10s, we get

(OC) rotates in anticlockwise sense with constant angular

velocity (7.)1' and the cone itself about it's own axis (OC) in clockwise sense with angular velocity ZO*0 (Fig.). Then the resultant angular velocity of the cone. -0 --., -. w (1) 1.1

CO

+

(0 o

As the rolling is pure the magnitudes of the vectors ZIP and (7.:0 can be easily found from Fig. v (2) co' = c = v/R R cot a ' o° As w 1 Z,70, from Eq. (1) and (2)

34

co . f7772.I-wo V

2

1/ (R cot a )2 "I' ( Rv ) 2 -I R cos a=

2.3 rad/s

(b) Vector of angular acceleration --IV

d (7;1 d (0

ir. ---

--1.

+ wo )

=

dt

dt

(as w = constant.) --I.,

--I. T

about the 00' axis with the angular velocity

h e Thus v e c t o r

c o o

w h i c h

r o t a t e s

,..4. --00 --1. p = co x coo

The magnitude of the vector 'is equal to

co , re

13 = co' coo (as al 1 o—io ) v

13 -

So,

v

-

v

2 tan a = 2.3 rad/s

R cot a R R2

1.58 The axis AB acquired the angular velocity CO

al

rt

(1)

a

Using the facts of the solution of 1.57, the angular velocity of the body

--).

if2-71" co = v coo + co a .402 4. 020 t 2

=

0.6 rad/s

Vo

A

And the angular. acceleration.

fr

.

di4 But

So, As,

gi Z1).0

dt

—3.0 di:1-i. d(0) +GO dt dt

.--110

....IP.

. co x coo , and

dol dt

(lioo dt

+ --wr

d Zir dt = fl: t

13= ( ro t x iiro ) + 137,1 so, it = 1/ (coo Po t)2 + fill = fill

--*

0' --= .4 t

V 1 + (coo t)2 = 0.2 rad/s2

-

• 35

THE FUNDAMENTAL EQUATION OF DYNAMICS 139 Let R be the constant upward thurst on the aerostat of mass m, coming down with a constant acceleration w. Applying Newton's second law of motion for the aerostat in projection form

F = mw mg —

R mw

(1)

Now, if Am be the mass, to be dumped, then using the Eq. F = mw

R — (m — Am) g

(m — Am)

(2)

w,

2 mw From Eqs. (1) and (2), we get,

Am

MI

g + w

1.60 Let us write the fundamental equation of dynamics for all the three blocks in terms of projections, having taken the positive direction of x and y axes as shown in Fig; and using the fact that kinematical relation between the accelerations is such that the blocks move with same value of acceleration (say w) mo g — T2 —

and

=mow

(1)

w

(2)

M2 w

(3)

g =

T2 — km2 g

v2

> frz

The simultaneous solution of Eqs. (1), (2) and (3) yields, w= g

[ mo

k (m1+ m2)1

mo+ m 1+m2

V

(1 + k) mo and

T2 at

mo+ mi+ m2

M2

g

?nog

As the block m0 moves down with acceleration w, so in vector form

[mo — k (mi + m2)] W at

m0+ /721 + M2

1.61 Let us indicate the positive direction of x-axis along the incline (Fig.). Figures show the force diagram for the blocks. Let, R be the force of interaction between the bars and they are obviously sliding down with the same constant acceleration w.

36 Newton's second law of motion in projection form along x -axis for the blocks gives : mi g sin cc -

mi g cos a + R

mi w

(1)

m2 g sin a - R - k2 m2 g cos a= m2 w

(2)

Solving Eqs. (1) and (2) simultaneously, we get mi + k2 m2 and

w= gsina-gcosa m1 ÷ m2 m1m2(k1 - k2) g cos a R

(3) m1+ m2

(b)

when the blocks just slide down the plane, w

0 , so from Eqn. (3)

mi + k2 m2 g sin a-gcos a

= 0 mi +

ir-2 m2) cos a

(M 1 + m 2) sina (ki mi

Or,

(k1 m1 + k2 m2) Hence

tan a mi +

1.62

Case 1.

When the body is launched up :

the velocity of projection and 1 the distance traversed Retarding force on the block - mg sin a + k mg cos a and hence the

Let k be the coefficeint of friction, u along the incline.

retardation 7 g sin a + kg cos a. Using the equation of particle kinematics along the incline, 0= u2- 2(g sin a + k g cos a )1 2 U

or,

1-

2

and

0= u - (g sin a + k g cos a) t

or,

u= (g sin a+ kg cos a) t

Using (2) in (1) 1Case (2).

(g s i n a + k g c o s a)

1 2 - (g sin a + k g cos a) t 2

When the block comes downward, the net force on the body =

mg sin a - km g cos a and hence its acceleration = g sin a - k g cos a Let, t be the time required then, 1 1=

-

(gsina—kgcosa)tI2

(4)

From Eqs. (3) and (4) t2

sin a - k

e2

But

cos a

sin a +kc os a t=

1 -

Hence on solving we get

k=

tan a = 016 (r12 +1)

(according to the question),

37

1.63 At the initial moment, obviously the tension in the thread connecting m1 and m2 equals the weight of m2. (a) For the block m2 to come down or the block m1 to go up, the conditions is m2g-Tz0 and T-migsina-fra 0 where T is tension and f, is friction which in the limiting case equals km1g cosa. Then or

m2 g - mi sin a > k mi g cos a m

Or

-

2

m1

> (k

cos a + sin a)

(b) Similarly in the case mlgsina-m2g>frjlm

or, mi. g sin a - tn2g > kmi g cos a or,

(sin a - k cos a) m1 (c) For this case, neither kind of motion is possible, and fr need not be limiting. -1712-
— > (sin a m1

Hence,

- k cos ot)

1.64 From the conditions, obtained in the previous problem, first we will check whether the

mass m2 goes up or down. Here, m2/m1 = -n > sin a + k cos a, (substituting the values). Hence the mass m2 will come down with an acceleration (say w). From the free body diagram of previous problem,

and

T-

m2-g

m2 w

g sin a - k

g cos a

(1) w

(2)

Adding (1) and (2), we get, m2 g w-

g sin a - k mi g cos a

(m2/m1 - sin a - k cos ct) g

(m1+ m2) w - sin a - k cos a) g

(1 + m2/m1)

1 + 11

Substituting all the values, w a. 0.048 g .0.05 g As m2 moves down with acceleration of magnitude w

0.05 g > 0, thus in vector form

acceleration of m2 : w2

(ri - sin a - k cos a) el 1+11

0.05

1.65 Let us write the Newton's second law in projection form along positive x-axis for the plank and the bar fr ml w1, fr m2 w2

(1)

38 At the initial moment, fr represents the static friction, and as the force F grows so does the friction force fr, but up to it's limiting value i.e.

fr = fro.) = k N = k m2 g. Unless this value is reached, both bodies moves as a single body with equal acceleration. But as soon as the force fr reaches the limit, the bar starts sliding over the plank i.e. w2 x w1. Substituting here the values of w1 and w2 taken from Eq. (1) and taking into account that

f, = k m2 g, we obtain, (at — km2 g)/ m2 a

—km2 g, were the sign "=" corresponds to the moment m1

t = to (say) k g m2 (mi + m2) Hence,

If

to =

a m1

- k m2 g t s to, then w1= m1 (constant). and

w2 = (at - km2 g)/ m2 On this basis w1(t) and w2 (t), plots are as shown in the figure of answersheet.

1.66 Let us designate the x-axis (Fig.) and apply Fx = m wx for body A : mg sina—kmgcosa= mw or,

w= gsina—kgcosa

Now, from kinematical equation : / sec a = 0 + (1/2) w t2 Or,

t= V 21 sec a/(sin a— k cos a) g = V 21 / (sin 2 a/2 — k cost a) g

(using Eq. (1)).

d

( sin 2 a — k cost a) 2

for trnin ,

i.e. or,

da 2

=0

s 2a "I + 2 k cos a sin a — 0 2

tan 2 a = — — et a= 49°

and putting the values of a, k and

/ in Eq. (2) we get t .

- ls.

1.67 Let us fix the x —y co-ordinate system to the wedge, taking the x — axis up, along the incline and the y — axis perpendicular to it (Fig.).

39 Now, we draw the free body diagram for the bar. Let us apply Newton's second law in projection form along x and y axis for the bar : T cos 13 — m g sin a — fr = 0

(1)

T sini3+N—mgcosa= 0 or, N= m g cos a— Tsin f3 But f; = kW and using (2) in (1), we get

(2)

T m g sin a+ k mg cos a/(cos 13+ k sin 13)

(3)

For T the value of (cos 13 + k sin 13) should be maximum d (cos 13 + k sin (3) — 0 or tan 13 = k d13

So,

Putting this value of (3 in Eq.

(3) we get,

m g (sin a + k cos a) mn

m g (sin a + k cos a)

hri71c2 + k2/V7+7:2

1 442

1.68 First of all let us draw the free body diagram for the small body of mass m and indicate x — axis along the horizontal plane and y — axis, perpendicular to it, as shown in the figure. Let the block breaks off the plane at t = to i.e. N = So,

0

N= m g — ato sin a .. 0

mg to — a sin a (1) From F. = m wx, for the body under or,

investigation : m d N/dt at cos a ; Integrating within the limits for v (t)

m f dvx= a cos aft dt

ds

So,

vg• dt

(using Eq. 1)

a cos a 2 t 2m

Integrating, Eqn. (2) for s (t) a cos a t3 2m 3 Using the value of t = to from Eq. (1), into Eqs. (2) and (3) v=m g2 cos a 2 a sin2 a

and s

2 _ m g3 cos a

6 a2 sin3 a

40 1.69

Newton's second law of motion in projection form, along horizontal or x - axis i.e. Fx = m wx gives. dv F c,os (as) my —ds.

or, F cos

(asa =

as)

(as) ds m v dv

Integrating, over the limits for v (s) -fC

Or

v==

OS (as)

ds=

X

2 v

2

177n .a ma

= V 2 g sin a/3 a (using F= 21-311 ) which is the sought relationship. 1.70 From the Newton's second law in projection from : For the bar, For the motor, T — 2 king = (2m) w T— long m w' Now, from the equation of kinematics in/the frame of bar or motor : 1 1= —

(3)

t2

+

2 From (1), (2) and (3) we get on eliminating T and w' t= V 21/ (kg + 3 w) 2171

(zr 1/11////////////////////112/ J

r

7 ;7)///17--).

fr

1.71 Let us write Newton's second law in vector from 1?= mg, for both the frame ofjround). T + mi 1%. mi WI (1) 71' 'n2 r= m2172 (2) These two equations contain three unknown quantities iic , ii-r2 and TT . The third equation is provided by the kinematic relationship between the accelerations : 3. 3., I. -1. -3., wi - wo + w , w2 - wo w (3)

blocks (in the

where ;V.. is th acceleration of the mass nt1 with respect to the pulley or elevator car. Summing up termwise the left hand and the right-hand sides of these kinematical equations, we get

41

+ = 2 IT,' W1 W2

(4)

The simultaneous solution of Eqs. (1), (2) and (4) yields -30 ("11

1112)

e+ 2 m2 wo

wi

ml+ m2 Using this result in Eq. (3), we get, 1711 -

m2

--I.

-,. 2 m1 m2 T(wo

--I.

(g -wo)

and

M1 + M2

m + M2 Using the results in Eq. (3) we get w

-I., M1 - M2 = m2 M1 +

wo)

(b) obviously the force exerted by the pulley on the celing of the car 4 mi

4

F= - 2 T -

-÷

(g wo)

m1 +m2 Note : one could also solve this problem in the frame of elevator car. 1.72 Let us write Newton's second law for both, bar 1 and body 2 in terms of projection having taken the positive direction of x1 and x2 as shown in the figure and assuming that body 2 starts sliding, say, upward along the incline T1-mlgsina= m1 w1

(1)

m2 g - T2

(2)

M2 W

For the pulley, moving in vertical direction from the equation Fx = m wx 2 T2 -

= (me )

(as mass of the pulley mP = Or

=0

0)

T1= 2 T2

(3)

As the length of the threads are constant, the kinematical relationship of accelerations becomes w = 2 wi (4) Simultaneous solutions of all these equations yields : m2

2 g (2-- SM a m1

2 g ( 2 ri - sin a) (41 + 1)

w= ( 4 -m2- + 1) m1

As 1 > 1, w is directed vertically downward, and hence in vector form w

4

+ 1 2 e(2

- sin a)

1n2

42 1.73 Let us write Newton's second law for masses m1 and m2 and moving pully in vertical direction along positive x - axis (Fig.) : mig-T=

miwa

(1)

m2 g

m2 wa

(2)

T1- 2T = 0 (as m = 0) or

(3)

T1= 2 T

Again using Newton's second law in projection form for mass mo along positive x1 direction (Fig.), we get

T1= mo wo

(4)

The kinematical relationship between the accelerations of masses gives in terms of projection on the x - axis wix + w2x = 2 Ivo

(5)

Simultaneous solution of the obtained five equations yields : [4 m1 m2 + mo (m1 - m2) ] g w -

4 mi m2 + mo (mi + m2)

In vector form [4 mi m2 + mo (mi - m2) ] r wi = 4 mi m2 + mo (mi + m2) 1.74 As the thread is not tied with m, so if there were no friction between the thread and the ball m, the tension in the thread would be zero and as a result both bodies will have free fall motion. Obviously in the given problem it is the friction force exerted by the ball on the thread, which becomes the tension in the thread. From the condition or language of the problem wm> wn, and as both are directed downward so, relative acceleration of M = wm - wn, and is directed downward. Kinematical equation for the ball in the frame of rod in projection form along upward direction gives : 1 1= — (w (1) ) 2 m w'" t2 NeWton's second law in projection form along vertically down direction for both, rod and ball gives, Mg-fr= M wm

(2)

mg - fr

(3)

mwm

Multiplying Eq. (2) by m and Eq. (3) by M and then subtracting Eq. (3) from (2) and after using Eq. (1) we get 21Mm (M - m) t2

T

43 1.75 Suppose, the ball goes up with accleration w1 and the rod comes down with the acceleration w2. As the length of the thread is constant, 2 w1 = w2

(1)

From Newton's second law in projection form along vertically upward for the ball and vertically downward for the rod respectively gives, T—mg= mw1 (2) and Mg—T'= Mw2 (3) but T=2T From Eqs. (1), (2), (3) and (4) w1

(2M— m) g m + 4M

and w2 —

(2

(because pulley is massless)

(4)

g (in upward direction)

11 + 4

2 (2 —

g

(downwards) + 4) From kinematical equation in projection form, we get 1 ( + )t2 1= —2 w1 w2 as, w1 and w2 are in the opposite direction. Putting the values of w1 and w2, the sought tune becomes t= V 2 / (ri + 4) / 3 (2 —

g = 1.4s

1.76 Using Newton's second law in projection form along x — axis for the body negative x — axis for the body 2 respectively, we get m g — T1 =

1 and along

`01111/11,11h

(1)

T2 -- M2 g = m2 w2 (2) For the pulley lowering in downward direction

T2 A

T2

from Newton's law along x axis, T1 — 2 T2 = 0 (as pulley is mass less)

or,

T1= 2 T2

(3)

As the length of the thread is constant so, = w2 2 wi (4) The simultaneous solution of above equations yields, 2 (mi — 2m2) g 2 (Ti _ 2) m1 w2 (as 7= YI) (5) 4 m2 + m i +4 2 1 comes to the horizontal floor Obviously during the time interval in which the body covering the distance h, the body 2 moves upward the distance 2h. At the moment when the body 2 is at the height 2h from the floor its velocity is given by the expression : u2 •2

[

IA,

2 (-h.) =

2 (ri — 2) g

2h

8 h (T) — 2) g

+4 After the body m1 touches the floor the thread becomes slack or the tension in the thread 11+4

zero, thus as a result body 2 is only under gravity for it's subsequent motion.

•

▪

44

Owing to the velocity v2 at that moment or at the height 2it from the floor, the body 2 further goes up under gravity by the distance, L

=

=

4h

2g

— 2) +4

Thus the sought maximum height attained by the body 2 : -2) 670 (TI (1) + 4) ri + 4 137 Let us draw free body diagram of each body, i.e. of rod A and of wedge B and also draw the kinemetical diagram for accelerations, after analysing the directions of motion of A and B. Kinematical relationship of accelarations is :

If= 2h+h'= 2h+

4h

W

tan a =

A

(1)

wB

Let us write Newton's second law for both bodies in terms of projections having taken positive directions of y and x axes as shown in the figure.

mA g — cos a = mA wA

(2)

N sin a = mB wB

(3)

and

Simultaneous solution of (1), (2) and (3) yields :

mA g sin a 2

WA

mA sin a + mB cot a cos a W

-

and

(1 + ri cot a)

W A

tan a = (tan a + cot a)

' W8 Note : We may also solve this problem using conservation of mechanical energy instead of Newton's second law. 1.78 Let us draw free body diagram of each body and fix the coordinate system, as shown in the figure. After anAlysing the motion of M and m on the basis of force diagrams, let us draw the kinematical diagram for accelerations (Fig.). As the length of threads are constant so,

ds.m=dsm and as

andd vm do not change their directions that why I iiram = I iirm I = w (say) and

45

1wmM 1As wm = Wnim 1- Wm so, from the triangle law of vector addition =VGw From the Eq. Fx m wx , for the wedge and block :

T—N= Mw, N mw

and

Now, from the Eq. Fy = mwy , for the block

mg — T — kN = mw Simultaneous solution of Eqs. (2), (3) and (4) yields

:

mg w— (lan + 2m + M)— (k + 2 +M/m) Hence using Eq. (1)

wm

I=

g (2 + k +M/m)

1.79 Bodies 1 and 2 will remain at rest with repect to bar A for wmin 5 w 5 w

, where wmmis

the sought minimum acceleration of the bar. Beyond these limits there will be a relative motion between bar and the bodies. For 0 5 w 5 wmin, the tendency of body 1 in relation to the bar A is to move towards right and is in the opposite sense for w a w

. On the

basis of above argument the static friction on 2 by A is directed upward and on 1 by A is directed towards left for the purpose of calculating wmin. Let us write Newton's second law for bodies 1 and 2 in terms of projection along positive x — axis (Fig.). T—fri. mw or, fri = T—mw (1) N2

mw

(2)

As body 2 has no acceleration in vertical direction, so mg fr2= —T (3)

46 From (4) and (5)

m (g — w) s mk (g + w) , or w Hence

g(1 — k) (1 + k)

g(1 — k) ( 1 +k)

wmin

1.80 On the basis of the initial argument of the solution of 1.79, the tendency of bar 2 with respect to 1 will be to move up along the plane. Let us fix (x — y) coordinate system in the frame of ground as shown in the figure. From second law of motion in projection form along y and x axes : mgcosa—N= mwsina Or,

N= m ( g cos a — w sin a ) m g sin ct+fr = mw cos a

(1)

or,

fr= m(w cos a — g sin a )

(2)

but fr s kN, so from (1) and (2) ( w cos a — g sin a)sk(g cos a + w sin a ) or, w( cos a —ksin a )sg(kcos a + sin a )

( cos a + sin a ) cos a — k sin a So, the sought maximum acceleration of the wedge ( k cos a + sin a ) g ( k cot a + 1) g where cot a > k wma X cos a — k sin a cot a — k 1.81 Let us draw the force diagram of each body, and on this basis we observe that the prism moves towards right say with an acceleration w 1and the bar 2 of mass m.z.moves down or,

w s g

-30

the plane with respect to 1, say with acceleration w21 , then ,

w2

=

w21 wi (Fig.)

Let us write Newton's second law for both bodies in projection form along positive y2 and x1 axes as shown in the Fig.

m2 g cos ct — N = m2

W2

(

y2

)

M2

[ W

21

( y2 )

y2 )

M2 [ 0 + Wi

sin a I

m2 g cos a — N = m2 w1 sin a

or,

N Sill a = miwi

and Solving (1) and (2), we get

m2 g sin a cos a •

mi

g sin a cos a

2 2

N

U./2

47 1.82 To analyse the kinematic relations between the bodies, sketch the force diagram of each body as shown in the figure. On the basis of force diagram, it is obvious that the wedge M will move towards right and the block will move down along the wedge. As the length of the thread is constant, the distance travelled by the block on the wedge must be equal to the distance travelled by the wedge on the floor. Hence d snof - d sm. As i;°' and vmI' do not change their directions and acceleration that's why 1;,„m t t

1,m and Tirm t t 17m. and wnim= wm= w

(say) and accordingly the diagram of kinematical dependence is shown in figure.

--,

-' ,

--91.

As wn, - wrnm + wm, so from triangle law of vector addition. wn, = 1/Wm2 1- Wmm - 2 wmm wm cos a

= wV2(1 - cos a)

(1)

From Fx = m wx , (for the wedge), T. T cos a +Nsin a . Mw

(2)

For the bar m let us fix ( x - y) coordinate system in the frame of ground Newton's law in projection form along x and y axes (Fig)) gives

mg sina - T - m wn.(x). m {wnaf (x)+ wm (x)] . m [wmat + w cos (a - a)]. m w (1 - cosa)m mgcos a-N= m wm(y)= m[w.m(y)+wm(y)1= m [ 0 + wsin a.]

(3) (4)

Solving the above Eqs. simultaneously, we get m g sin a w-

M+ 2m ( 1 - cos a )

Note : We can study the motion of the block m in the frame of wedge also, alternately we may

solve this problem using conservation of mechanical energy. 1.83 Let us sketch the diagram for the motion of the particle of mass m along the circle of radius R and indicate x and y axis, as shown in the figure. -I. (a) For the particle, change in momentum Ap = my (- i) - m v (j ) -1. so, I Ap I = 1/2- m v and time taken in describing quarter of the circle,

4-8 R 2v

At=

I AP I At

Hence, -

mv R/2v

2{2- m v 2 nR

(b) In this case --30

pi= 0 and pf= mwtt(- i), —30 so

I Ap I

m wt t

Hence, I < F >1 -

t

m wt.

1.84 While moving in a loop, normal reaction exerted by the flyer on the loop at different points and uncompensated weight if any contribute to the weight of flyer at those points. (a) When the aircraft is at the lowermost point, Newton's second law of motion in projection form Fn = m wn gives 2

N - mg or, N= mg +

mv

R

2

- 2 09 IN

(b) When it is at the upper most point, again from F,, = m wn we get my2 N" + mg - R

N"=

mv

2 -

m g = 0.71N

(c) When the aircraft is at the middle point of the loop, again from F„= m w,, N'

mv2

- 1.4 kN

The uncompensated weight is mg. Thus effective weight obliquely.

1IN'2+

m2 g2 =1.56 kN acts

1.85 Let us depict the forces acting on the small sphere m, (at an arbitrary position when the --0 thread makes an angle 0 from the vertical) and write equation F = m w via projection on A

A

the unit vectors ur and un. From Fr= m wt we have

m I.

dv mg sin Dim — dt vdv _ vdv ds m 1 (-d0)

(as vertical is refrence line of angular position)

49 vdv = - gl sin 0 d 0

or Integrating both the sides :

v 0 fvdv= -glf sin0d0 o

xr2 V

2

—= g 1 cos 0 2

or, 2

V

Hence — = 2 g cos 0 = wn

.(1)

1

(Eq. (1) can be easily obtained by the conservation of mechanical energy). From Fn = m wn

T-mgcos0-

in V 2

1

•

Using (1) we have

775,

T = 3 mg cos 0

(2)

Again from the Eq. Ft - m wt :

mg sin 0 = m wt or wt = g sin 0 Hence w =

(3)

Tv. 2 + wt = V( g sin 0 ) 2 + ( 2g cos 0) 2

( using

= g 1/1 + 3 cos 2 0 (b) Vertical component of velocity, vy = v sin 0 So,

v Y2=V2s

For maximum

in

v, or Y

2g / cos 0 sin 2 0 (using 1) 20= 2 Vy

,

d (COS 0

sin 2 0)

=

0

d0

cos 0 = Ty-

which yields

v--r mg

Therefore from (2) T = 3 m g - --- =

3

(c) We have w = wt ut + wn un thus wy = wo)+ wn(y) But in accordance with the problem wy = So, or, or,

Wo )

+

Wn(y) 312

0 0

g sin 0 sin 0 ÷ 2g cos 2 0 (- COS 0) = cos 0 = Ts- or,

0 = 54.7°

0

1 and 3 )

SO 1.86 The ball has only normal acceleration at the lowest position and only tangential acceleration at any of the extreme position. Let v be the speed of the ball at its lowest position and 1 be the length of the thread, then according to the problem 2 V — ill

(1)

sin cc i g where a is the maximum deflection angle

From Newton's law in projection form : Ft= mwt

dv — mg sin 0 . mv or,

1d0 —glsin0d0. vdv

On integrating both the sides within their limits.

Or,

2 V = 2g1

(1 — cos a)

(2)

Note : Eq. (2) can easily be obtained by the conservation of mechanical energy of the ball in the uniform field of gravity. From Eqs. (1) and (2) with 0

=a

2g1(1 — cos a) = lg cos a or,

cos a = 2- so a = 53° 3 ' 1.87 Let us depict the forces acting on the body A (which are the force of gravity mga nd the normal reaction N ) and write equation F = mw via projection on the unit vectors A A lit and un (Fig.) From F t. mwt

dv mg sin 0 = m— dt vdv vdv —m .m ds Rd0 or,

gR sin 0 d 0 = vdv

Integrating both side for obtaining v (0) 0

v

N

R sin 0d0 - fvdv

fg

(i) v2 mg cos 0 —N — m—R

ing

(2)

At the moment the body loses contact with the surface, N — 0 and therefore the Eq. (2) becomes

v2= gR cos 0

(3)

51 where v and 0 correspond to the moment when the body loses contact with the surface. Solving Eqs. (1) and (3) we obtain cos 0 = 3 or, 0 = cos -1(2/3) and v =V2,11?/3 1.88 At first draw the free body diagram of the device as, shown. The forces, acting on the sleeve are it's weight, acting vertically downward, spring force, along the length of the spring and normal reaction by the rod, perpendicular to its length. Let F be the spring force, and Al be the elongation. From, Fn = mwn :

(1)

Nsin 0 +F cos 0 = m to2r where r cos 0 =

+ Al).

Similarly from F1= mw,

N cos 0- F sin 0= 0 or, N= F sin 0/cos 0

(2)

From (1) and (2) m co2 r

F (sin 0/cos 0)•sin 0 + F cos 0 = m w2 (10 + A1)/cos On putting F = K A 1,

KA/sin20+KA/cos20= moi2(10+A1) on solving, we get, 10

Al= m (.02

2

(K/M CO2 — 1)

K — M 0)-

and it is independent of the direction of rotation.

1.89 According to the question, the cyclist moves along the circular path and the centripetal force is provided by the frictional force. Thus from the equation F,1= m wn 2

2 v

Jr m = — or

ko (1

or kmg

M

V

2 — )g = v — or v2= ko (r - r2 /R) g R r

d(r For v., we should have

(1)

r2 =0

dr 2r

or, Hence vmax= Nrro

1-

R = v0, so r = R/2

TR

1.90 As initial velocity is zero thus 2 V as

2 wt s

As wt > 0 the speed of the car increases with time or distance. Till the moment, sliding starts, the static friction provides the required centripetal acceleration to the car. Thus

fr = mw, but fr s king

(1)

52 So,

Hence

v max

(k2 g2 w21 R ti v2 1

V

r

— -11 = 60m. 2 wt 2 w tr 1.91 Since the car follows a curve, so the maximum velocity at which it can ride without sliding at the point of minimum radius of curvature is the sought velocity and obviously in this case the static friction between the car and the road is limiting. so, from Eqn. (1), the sought distance s

2

Hence from the equation Fn = mw m v2

king so

v

or v s 1/1/tg

=3/1c17-W .

(1)

We know that, radius of curvature for a curve at any point (x, y) is given as, R

[ 1 + (dy/dx)2 13/2

(2)

(d2Y) M1'2 For the given curve, x d2y. -a cos (—x) and = —2 sin a a dX2 a Substituting this value in (2) we get, 2/ 2 (x/a) j [1 + (a a ) cos2 3/2 R (a/a2) sin (x/a) It X For the minimum R, dx

a

and therefore, corresponding radius of curvature Rmin.

a2 a

Hence from (1) and (2) vmax - a V kg/a 1.92 The sought tensile stress acts on each element of the chain. Hence divide the chain into small, similar elements so that each element may be assumed as a particle. We consider one such element of mass dm, which subtends angle d a at the centre. The chain moves along a circle of known radius R with a known angular speed w and certain forces act on it. We have to find one of these forces. From Newton's second law in projection form, Fx = mwx we get 2 T sin (da/2) - dN cos 0 = dm w2 R and from Fz = mwz we get dArsin0 = gdm Then putting dm = mda/2 n and sin (da/2) = da/2 and solving, we get, m ((o2R +gcot0) T= 2n

(3)

53

cl-mg1.93 Let, us consider a small element of the thread and draw free body diagram for this element. (a) Applying Newton's second law of motion in projection form, Fn. mwn for this element, (T +

dT) sin (d 0/2) + T sin (d 0/2) — d N = dm (o2 R = 0 2T sin (d 0/2) = dN, [negelecting the term(dT sin d 0/2) ]

Or,

TdO= dN, as s

or,

in

0

=d

0

(1)

2

dfr= kdN= (T+dT)—T= dT

Also,

(2)

In this case Q = n so, T2

Or,

So,

or, In — = kn Ti. k=

(3)

T2 1 -- = _ In 11 0 T1 It

T2 as

M2 g m2 = — = rio g Ti I. m1 m1

m2 (b) When — az ri, Which is greater than Tio, the blocks will move with same value of ml acceleration. (say w) and clearly m2 moves downward. From Newton's second law in projection form (downward for m2 and upward for m1) we get :

and

M2 g — T2 = M2 W

(4)

T1 — mi g = mi w

(5)

-•

54

T2 Also

(6)

Ti = 71°

Simultaneous solution of Eqs. (4), (5) and (6) yields : (m2 - 110 mi ) g

w-

(m2+10mi)

m2 Ti g as — = ) m1 01+10)

(11 - lio )

1.94 The force with which the cylinder wall acts on the particle will provide centripetal force necessary for the motion of the particle, and since there is no acceleration acting in the horizontal direction, horizontal component of the velocity will remain constant througout the motion. So

vx . vo cos a

Using, F„ = m wn , for the particle of mass m,

N-

2 2 M Vo cos a x , R R

mV

2

which is the required normal force. 1.95 Obviously the radius vector describing the position of the particle relative to the origin of coordinate is

r= xt+yi= asincotti - bcoscotj Differentiating twice with respect the time :

Thus 1.96

(a) We have

2 --D. -3. -3. r w = —,, - - (02 (a sinoui+bcoscotj)= d t` -)' F= mw= -mw 2rr AFA.

f

-(027.4b

ldt t = f m rdt = m et 0

(b) Using the solution of problem 1.28 (b), the total time of motion, t Hence using

(1)

(1)

-3.-3. 2 (vo.g) 2 g

t = -c in (1) I I V =mgt

1.97 From the equation of the given time dependence force F = a t(t-t) at t= t, the force vanishes,

-t (a) Thus

Ap=p= f Fdt 0

55 t

--* ,

or,

,,

p= f atut—t j at 0 -•

-I.

but

--o

p. my so v.

n

3

6

--• 3

at 6m

(b) Again from the equation F miir at(t —t)-m-dt iibttt —t2)dt. mdr

or,

Integrating within the limits for irtt ), I o

f

V

Gilt T - t

2

) dt = mf dr o --, 2

Or,

V is

5+( X t2

t3 )

m

3

vi•

Thus

2

a t ( "C m 2

t 3

a t2 ( t ) — L for t s t m 2 3

Hence distance covered during the time interval t - t, I

vdt

S IMI f

0 T a t2

-f 0

t

(

— t d t — a

On integrating, --)

m

— F0

T 4 2 3 m

When

1 2 my - 70- cos cot + C, (where C is integration constant) ce, rH e n

—F0 m o)

4

t - 0, v - 0, so C-

m w

cos cot + F0 m co F0 As 1 cos cot s 1 so, v -

m co

( 1 — cos cot )

56 t s= f v dt

Thus

o =

Fo t Fo sin cot — 2 = m co m (0

F0 PI w2 ( Wt — s

in cot )

(Figure in the answer sheet).

—*

1.99 According to the problem, the force acting on the particle of mass m is, F = Fo cos wt

dr —0 • Fo m -- = Fo cos cot or d v —. — cos cot dt dt m

So,

Integrating, within the limits. ■II.

v

-.. I , Fo --I. F0 f d v = -77-1 cos wt dt or v — — sin cot m co 0

0

It is clear from equation (1), that after starting at t — 7r the first time at t = co —3. F0 From Eq. (1), v = I vl = — sin wt for t s —7t M0)

0, the particle comes to rest fro

0)

Thus during the time interval t = 31/co, the sought distance xlw

2F s = F° f sin cot dt = mw 0

m 0)2

From Eq. (1) v 1.100 (a) From the problem Thus

or, On integrating But at or, Thus for

max

F0 .333 as I sin cot I s

1

M 0)

—1. --, d r -, F = — ry so m —= - ry dt dv m—= [ as dv ti v] dt ry dv r - — dt v m r In v=-- t+C m t = 0, v = vo, so, C = In vo

v r In — = — —t or, v= vo e me r vo m t --, co, v = 0

dv (b) We have m — = —ry so d v = dt

—r — ds

m

r

(2)

57 Integrating within the given limits to obtain v (s):

f

dv

Or,

Vo

MI

m (1)

-

f d s o

r

r v a m V o r s

0 P i V

Thus for

v

0, s =

m dv

(c) Let we have

stow=

- r v or

v

dv

r -r, —at m

— v

vin or,

vo

dv J v

=-7-1 dt , or, In m

0

—t 1V°

1/71 ) r Now, average velocity over this time interval,

So

m

0

t

— M ln(

m In n r

—Inn rt

voe-7; dt f v dt

dt

1.101 According to the problem

vo(rl dv

—m In ri

1)

In

dv m dt = -kv2 or, n12""-kdt v Integrating, withing the limits, v2

k

(vo - v ) f dt or, t

m

m 0 o To fine. the value of k, rewrite dv dv =-kv2 or, — my ds v On integrating

vo v

(1)

k - —as

ffL. J.c_f ds J

V

Mel

0

0

m vo k = — In — So, h v Putting the value of k from (2) in (1), we get h(vo-v) t=

Vo

v In —

r

(2)

58 1.102 From Newton's second law for the bar in projection from, Fx = m wx along x direction we get

mg sin a - kmg cos a = mw

Hence, the maximum velocity will be at the distance, x .. tan a/a Putting this value of x in (1) the maximum velocity,

Al g sin a tan a V

MB

max a 1.103 Since, the applied force is proportional to the time and the frictional force also exists, the motion does not start just after applying the force. The body starts its motion when F equals the limiting friction. Let the motion start after time to , then

mg F = ato = kmg or, to - k a So, for t = s to, the body remains at rest and for t > to obviously mdv

dt

- a (t - to) or, in dv = a (t - to) dt

Integrating, and noting v v

= 0 at t = to , we have for t > to r

a

f m dv = a f (t - to) dt or v= 0

12 (t - t

2m

°'

to

I Thus to )3

s= 5 v dt = -

Co

2

712- f ( t - to)2 dt =

om

(t-

59 1.104 While going upward; from Newton's second law in vertical direction :

v dv m — ds

- ( mg +

kv

v dv kv 2

2 ) or

- ds

m g At the maximum height h, the speed v = 0, so 0

vdv g + ( kv

ds 2/flt ) 0

0

Integrating and solving, we get, kv2

h=

(1) (1+ 2k In mg When the body falls downward, the net force acting on the body in downward direction equals ( mg k v 2 ), Hence net acceleration, in downward direction, according to second law of motion v dv kv 2 = g--or, ds

v dv kv2

ds

a

in

a

v' f

Thus

V dV

112 f d s

kv 2/ 2/m

o Integrating and putting the value of h from (1), we get, v'=

111+kvo2/mg.

Vo

1.105 Let us fix x - y co-ordinate system to the given plane, taking x-axis in the direction along which the force vector was oriented at the moment t = 0, then the fundamental equation of dynamics expressed via the projection on x and y-axes gives,

d vx F cos t m

and

Fsincot=

dt

(1)

dv d

(2)

(a) Using the condition v(0) - 0, we obtain vv =

m

sin w t

(3)

and F V Y

Hence,

v

mw

(1 -coswt)

(4)

1I v2 + v 2x

M

2 )1

60 (b) It is seen from this that the velocity v turns into zero after the time interval A t, which can be found from the relation, At co — = 1E. Consequentely, 2

F

the sought distance, is

0 =wt

At

s

v dt

8F 771 0) 2 ••••

f v dt

0

Average velocity, < v > =

0

X

fdt

2n/au

So, = f

•••

2F ._ co t , / ,, — sm — ) at / k Z 31 0) ) — in 0) 2

4F /I M 0)

1.106 The acceleration of the disc along the plane is determined by the projection of the force of gravity on this plane Fx. mg sin a and the friction force fr . kmg cos a. In our case

k = tan a and therefore fr= Fx mg sin a, Let us find the projection of the acceleration on the derection of the tangent to the trajectory and on the x-axis :

m

= Fx cos cp — fr = mg sin a, ( cos q) — 1 )

m

=

— fr cos cp = mg sin a ( 1— cos cp )

It is seen fromthis that wt =

— wx, which means

that the velocity v and its projection vx differ only by a constant value C which does not change with time, i.e. v = vx + C, where vx = v cos cp. The constant C is found from the initial condition v = v0, whence

C = v0 since (Ft — —initially. Finally we obtain 2

v= vo

( 1 + cos tp ).

In the cource of time p 0 and v v0/2. (Motion then is unaccelerated.) 1.107 Let us consider an element of length ds at an angle p from the vertical diameter. As the speed of this element is zero at initial instant of time, it's centripetal acceleration is zero, and hence, d N — ds cos p = 0, where k is the linear mass density of the chain Let T and T + dT be the tension at the upper and the lower ends of ds. we have from, Ft = m wt

61 If we sum the above equation for all elements, the term f dT = 0 because there is no tension at the free ends, so 1/R

k gR f sin cp d cp = Xwt f ds =XI wt

o ( 1 — cos

Hence wt =

R

As wn = a at initial moment

, ei ,

So, w = I w 1= (I — cosR1 ' 1.108 In the problem, we require the velocity of the body, realtive to the sphere, which itself moves with an acceleration wo in horizontal direction (say towards left). Hence it is advisible to solve the problem in the frame of sphere (non-inertial frame). At an arbitary moment, when the body is at an angle 0 with the vertical, we sketch the force diagram for the body and write the second law of motion in projection form Fn = mwn mv 2 or, mg cos 0 — N — mwo sin 0 = R At the break off point, N = 0, 0 =

(1) 00

and let

v = voso the Eq. (1) becomes, 2 V0 —r e g

(2)

cos 00 - wo sin 00 From,

Ft = mwt

v dv ds 21 m R d 0 or, v dv = R ( g sin 0 + wo cos 0 ) d 0 v0 o0

mg sin 0 - mwo cos 0 - m

v dv

Integrating, f v dv = 5 R (g sine + w0 cos0) d 0 o

o 2

V0= g(1 — cos00) + w0 sin 00 2R Note that the Eq. (3) can also be obtained by the work-energy theorem A = A T

frame of sphere) 1

therefore,

mgR (1 — cos 00) + mwo R sin 00 = .f mvo2

[here mwo ) sin 00 is the work done by the pseudoforce (— mi.)] Or,

2 V0 —2if = g

( 1 - cos 00) + wo sin 00

(3) (in

the

62

Hence

1.109

This is not central force problem unless the path is a circle about the said point. Rather here

Ft

(tangential force) vanishes. Thus equation of motion becomes, vt = v0= constant

and,

2 MVO

A

r

r2

for

r = ro

22

We can consider the latter equation as the equilibrium under two forces.

r=

is perturbed, we write 2 A

+

— ( r0 +

x)n

r0+

x

—A (

mvo

= To+ x

When the motion

and the net force acting on the particle is,

1—

r;

2

nx )

+

m vo

ro

To

1— (

x ro

2

)= —

m vo r0

(1—n

)x

2

2 mv„ (--r-1--is an outward directed centrifugul

This is opposite to the displacement x, if n < F

- A -A is th8 inward directed external force).

force while

r 1.110

There are two forces on the sleeve, the weight F1 and the centrifugal force F2.

We resolve

both forces into tangential and normal component then the net downward tangential force on the sleeve is, 2

mg

sin() (1 —

R cos 0)

co

g This vanishes for 0,=

0 and for

0 = 00 = cos -1 (--g--, w hic h is re al if (0.4R 2 2 (.0

2

R > g. If w R < g, then 1 —

W

R cos 0

g

is always positive for small values of hence

the

net

tangential

force

0

and

near 0 =

0

0'

opposes any displacement away from it. 0 = 0 is then stable.

2 If w

2

R > g, 1 — (°

0 near 0 =

0

However 0 =

and

R

cos()

0=0

is negative for small

is then unstable.

00 is stable because the force tends

to bring the sleeve near the equilibrium

position 0 = 00. If w2R =

g, the two positions coincide and becomes a

stableequilibrium point.

63

1.111

De f i n e

the

assum e

axes

w e

as

are

s h o wn

in

the

wi t h

z along

northern

the

local

hem i sphere).

vertical,

Th e n

the

x

d u e e a s t a n d y d u e n o r t h . ( We

Co r i o l i s

force

has

the

c o mp o n e n t s .

--11., Fc0r

=

2 mw [

since

vx

x

=

s ma l l

of

2 mw

and

mo t i o n

(v

Integrating

x=

wh e n

we

get

the

203 v silly

v

(cTi x

il

•

=

0

(constant),

70

j + vx cos0 k 1 = 2mw(vy cos() — vz sin0) t in

wh i c h

centrifugal

vz coscp), y

Y =

m

---,

vx cos0

direction

(neglecting

sing) -

Y

-

2

-

:-• vy cos0 - vz sin0) t

is

equation

=

de

forces)

and

z

=

i =

gun

is

fired

is

due

north.

Th u s

the

are

_NO Trh

It)

—g - gt

Z -Ve r t j c

t + wg t 2 c o s c p

al

x_ E a S t

Finally, 2_

1

r2 singe +

X =

Now

v

»

gt

cogt 3 CORI)

3

in the present

case.

so,

2

2

s

x=

w v sincp

7 cm

1.112

Th e

disc

1.113

exerts

mg,

the

w eight,

the

vertical

force

=

w

sing)

I-

v

and

(to

the

three

east).

forces

wh i c h

v e r t i c a l l y u p wa r d , along

the

are

mu t u a l l y

the

d i a me t e r ,

Co r i o l i s and

force

mw 2 r

Th e y

2 mv ' w

outward

along

are

the

reaction

sleeve

is

Th e

equation

free

to

slide

along

the

rod

A.B.

Th u s

only

the

the

dia

me t e r .

centrifugal

is,

my = mw2 r

wh e r e

d

dv B ut I., =

1

dr at

1 2

v

dr = dr SO,

v =

1 V2 = —2

w

22 r +

2

v

constant

2 2 2 V = Vo+ C.0 r2

Or, being

the

initial

2m

velocity

col/v02 +

=

w

hen

r =

0 . Th e

Co r i o l i s

force

co2 r2 = 2=1)2 r 111 + vo2/(o2 r2

2.83

N

on

putting

the

values.

is

of

perpendicular to the plane of Th e

resultant

is,

Th e

vo

perpendicular.

then,

force

acts

on

it.

64 1.114 The disc OBAC is rotating with angular velocity w about the axis 00' passing through the edge point 0. The equation of motion in rotating frame is, mw = F + mco2 R+2mv xw= F+Fy where Fin is the resultant inertial forc,.., kpseudo force) which is the vector sum of centrifugal and Coriolis forces. — 2mco2 R n + 2mt, co nA

(a) At A, Fin vanishes. Thus 0 =

where n is the inward drawn unit vector to the centre from the point in Thus,

question,here A.

wg (OR _V,2

►2 _V

(02 R.

so, p R Fin= mw2 OC + mw2 BC

(b) At B its magnitude is mw2

V4R2 — r2 , where r = OB.

1.115 The equation of motion in the rotating coordinate system is, mw = F + mco2 R + 2m Now, ' and

v w

1 = 2m F'

R

x (1.

e 4+ R sin()

--a , w' cos 0 er — w sin 0 ee e,,

4

ev R sin 0 ip 0

0 R0 co cos 0 — co sin A

= Fr°((oR sin20 4)) + UDR sin 0 cos 0 ip

coR 0 cos 0

4

Now on the sphere, i7= (— R 62 - R sin2 0 cp2) + (R 6. — R sin cos cp.

2)

+ (R sin

04). + 2R cos 0 c.p ) Thus the equation of motion are, m (— R 62 R sin2 0 c.p 2 ) = N — mg cos 0 + mw2 R sin2 0 + 2mw R sin2 0 4 m (R O. — R sin 0 cos 0 cp 2 ) = mg sin 0 +

R sin 0 cos 0 + 2mco R sin 0 cos 0 c.p m (R

sin 0 cp.. + 2R cos 0 ci)) = — 2 mai R 0 cos 0 From the third equation, we get, q) = — A result that is easy to understant by considering the motion in non-rotating frame. The we get, eliminating 'p mR 62 = mg cos 0 — N m R 0. = mg sin 0 Integrating the last equation, 1 -

2

•2

m R 0 = mg (1 — cos 0)

65 N=

H e nce

So

t he

b

o d y

m u st

fl

y

o ff

fo r

(3

0

-

=

2

0 0

co s

=

mg

0)

co s

-

1

e x a ct l y

as

if

sphere

th e

w ere

n o nr o t a t i n g

5

N o w,

at

t h is

po i nt

Fror

Fc 1

=

c e n tr i

fu g a l

V 0)2 R2 02

fo r ce

si n

0 0

_ 5 (0)2 R

sine0

+ 0)2 R2 x _4 x 2g

V 9

9

x

x

2m

2m

Wh e n

th e

co m po ne nt

2 3

tr a in

and

is

i ts

R I +

m c o2

3

3R

nl (a)

mco2 R

=

0 + (0)2 R2)2 cost

1.116

mw2 R

=

m o v in g

al

a g ni t u d e

m

o ng

(see

2 m

co

t he

v

(He r e

a

dian

p r ev i

co s

0

w e

mer i

o u

=

have

o)

i

is

fo r ce

h as

a

lat e r al

s,

v)

Irs

3 2

ol

k

R

p ut

C o r i

p r ob l e m)

si n

R

th e

o

s

2m

y

co

7t

5

40

0

0

x 1 S

o

,

Fi

a

t

erd

=

2

x

2

00

0

x

8

676

,

x

0

u

=

(b )

T he

o n

res u l ta n t

t he

of

trai n

kN ,

(w e

i n ert i

th e

al

_ 2„ 0

co

(mo)2 R sin2

s

7 7

w r it e

k

fo r ce s

fo r

2

th e

l a ti

tu

de)

a c t i ng

is ,

+ (mo)2 R sin

Th i

3

3 6 00

s

0

+

v a n is h e s

i

+

2m

R

R

co

f

=

For

R 0 cos 0 e:

2m

s

i

n

0

co

s

Ci9 )

0

Ci) )

s in 2

-

0,

co

(

v

=

v

1

=

e

Th u s

cp

(4)

'

"

R

e

tr a i

n

m u st

m o v e

fr o m

4

8

R

co s

fo r

th e

g o

to

t he

t o

east

l

at it u d e

w es t

al

here)

o n g

60th

t he

m o tio n

t he

in

6 -3 7

0

x

g ive n

in

1

1 1 1

H ere

x

1 1

5

8

m /s

4 1

vy

0

=

so

w e

can

plane

co =

-

vz

cos

g

-

=

ng ,

z

0

2

and

I n te gr a t i

as

6

1 x

e

th e

x

=

1 2 —gt

-

2

g cos cpt 2

=

3/2 x=

So

3

1

-

co

g

c os

cp

t

= —

2h

1

g

cos

cp

3

3

2

co h

cos cp /27ji

3

t Th e r e

k

i

s

t h us

a

displ

to

th

e

e

a

s

o

t

f

a ce m e n

3

7

p ar a ll

el

w it h

a

s pee d ,

km / h r

64

q u at i o n

t he

-

2

w r it e

-4 1 R cos A. = — x —21( x 10

2

We

-

t h

co

1.117

0

v We

Th u s

s in

\F0- 0

2 7t

x

-8-

6 4

x

5 00

x

1

x

2 6

cm

ta k e

y

=

0,

t hus

w e

get

fo r

66

1.3 1.118

Laws of Conservation of Energy, Momentum and Angular Momentum. As

A

s constant so the sought work done A=

or,

1.119

A= (3

(7:2b--10

i+4j )•[(2i-3j

)-(i+2j )]..

(3

i+4j )•(i-5j

17J

Differentating v (s) with respect to time dy

dt

we

a ds — 2 Ifs dt

a

a2

_r-

w

CIVS

2

2 is-

(As locomotive is in unidrectional motion) Hence force acting on the locomotive

ma2

F=mw=

2 Let,

at v =

0 at t = 0 then the distance covered during the first 1

S=

, wt

2

a 2 —t = z 2

=

ma2 0721.2) Hence the sought work, A =

seconds

aa t 2 2 4 m aat 2

Fs 2

1.120

t

2

4

8

We have 1

2

2

T- —my = as

2 or,

2 as2

v

2

Differentating Eq. (1) with respect to time 2 vw = r

— 4 as v or, Wr = m

2

as

Hence net acceleration of the particle

Hence the sought force,

1.121

Let

F

F = mw = 2as

\/1 + (s/R)2

makes an angle 0 with the horizontal at any instant of time (Fig.). Newton's second

law in projection form along the direction of the force, gives :

F = kmg

cos 0 +

mg

sin 0 (because there is no

acceleration of the body.) As F tt d rthe differential work done by the force F ,

dA=I•dr" Fds,

(where

ds

= long ds (cos 0) + mg ds kmg dx + mg dy.

sin 0

1 Hence, A

long dx + mg J dy

= lcmg 1 + mgh= mg(k 1+ h).

=

67

1.122 Let s be the distance covered by the disc along the incline, from the Eq. of increment of M.E. of the disc in the field of gravity : AT + AU = Afr 0 + (- mgs sin a) = - lcmg cos a s - lang 1 k I or,

s

(1)

sin a - k cos a Hence the sought work Afr = - kmg [s cos a + / ]

Afr =

k 1mg 1 - k cot a

[Using the Eqn. (1)]

On puting the values Afros -0.05 J 1.123 Let x be the compression in the spring when the bar m2 is about to shift. Therefore at this moment spring force on m2 is equal to the limiting friction between the bar m2 and horizontal floor. Hence K.Xlm k m2 g [where K is the spring constant (say)] (1) For the block m1 from work-energy theorem : A = AT = 0 for minimum force. (A here indudes the work done in stretching the spring.) so,

Fx - 2 From (1) and (2),

- long x = 0 or

F = kg (nt

1

K

2 F

+n12)

(2)

- km g

•

1.124 From the initial condition of the problem the limiting fricition between the chain lying on the horizontal table equals the weight of the over hanging part of the chain, i.e.

N

X rl lg = k (1 lg (where X, is the linear mass density of the chain) So,

k=

(1) 1 -II fr Let (at an arbitrary moment of time) the length . 'of the chain on the table is x. So the net friction force between the chain and the table, at this moment : kN= kkxg (2) The differential work done by the friction forces : dA = r. • dr= - f,. ds-

(--1—)x dr (3) 1 (Note that here we have written ds = - dx., because ds is essentially a positive term and as the length of the chain decreases with time, dx is negative) Hence, the sought work done - kkxg (- dx) = kg

0

A=f

( 1

1 1-1

1 ) 1

x dx

- (1 -

i) rl

2

= - 1.3 J

68 1125

The velocity of the body, t seconds after the begining of the motion becomes v = vo + g t. Thy power developed by the gravity (m g) at that moment, is

P

( T;+ g2t)

mg g -vo a)

(1)

As me is a constant force, so the average power

A mg - Or

= = where 67 is the net displacement of the body during time of flight. As,

= 0

mei. di* so 2 v = — am at 2, Wn

1.126 We have

or, v = wiT? t,

t is defined to start from the begining of motion from rest. dv So, IV iV

dt

Instantaneous power, P = F • v m (wt A A

A

+w

) • KIT? tA ),

A

(where ut and ut are unit vectors along the direction of tangent (velocity) and normal respectively) So, P = mw IfirK t = ma Rt Hence the sought average power t

t

f P dt f ma Rt dt 0

0 t

dt 0 Hence

ma R t 2 2t

ma Rt

2 1.127 Let the body m acquire the horizontal velocity vc, along positive x - axis at the point 0. (a) Velocity of the body t seconds after the begining of the motion, —1, V ni

—10

Vo "I'

Instantaneous power P = F • 17.

w t . (v0-kgt) (- long

) • (vo - kg t) i =

(1)

- long (vo - kgt )

From Eq. (1), the time of motion i = v0/ kg Hence sought average power during the time of motion

king vc, = - 2 W (On substitution) 2

dvx - fang = mwx = mvx or,

dx dvx = - kg dx = - ag x dx

69

To fmd v (x), let us integrate the above equation

J vxdvx. - agf x dx0

or, v2

vii-agx2

(1)

0

P= F• i7= -maxg1/4-agx2

Now,

For maximum power,

dt (142 °x2 - kgx4 Putting this value of x, in Eq. (2) we get, 1

P

(2)

= 0 which yields x.

vo Nr2TIg

2

- — m vo licrg2 1.128 Centrifugal force of inertia is directed outward along radial line, thus the sought work r2

A = f mw2 r dr = 12 mw2

-

= 0.20 T

(On substitution)

1.129 Since the springs are connected in series, the combination may be treated as a single spring of spring constant. K1 K2 K +K2 K1 AT+AU.. From the equation of increment of M.E., 1 K K2 1 2 2 0 + — K AI = A, or, A= M 2 1C 1 K2 ) 1.130 First, let us fmd the total height of ascent. At the beginning and the end of the path of velocity of the body is equal to zero, and therefore the increment of the kinetic energy of the body is also equal to zero. On the other hand, in according with work-energy theorem AT is equal to the algebraic sum of the works A performed by all the forces, i.e. by the force F and gravity, over this path. However, since AT = 0 then A = 0. Taking into account that the upward direction is assumed to coincide with the positive direction of the y — axis, we can write

A= f

+

d r f (Fy - mg) dy

0

0 h

sr. mg f (1 - 2 ay) dy

mgh (1 - ah) = 0.

whence h = 1/a. The work performed by the force F over the first half of the ascent is h/2

AF

h/2

Fy dy = 2mg

(1 - ay) dy =

3 mg/4a.

The corresponding increment of the potential energy is A U= mgh/2 = mg/2a.

70

1.131 From the equation Fr=

-

2a b dU we get Fr= [- ---i- + --i. dr r r

2a b To check, whether the position is steady (the position of stable equilibrium), we have to satisfy d2 U> (a) we have at r = r0, the particle is in equilibrium position. i.e. Fr is 0 so, ro =

0

dr d2U

We have

6a _ 2b

-

dr2

r4

r3

2a

7, , we get

Putting the value of r = ill= d2 U

b4 - — (as a and b are positive constant) dr2 8a3 ' d2U b2 n So, dr2 8a3 > which indicates that the potential energy of the system is minimum, hence this position is steady. dU [- 2a + (b) We have Frm - dr 2. r3 r2 dF For F, to be maximum, SO r

drr= 0

a -L and then F ma b r( x) INE

b3 '

27a As Fr is negative, the force is attractive. 1.132

(a) We have

Fx= - au So,

-

2ax and F

-au dy

Y

-2py

F= 2axi-2 13yi and, F= 2ila2x2+f12y2

For a central force, r x F= 0 Here,

rxF= (xi+y)

)x(-2ctxT:213yr)

= -213xyic-2axy(r)• 0 Hence the force is not a central force. (b) As U= ax2 +13y2 OU So, = — = - 2 a x and F Y ax So,

F=111!+F;=

- au

4a2x2+4 f3

ay

2 y2

According to the problem F = 2 a2 x2 + 12 y2 = C (constant)

-

f3 y.

(1)

71

c2 2

a2 X2 + 02 y2=

Or, 2

2

2 C k (say) 2+ 2 = 22 13 a 2a Therefore the surfaces for which F is constant is an ellipse.

Or,

x

y

(2)

For an equipotential surface U is constant. So,

a x2 + 13 y2 = Co (constant) 2 x

or,

y

Vi3T.

2

+

CO

a 13

Ko (constant)

Hence the equipotential surface is also an ellipse. k.133 Let us calculate the work performed by the forces of each field over the path from a certain point 1 (x1, y1) to another certain point 2 (x2, y2) x2

(i) d A . F cli:1 ay r d F"t aydx or, Ax a f y dx x1

(ii) d A

+ byil.) • dit axcbc + bydy X2

Hence

y2

A= f a xdx + f bydy xi

yl

In the first case, the integral depends on the function of type y (x), i.e. on the shape of the path. Consequently, the first field of force is not potential. In the second case, both the integrals do not depend on the shape of the path. They are defined only by the coordinate of the initial and final points of the path, therefore the second field of force is potential. 1.134 Let s be the sought distance, then from the equation of increment of M.E. r AT+ AU= Af 0- -a niq

+ (+ mg s sina) = - lang cos a s v,2i

or,

/ (sin a + k cos a)

s 2g

- km 4, Hence

Afr = -king cos as-

2 (k + tan a)

1.135 Velocity of the body at height h, vh = V 2g (H - h), horizontally (from the figure given in the problem). Time taken in falling through the distance h. t=

Now

(as initial vertical component of the velocity is zero.)

s = vh t = 1/2g (H + h) x

= 114 (Hh - h2)

72

For sue, --- (Hh — h2) = 0, which yields h =

2

Putting this value of h in the expression obtained for s, we get, smax

=H

1.136 To complete a smooth vertical track of radius R, the minimum height at which a particle 5 starts, must be equal to —R (one can proved it from energy conservation). Thus in our 2 problem body could not reach the upper most point of the vertical track of radius R/2. Let the particle A leave the track at some point 0 with speed v (Fig.). Now from energy conservation for the body A in the field of gravity : mg [h — Of,

2

(1 + sin 0) 1= —1 mv2 , 2

v2 = gh (1 — sin 0)

(1)

From Newton's second laW for the particle at the point 0; FR = mw, , N + mg sin 0 —

my (h/2)

But, at the point 0 the normal reaction N = So,

V2 =

eg

sin 0

From (3) and (4), sin 0 =

0 (2)

2 — and v 3

3

After leaving the track at 0, the particle A comes in air and further goes up and at maximum height of it's trajectory in air, it's velocity (say v') becomes horizontal (Fig.). Hence, the sought velocity of A at this point.

1.137 Let, the point of suspension be shifted with velocity vA in the horizontal direction towards left then in the rest frame of point of suspension the ball starts with same velocity horizontally towards right. Let us work in this, frame. From Newton's second law in projection form towards the point of suspension at the upper most point (say B) : 2 MVB

mg + T =

2 MVB

I

or,

T-

mg

1

Condition required, to complete the vertical circle is that T2 0. But mv2 sw mg (21) +1 mv2 So, 2

A

2

13

(1)

v2= v2 — 4 gl 3

A

(2) (3)

73 From (1), (2) and (3) m (vd2i

4 gl)

T=

- mg

1

k

0 or, vA

B 1:.--"'----.,, VB

2 Arcir

1 / I I I

Thus vA (min) .. 1157gi From the equation Fn = mivn at point C

mv

2

c T=

(4)

/

1 1 1 % ■ \

Again from energy conservation mv2 = 1 mv2 + mg1 2 A 2

\

vT

\ I 112.

V 771g 0, the corresponding vector Fo.coincides with the poitive direction to the z axis, and vice versa. As both discs rotates about the same vertical axis z, thus in vector form. (.0m

--11, --• / li(01÷.12W2i(li+/2)

-1111■

However, the problem makes sense only if col

11 co2

or

wlT �w2

(b) From the equation of increment of mechanical energy of a system: Afr. AT. 1

(il+

) (01 -

W i2z

/2 CO 22z

Using Eq. (1)

1.279 For the closed system (disc + rod), the angular momentum is conserved about any axis. Thus from the conservation of angular momentum of the system about the rotation axis of rod passing through its C.M. gives : mv — 2

,

/

M1

my

2

12

2 CO

(1)

140

(v' is the final velocity of the disc and w angular velocity of the rod) For the closed system linear momentum is also conserved. Hence my

(2)

= my'

4

1 1

ni Vc

(where 1,, is the velocity of C.M. of the rod) From Eqs (1) and (2) we get 1

vc =

3

v—

aandnu

= rive

Applying conservation of kinetic energy, as the collision is elastic 1 2 1 —MV = —MV, 2

2

2

1 2 + —Timv

or

1

2

+— w (3) 2 2 12 2 v — I/2 = 41vc2 and hence v +

4vc

Then 4 v1— 4

i

12 v v and w= (4 + TO / Ti

•

C

Vectorially, noting that we have taken 17* parallel to ii* (4 4+

0-+

So, u= 0 for rl = 4 and /7* 1l t 17* for ri > 4

1.280 See the diagram in the book (Fig. 1.72) (a) When the shaft BB' is turned through 90° the platform must start turning with angular velocity S2 so that the angular momentum remains constant. Here /o o ( / + /0 ) C2 = I° w° or, Q= 1+ 4 The work performed by the motor is therefore

2

(/+

I2 1

4 ) S22 =

0 w2wo

2 / + /0

If the shaft is turned through 180°, angular velocity of the sphere changes sign. Thus from conservation of angular momentum, Q —10 wo = 10 coo (Here

— 4, coo

is the complete angular momentum of the sphere i. e. we assume that the

angular velocity of the sphere is just — coo). Then Q

IT(1)0

and the work done must be, —

2

1

/

1 + — / co

2 o o

2 /

2 0 o

—

r2 L'10

2

141

(b) In the case (a), first part, the angular momentum vector of the sphere is precessing with angular velocity Q. Thus a torque, 2 2 /0 Wo is needed. c Q I 4.10 Io o0 = 1.281 The total centrifugal force can be calculated by, to 2 _i_ i 4-' 0.)1 Xtu- = m i co 2 ° 2

J 0

-

Then for equilibrium,

1

10 t

(T2—T1)2= mg

1 2 and, T2+ T1= — m 100) Thus Ti vanishes, when (

0 2=

(0. IF,

a

/ '

/

= 6 rad/s

0

10

Then T2 = mg— . 25 N / 1.71).

1.282 See the diagram in the book (Fig.

(a) The angular velocity Oihabout 00' can be resolved into a component parallel to the rod and a component co sin() perpendicular to the rod through C. The component parallel to the rod does not contribute so the angular momentum

M= I w sin() = —1 m /2 co sin() 12

Mz = Msin0 = —121

Also,

mi2cosin20

This can be obtained directly also, (b) The modulus of M does not change but the modulus of the change of M is I A M I. I Aril = 2Msin(90-0)= im/2cosin20 12 (c) Here M1= M cosO = 1 w sin° cos° (0 dt

Now

1

2

2

• 2

— — — ml o) sm 0 dt 24 as M precesses with angular velocity w.

dt

= /co sin() cos0

142

1.283 Here M = Ro is along the symmetry axis. It has two components, the part /co cos0 is constant and the part Af.c. /co sine presesses, then dM = / co sine a = mgl sin() dt or,

co' = precession frequency = jn1g013 = 0.7 rad/s

(b) This force is the centripetal force due to precession. It acts inward and has the magnitude --* IF' ..

/2 12 Mi W

M CO' 2 / 1

sin() = 12 mN.

=

517 is the distance of the i th element from the axis. This is the force that the table will exert on the top. See the diagram in the answer sheet

1.284 See the diagram in the book (Fig.

1.73).

The moment of inertia of the disc about its symmentry axis is —

1 2

m R 2. If the angular

1 velocity of the disc is w then the angular momentum is — m R . The precession frequency 2 2w being 2n n, we have

dM dt

1 = — mR 2 co x 2nn

This must equal m ( g + w ) 1, the effective gravitational torques g + (g being replaced by w in the elevator). Thus, co = ( g + w )1 — 300 rad/s 7rR2n

143

1.285 The effective g is 11g 2 + w2 inclined at angle tan- I Iv with the vertical. Then with reference to the new " vertical" we proceed as in problem F283. Thus niv g2T. w2 w=

= 0.8 rad/s.

/co

= tan-i —w - 6° with the normal vertical.

The vector cTi' forms an angle 0

2

2

1.286 The moment of inertia of the sphere is --mR and hence the value of angular momentum

5

2 is - mR 2w. Since it precesses at speed w' the torque required is 5 2

=

- R co 5m

2 2 F' = - mR

So,

5

1

coin = 300 N

(The force F must be vertical.) 1

2

1.287 The moment of inertia is -mr and angular momentum 2 about a horizontal axis making an instantaneous angle.

1

2

- mr co. The axle oscillates 2

iss

27tt cip = (Pm sin T This means that there is a variable precession with a rate of precession value of this is

dc

dt . The maximum

2.7tcp „, . When the angle between the axle and the axis is at its maximum

value, a torque Ico Q 1 =

2

2

2a(1).

mr

7t mr 2 T mo'"

acts on it.

mr 2 coy,. The corresponding gyroscopic force will be

1T

- 90 N

1.288 The revolutions per minute of the flywheel being n, the angular momentum of the flywheel

is 1 x 27tn. The rate of precession is R1'• Thus N = 2 nINV/R = 5.97 1(1■1. m. 1.289 As in the previous problem a couple 27rInv/R must come in play. This can be done if a 23t/nv force, acts on the rails in opposite directions in addition to the centrifugal and other

RI

forces. The force on the outer rail is increased and that on the inner rail decreased. The additional force in this case has the magnitude 1.4 kN. m.

144

1.6

ELASTIC DEFORMATIONS OF A SOLID BODY

1.290 Variation of length with temperature is given by A/ 4= /0( + aAt ) or — = a At = e

(1)

to

But

e=

E, Thus a = ccAtE, which is the sought stress of pressure. Putting the value of a and E from Appendix and taking At =

100°C, we get

= 2.2x 103 atm. 1.291

(a) Consider a transverse section of the tube and concentrate on an element which subtends an angle AT at the centre. The forces acting on a portion of length Al on the element are (1) tensile forces side ways of magnitude aArAL The resultant of these is 2aArA/sin — aArA/Aq5 2 radially towards the cente. (2) The force due to fluid pressure — pr&pAl Since these balance, we get

Ar pma. = am7

where am is the maximum tensile force. Putting the values we get pma.

= 19.7 atinos.

(b) Consider an element of area c1.5

it (

r A0/2 )2 about z —axis chosen arbitrarily. There

are tangential tensile forces all around the ring of the cap. Their resultant is a [ 27t r

AO

Ar

sin

AO 2

Hence in the limit 2 rA0 ) = am n rA0 Ar AO 2 Pm 2 2a. Ar Or

pm =

= 396 atmos.

1.292 Let us consider an element of rod at a distance x from its rotation axis Newton's second law in projection form directed towards the rotation axis — dr = (dm) CO 2 X gm -M W 2x cLx On integrating 2

—T

M O)

2 X

— — + C ( constant ) 1 2

(Fig.). From

145

/

x= ± — or free end, T= 0

But at

2

2

Thus

2

MO)

0-

1

2

Hence

T—

T

Thus

—

MCO 2

+C or C .

4

/

8

mw2 (1 2

MO) 2i

_m8

x2)

4

1

(at mid point)

Condition required for the problem is

Tom= Sam 2

7213 r 2 — S am or co = — — O' 8 / P Hence the sought number of r p s MCO

/

S

n=

a)

1 1/2a,„

2n

n/ •

—=

[using the table n . 0.8 x 102rps ]

v

p

1.293 Let us consider an element of the ring

(Fig.). From Newton's law Fn = mw„ for this

element, we get,

Td0 = (—m dO)co 2 r

[see solution of 1.93or 1.921

2n nl 2 AT =Mg .-.-- '€0 r Lit

So, Condition for the problem is :

T nr

or,

2

MO) r, 2 S 0m o

It 2

2

r

r 2 S aamm

2Jt 2 an, r

wymix =

nr 2 ( 27trp

am )

Pre

Thus sought number of rps comax n=

1 _2y

2n 2nr p Using the table of appendices n -

23rps

1.294 Let the point 0 desend by the distance x (Fig.). From the condition of equilibrium of point 0.

— V( 1/2 )2 + x 2

2T sine . mg or T= --1-1111--= 2sin0

T

Now, n

( d/2

a -

el

2x

or T.

)2

( a here is stress and

E

ez

d2 n — 4

is strain.)

146

In addition to it, V(//2)2+x2 E_

1/1 +

1/2

g-X

1 (3)

-

From Eqs. (1), (2) and (3) x

xV 4x3 So, — 212

T )2 2x

32f21_ as x « / 7tEd 2

mgl nEd 2 m

1/3

or, x= 1 (

g = 2.5 cm 2irEd1.295 Let us consider an element of the rod at a distance x from the free end (Fig.). For the considered element 'T - T ' are internal restoring forces which produce elongation and dT provides the acceleration to the element. For the element from Newton's law : F (-171d,r)F1 dx 1 m 1 As free end has zero tension, on integrating the above expression, dT = (dm) w

dr =

T

0

dx

F or T= °x

0

Elongation in the considered element of lenght dx : a" EL (x)

T —SE dx SEL F.1

Thus total elengation

=

xdx=2SE

SE1

T

Tilfr

Hence the sought strain z F. 0 1 2SE 1.296 Let-us consider an element of the rod at a distance r from it's rotation axis. As the element rotates in a horizontal circle of radius r, we have from Newton's second law in projection form directed toward the axis of rotation : T - (T + dn. (dm) w2 r or,

- dr =

1

dr) w2 r =

1

w2 r dr

147 At the free end tension becomes zero. Integrating the above experession we get, thus

0

1

— f dT = L?i 0)2 f r dr T

Thus

T=

m

(02

r

12

- r2 ) . m (02 / (1 _ r2

2

1

2

/2

Elongation in elemental length dr is given by :

a = al.l dr ..• --L dr E SE (where S is the cross sectional area of the rod and T is the tension in the rod at the considered element) 2 m (o2 1 Or, a = SE 1- )dr 2 12 Thus the sought elongation

I

mw2ir,

-. .fd-or,

2SE

2SE

2\

dr

J ri-2) 0

m w21 21 -

r

3

(S/ p) =

2 (1)

13

3 SE 1 2 3 ' - =''' --- (where p is the density of the copper.) m 3 E

1.297 Volume of a solid cylinder V = TE r2 1 So,

AV it2rArl V ic

n r2 1

xr2 Al

.

2Ar

+ I t r2 1

r

4.

A/ 1

But longitudinal strain A1/1 and accompanying lateral strain A r/r are related as Ar Al r 2E —II

7-

Al 1

But

— Fh t r2 E

(Because the increment in the length of cylinder Al is negative) So,

AV , v

—F it r2 E

(1 — 2 ii)

(1)

148

Thus,

- Fl

AV

(1 - 2 it)

Negative sign means that the volume of the cylinder has decreased. 1.298

(a) As free end has zero tension, thus the tension in the rod at a vestical distance y from its lower end -7.gy

(1)

Let al be the elongation of the element of length dy, then al Td = SE 1"-Y =

SlE

-

a

1) 0 dy

- p gydy/E (where p is the density of the copper)

Thus the sought elongation Al=

p ra

ydy. ln ai2/E 'J E 2"

f

A/ (b) If the longitudinal (tensile) strain is e = —

(2)

the accompanying lateral (compressive)

strain is given by , E

Ar -

(3)

Then since V = n r21 we have AV V

2A r A/ +— r /

A where — is given in part (a), it is the Poisson ratio for copper. / 1.299 Consider a cube of unit length before pressure is applied. The pressure acts on each face.

The pressures on the opposite faces constitute a tensile stress producing longitudianl compression and lateral extension. The compressions is

aa nndd the lateral extension is it, E.

The net result is a compression 2-- (1 - 20 in each side.

Hence

AV AV =(1 - 2 .t) because from symmetry v V E

Al =3T

149 (b) Let us consider a cube under an equal compressive stress a, acting on all its faces. Then,

volume strain = -

AV — = k' V

(1)

where k is the bulk modulus of elasticity. a

3a

So

- 2 IA)

or ,

E= 3 k (1 -

3 = -(1 - 21.4)(as k=

1 s - i if E and 3 are both to remain positive. 2 1.300 A beam clamped at one end and supporting an applied load at the free end is called a cantilever. The theory of cantilevers is discussed in advanced text book on mechanics. The key result is that elastic forces in the beam generate a couple, whose moment, called the moment of resistances, balances the external bending moment due to weight of the beam, load etc. The moment of resistance, also called internal bending moment (I.B.M) is given by I.B.M. = BUR Here R is the radius of curvature of the beam at the representative point (x, y)./ is called the geometrical moment of inertia

fz2ds of the cross section relative to the axis passing through the netural layer which remains unstretched. (Fig.1.). The section of the beam beyond P exerts the bending moment N (x) and we have, El — - N(x) R If there is no load other than that due to the weight of the beam, then 1 N (x) = - p g (1 - x)2 bh 2

Here b = width of the beam perpendicular to paper. h/2 3

Also,

If

ttiz

bh 12

-h/2

Hence,

(1) k)(1

6 p g12 Eh2

- (0.121 km) -I-

TIEDIFI:12727:011.=

t

150 1.301 We use the equation given above and use the result that when y is small d 2y dx2

R

Thus

d2y ' dx2

N (x) E

(a) Here N (x) = No is a constant. Then integration gives, dy No r dr E + s-1 But

= 0 for x = 0, so C1 - 0. Integrating again, 2 No x

Y = 2 EI where we have used y = 0 for x = 0 to set the constant of integration at zero. This is the equation of a parabola. The sag of the free end is No /2 X

—

y

(x

In

2EI (b) In this case N (x) = F (1— x) because the load F at the extremity is balanced by a similar force at F directed upward and they constitute a couple. Then d2y F (1 — x) dr2 EI F (lx — x2/2) + L-1 dx EI As before C1 = 0. Integrating again, using y = 0 for x = 0 dy

Integrating,

/x2 x3\ 2 6

F(

y=

Here for a square cross section

F/3 3 EI

here k EI a/2

I= f 2 a dz = a4/12. - a/2

1302 One can think of it as analogous to the previous case but with a beam of length 1/2 loaded

F/2

upward by a force F/2. Thus

k

T

F /3 48E1'

1.

••• ` ...

s' ... ••• ... .........' .... .1•1•

MEM ...............

On using the last result of the previous problem. 1.303 (a) In this case N (x) = Also

1

pg b h (1 — x)2 where b = width of the girder. I = b h3/12. Then,

..• ...,

151

Ebh2d2y.pgbh 12

dx2

(12

21x +x2).

2 3 6 pg(g x_ix2+ 1

Integrating,

dr using

E h2

3

-c1X .. 0 for x = 0. Again integrating

dr 6

pg (/ 2 X2

Eh2

3

+12

1

1

3

12

2

6pg14(1 Thus

x4)

1X3

X

-+ -

Eh2

2

4

4

6 pg1

3

3 pg/

—Eh2—.1 3g —2E7r

3w

2

(b)

As before,

EI

d y

N (x)

where

N (x)

is the bending moment due to section

PB.

cbc2 This bending moment is clearly

N= f w

dg

-

-

(2/ -x)

x 2 W 212 -

(Here

w= p g b h

dy El —dx w ( T -x3

or since

= 0 for x =

As

again,

y = 0 for

1.304

1, co = w x4 (— 24

EIy=w

x = 0, c1 =

" Y(x-

- xl

is weight of the beam per unit length)

Now integrating,

Interggrating

wl (21 - x)= w

+ -x12

x2 / co

13/3

-

x3 /)

+

w/3 x +C1 3

0. From this we find

5"4 /EI 24

5Pg14 1 7

The deflection of the plate can be noticed by going to a co - rotating frame. In this frame each

element

of the plate experiences a pseudo force proportional to its mass. These

forces have a moment which constitutes the bending moment of the problem. To calculate this moment we note that the acceleration of an element at a distance a =

from the axis is

and the moment of the forces exerted by the section between x and

1 is

152

x

From the fundamental equation (13 — x3).

E/5j= tp 1 h +1./2 1 h3 12.

The moment of inertia / f z21 d z-

Note that the neutral surface (i.e. the surface which contains lines which are neither stretched nor compressed) is a vertical plane here and z is perpendicular to it. d 2y dx2

4 P13 (j3 — x3). Integrating Eh2 4 4 p (3 (13x x ) E h2

ay dX

Since

+c

A= 0 , for x = 0, c1= 0. Integrating again, 13x2 Eh

2

xs 20 +c2

Thus 5E h2 1.305

(a) Consider a hollow cylinder of length 1, outer radius r + Ar inner radius r, fixed at one end and twisted at the other by means of a couple of moment N. The angular displacement p, at a distance 1 from the fixed end, is proportional to both 1 and N. Consider an element of length dx at the twisted end. It is moved by an angle q as shown. A vertical section is also shown and the twisting of the parallelopipe of length 1 and area Ar dx under the action of the twisting couple can be discussed by elementary means. If f is the tangential force generated then shearing stress is f/Ardx and this must equal GO= G ?1. since 0= 1' Hence,

f G Ar dx

1•

/•

The force f has moment fr about the axis and so the total moment is N— G Ar

1

r

dx =

23tr3 Ar 1

G

153 (b) For a solid cylinder we must integrate over r. Thus r

dx

N=

f 2 a r3 dr cp G

1

0

7( r4 G

21

(1)

d2/2

3 1306 Clearly N =

f2nr

dr 9)

it G (p (d24 32 1 -4)

G

d1/2

G= 81 G Pa = 84 x 101

using

oN m2

d2= 5 x 10 -21111, d1= 3 x 10 -2 M

ozp = 2.0° = 90

radians,

n x 8.1 x it ( ,)c 32 x 3 x 90 "--'

N-

1= 3 m

- 81) x 102 N•m

= 0.5033 x 103 N•m

=, 0.5 k N•m

1.307 The maximum power that can be transmitted by means of a shaft rotating about its axis is clearly N w where N is the moment of the couple producing the maximum permissible torsion, cp. Thus

P.

itr4G cp 2/

co = 16.9 kw

1.308 Consider an elementary ring of width dr at a distant r from the axis. The part outside

d exerts a couple N + N—

dr

dr on this ring while the part inside exerts a couple N in the

opposite direction. We have for equilibrium

dN dr dr = - dip

154 where dl is the moment of inertia of the elementary ring, 3 is the angular acceleration and minus sign is needed because the couple N (r) decreases, with distance vanshing at the outer radius, N (r2) = 0. Now

m dl =

2ardr

r

2'

ri)

2m 13

Thus

dN

Or,

(r2i — r2i)

dr

(r1 — r4), on integration

m

N=2 1

3 r

N-i-del dr

1.309 We assume that the deformation is wholly due to external load, neglecting the effect of the weight of the rod (see next problem). Then a well known formula says, elastic energy per unit volume —

1 stress x strain = — 2 2

1 m This gives — — 2 c. 0.04 kJ for the total deformation energy. 2 p Ee 1310 When a rod is deformed by its own weight the stress increases as one moves up, the stretching force being the weight of the portion below the element considered. The stress on the element dx is p Jtr2 (1 - x) g/n r2 = p g (1 - x)

The

extension of the element is A dx = d &

=

p

g (I -

x) dx/E

1 Integrating A/ = — s the extension of 2 p g 12/E i the whole rod. The elastic energy of the element is 1

g(1--X) Pg(1 - X ) P E

2 r

cfr

Integrating Ai

2g2 /3/E AU= —1 70.2 n 6 r

_2 3

r21 E

r)

2

155 1.311 The work done to make a loop out of a steel band appears as the elastic energy of the loop and may be calculated from the same. If the length of the band is

1, the radius of the loop R =

/7t. Now consider an element 2 ABCD of the loop. The elastic energy of this element can be calculated by the same sort of arguments as used to derive the formula for internal bending moment. Consider a fibre at a distance z from the neutral surface PQ. This fibre experiences a force p and undergoes an extension ds where ds = Z d p, while PQ = s = R d cp. Thus strain —

ds s

Z = R. If a is the

cross sectional area of the fibre, the elastic energy associated with it is _1 E (_Z)2 R dcp a 2 R Summing over all the fibres we get EIS 2R

I Z-_

E 1 d (ID 2R

ds

For the whole loop this gives,

21

using f d cp = 21t, n it

2E/7t 2

6/2 I = f Z2 hdz

Now

h 63

12 -6/2 1 TE2Eh 83

So the energy is

6

1

k

— 0.08 kJ

1.312 When the rod is twisted through an angle 0, a couple

N

(0) —

r4 G 0 appears to resist this. Work done in twisting the rod by an angle pis

2/

then 7 J on putting the values.

IN(0)d0= / 4

4

r

0

G

c p 2 = 1.313 The energy between radii r and r + dr is, liy differentiation,

7cr 3 dr

/ 7C r3 dr

G cp2 27crdrl 1

Its density is

G

1 G cp2 r2 2

12

1/2 stress x strain. Stress is the pressure p g h. Strain is 1.314 The energy density is as usual p gh by defination of (I. Thus

px

=

p (p gh)2

=

23.5 kJ/m3 on putting the values.

156

1.7

HYDRODYNAMICS

1.315 Between 1 and 2 fluid particles are in nearly circular motion and therefore have centripetal acceleration. The force for this acceleration, like for any other situation in an ideal fluid, can only come from the pressure variation along the line joining 1 and 2. This requires that pressure at 1 should be greater than the pressure at 2 i.e. Pi >P2 so that the fluid particles can have required acceleration. If there is no turbulence, the motion can be taken as irrotational. Then by considering

f i

17t dr= 0

along the circuit shown we infer that V2 > V1

(The portion of the circuit near 1 and 2 are streamlines while the other two arms are at right angle to streamlines)

)

._-.-,

In an incompressible liquid we also have div i7=

0 By electrostatic analogy we then find that the density of streamlines is proportional to the velocity at that point. 1.316 From the conservation of mass V1 S1 V2 S2 (1) But S1 < S2 as shown in the figure of the problem, therefore vi > V2 As every streamline is horizontal between 1 & 2, Bernoull's theorem becomes 1 p+— v2 = constant, which gives 2 Pi < p2 as vi > v2 As the difference in height of the water column is Ah, therefore P2 -P1 m

(2)

P Oh

From Bernoull's theorem between points 1 and 2 of a streamline P1 + Of,

or

1 2 1 2 2 P vi =P2 + 2 PV2 1 p (v2, 2 (v1

P2 - P1 m

pgAh . 2

vi) -

(3)

p (vi — v22)

(using Eq. 2)

using (1) in (3), we get _____> Vi

... S i

1I

2gAh 2 4

_

— 4 -- - , '

Hence the sought volume of water flowing per see 2gAh

i

2

157 1.317 Applying Bernoulli's theorem for the point A and B, pA = pB 1

or,

v2 as, vA =

2p 2

0 A h Po g

P v = PA -PB

.1 /2 Ah po g v=v

So,

A V2Ahpog

Thus, rate of flow of gas, Q = S v = S The gas flows over the tube past it at B. But at A the gas becomes stationary as the gas will move into the tube which already contains gas.

1

In applying Bernoulli's theorem we should remember that p + v2 + gz is constant along p -2 a streamline. In the present case, we are really applying Bernoulli's theorem somewhat indirectly. The streamline at A is not the streamline at B. Nevertheless the result is correct. To be convinced of this, we need only apply Bernoull's theorem to the streamline that goes through A by comparing the situation at A with that above B on the same level. In steady conditions, this agrees with the result derived because there cannot be a transverse pressure differential. 1.318 Since, the density of water is greater than that of kerosene oil, it will collect at the bottom. Now, pressure due to water level equals h 1 p1 g and pressure due to kerosene oil level equals h2 p2 g. So, net pressure becomes h1 p1 g + h2 p2 g. From Bernoulli's theorem, this pressure energy will be converted into kinetic energy while

- - ro s e n Oi La■-

maL

lfowing through the whole A. i.e.

hi pi g + h2 p2 g =

1 2 2 pi v

1 --

P2

2 (h1 + h2 —) g Pi

Hence v =

= 3 m/s

A

1.319 Let, H be the total height of water column and the hole is made at a height It from the bottom. _ Then from Bernoulli's theorem

1

2

- pv 2

(H - h) pg

r

•••••

."1". •••••

or, v =

- h) 2g , which is direFted horizontally.

For the horizontal range, 1= v t = 1/2 g (H — h)

• \r"h

_

21I (Hh — h2)

ow/

HL

158

Now, for maximum 1,

d (H

h — h2) dh =0 h = 2= 25 cm.

which yields

1.320 Let the velocity of the water jet, near the orifice be v', then applying Bernoullis theorem, 1

1

2

p v 2 = ho p g + p v

= V v2 — 2g ho

Or,

(1)

Here the pressure term on both sides is the same and equal to atmospheric pressure. (In the problem book Fig. should be more clear.) Now, if it rises upto a height h, then at this height, whole of its kinetic energy will be converted into potential energy. So, 1 V12 — va = p gh or h 2g 2

V2

— ho g• 20 cm, [using Eq. (1)]

1321 Water flows through the small clearance into the orifice. Let d be the clearance. Then from the equation of continuity ( 27( Rid ) vi =

r d)v = (2arR2d) v2

(

R R v1 1 = v r = v2 2

or

(1)

where v1 , v2 and v are respectively the inward

radial velocities of the fluid at 1, 2 and 3. Now by Bernoulli's theorem just before 2 and just after it in the clearance 1 PO

h Pg

132 +

2 PV2

•we

1.11

••■• •■•• ••■• m• •• •

■■11.

sum. ..m.on

••■■•0

(2)

•••••

■•••■ ••■■■■ •••••• ■•• mmi•

..■■■••

.6*

...• •••■■■••

ma / •■•••••■•

■■■••

Applying the same theorem at 3 and 1 we find that this also equals 1

p + — pv 2

2

1

p0 + — pvi2

(3)

2

(since the pressure in the orifice is pc, ) From Eqs. (2) and (3) we also hence (4)

vi=

and

po +

= p0 +

1

2 pvi

hpg (1

—

[Using (1) and (4)]

159

1311 Let the tome acting on the *ton be F and the length of the cylinder be 1. Then, work done = Fl

(1)

iso441NAny„ lat.mealli's N)attittill for points 2 A and B,p . -1- p v where p is the density 2 and v is the velocity at point B. Now, force on the piston, F - pA. p v2A

/

-

(2)

...,

. . . w o w ,a ........................

4. ..---..

■■•■••

.www.

..■...•

_ ..........

-........... 1 ...................- - - - - 1

V V. Svt or v= -,-.,at From Eq. (1), (2) and (3) work done

1

■■•• •••

- V - ---- s - -----.*: - -__ --_- -z-- A-- .---a= ..._, -_ _ --_ .... - -,. a

2 where A is the cross section area of piston. Also, discharge through the orifice during time interval t = Svt and this is equal to the volume of the cylinder, i.e.,

1 i. -i pv 2Al-

,__ ■ ...ma.

(3)

A V2 /- /pv3/s31.2

(as Al - V) i P (St)2 2 1323 Let at any moment of time, water level in the vessel be H then speed of flow of water through the orifice, at that moment will be v -112-111-

(1)

In the time interval di, the volume of water ejected through orifice, dV - svdt On the other hand, the volume of water in the vessel at time t equals V.. S H Differentiating (3) with respect to time,

Eqs. (2) and (4)

dV e, dH -a — dt dt

or dV .. S d H

SdH= svdt or dt=

S dH

fdt-

S

,from (2)

t=

dh

siih— f VII k

0

Thus,

(4)

s Nrig1 7 0

r

Integrating,

(2)

S s

VT; g

1.324 In a rotating frame (with constant angular velocity) the Eulerian equation is , d v" - V p + p g'+ 2 p 67' x (7.)') + p co 2 F..= p dt In the frame of rotating tube the liquid in the "column" is practically static because the orifice is sufficiently small. Thus the Eulerian Eq. in projection form along 7*(which is

160 the position vector of an arbitrary liquid element of Tenth dr relative to the rotation axis) reduces to

p (r) = po +P-(2L) —2{r2

Thus

— (1— h)21

(1)

Hence the pressure at the end B just before the orifice i.e. (2) Then applying Bernoull's theorem at the orifice for the points just inside and outside of the end B

Po +

1

p (o2 (21 h — h2) = Po + 2

So, 1.325 The Euler's equation is

2 p v ( where v is the sought velocity)

v = wit

\r2 / — Ti- — 1

ctiT' --a --, P Tit- — f — VP I.

dv

av

—

V (p + p gz), where z is vertically upwards.

, --1 „b, —,..

--. — + k . v j V. dt at v•

Now

(1)

But

V)v V (1 v2 rx Curl r (2) 2 we consider the steady (i.e. a,",at = 0) flow of an incompressible fluid then p = constant. and as the motion is irrotational Curl So from (1) and (2)

or,

Hence

v

0

p V v2) = — V (p + p gz) V p+Z pv2+pgz

=0

2 p + —1 p v p gz = constant. 2

1.326 Let the velocity of water, flowing through A be vA and that through B be vB, then discharging rate through A = QA = S VA and similarly through B = S vB. Now, force of reaction at A,

161 Hence, the net force,

F=

pS (vA2 - vB2) as

Applying

Bernoulli's

F

A11

theorem

Fp

to

(1)

the

liquid

flowing out of A we get

1 PO +

PO +

Pgh

and similarly at

2

amIND •■•■••

em•

•••••

.1••■• ■•••

PvA

=WO ••■

ea .

B

Po + Pg

(11+ Ah)

=

(v8

Hence

1 pvB

Po +

B •mw

- vd2i)

2

=Ahpg

•■••• ••■

F = 2pgSAh =

Thus 1327

Consider an element of height

dy at

0.50 N

a distance y from the top. The velocity of the fluid

coming out of the element is

v 1/2-Ty The force of reaction = p

dF

due to this is

dF = p dA v2, as

in the previous problem,

(b dy) 2 gy h

F = p gb f 2y dy

Integrating

h-1 = p gb

[h2 - (h - 1)2]= p gbl (2h - 1) (The slit runs from a depth h -1 to a depth h from the top.) 1.328

Let the velocity of water flowing through the tube at a certain instant of time be

u.

--2-, where

Q

is the rate of flow of water and n

r2

u, then

is the cross section area of the tube.

r —

pQui-pQuj.

AN•1■0111.

■•■=1.

al•■•

■N.

Hence,

the

sought moment of force about 0

••••■•

mama

■••

■•

becomes 2

1 (-1)x(pQui-pQui)= pQulr=

2-2— 1i° 2

r

and

I/71 P

r2

Q21

0.70 Alin

▪

▪

▪

162

1329 Suppose the radius at A is R and it decreases uniformaly to r at B where S

= 7r.R2 and

s = 7tr2. Assume also that the semi vectical angle at 0 is a. Then L2 So

Y

r +

Li

X

R-r (x - L 1) L2 - Li

where y is the radius at the point P distant x from the vertex 0. Suppose the velocity with which the liquid flows out is V at A, v at B and u at P. Then by the equation of continuity ty2u 3tR2v . 7tr2v . 7 The velocity v of efflux is given by ..... - _ _ _ ... I-_ - -_

v = 112 i and Bernoulli's theorem gives

h .. _ _-..

_ _2 1 -2 pv2

1 2 pp + -f put = po +

B

"P.

_

where pp is the pressure at P and pc, is the atmospheric pressure which is the pressure just outside of B. The force on the nozzle tending to pull it out is then

...-

A Bt

0

0

F = f (pp - pa) sin() Ityds We have subtracted pc, which is the force due to atmosphenic pressure the factor sin 0 gives horizontal component of the force and ds is the length of the element of nozzle surface, ds = dx sec 0 and

-r tan 0 = R L2-L1 Thus L2

dx

4 Ttp f v2 (I

y dy

-

4

= 71 pv2 1 (R2 2

- rr2

n(R2 +—r -

p gh (

r

=

r2)2

2

R2

= pgh (S - s)2/S = 6-02N on putting the values. Note : If we try to calculate F from the momentum change of the liquid flowing out wl will be wrong even as regards the sign of the force. There is of course the effect of pressure at S and s but quantitative derivation of F fron Newton's law is difficult.

163

-4, 1.330 The Euler's equation is p — f - V p in the space fixed frame where f = - p g k dt downward. We assume incompressible fluid so p is constant. Then f = - V (pg z) where z is the height vertically upwards from some fixed origin. We go to rotating frame where the equation becomes 2

-V(p+ pgz) +pco r

-11 + 2p (i/ x ro)

the additional terms on the right are the well known coriolis and centrifugal forces. In the frame rotating with the liquid v = 0 so 1 2 7 p + pg z - -2 p co ror

2

=0

p+ pgz-

p co2r2 = constant 2 On the free surface p = constant, thus

z

r2 + constant 2g If we choose the origin at point r 0 (i.e. the axis) of the free surface then "cosntant" = 0 and f.12

z=

r2 (The paraboloid of revolution) 2g At the bottom z = constant So

p=

p (1)2 r2 + constant 2 If p = Po on the axis at the bottom, then 1 Po + -2 P

2 _2

1.331 When the disc rotates the fuild in contact with, corotates but the fluid in contact with the walls of the cavity does not rotate. A velocity gradient is then set up leading to viscous forces. At a distance r from the axis the linear velocity is w r so there is a velocity gradient co r both in the upper and lower clearance. The corresponding force on the element whose radial width is dr is cor dv ) Ti 2nrdr — (from the formular F = iiA h The torque due to this force is tor 23rrdr —r h ind the net torque considering both the upper and lower clearance is 2f T12 itr3 dr 51-) o = iciewri/h So power developed is ecozpii P v •n = 9.05 W (on putting the values). (As instructed end effects i.e. rotation of fluid in the clearance r > R has been neglected.)

164

1.332 Let us consider a coaxial cylinder of radius r and thickness dr, then force of friction or viscous force on this elemental layer, F =

2 It r 1 1 —dv. dr'

This force must be constant from layer to layer so that steady motion may be possible. F dr

or,

= 2n 1

(1)

dv.

Integrating, r

F f

= 27( I rl f dv R2

or,

F In

v

=

(2)

Putting

r = R1,

we get

R1 Fin-1.2

2nlrl vo

From (2) by (3) we get, In v = vo Note : The force velocity vo . 1.333

F

r/R2

In R1/R2

is supplied by the agency which tries to carry the inner cylinder with

(a) Let

us consider an elemental cylinder of radius r and thickness formula dw F= 2nrli

r

from Newton's

d(O

= dr

dr then

2n/Tir 2

dr

and moment of this force acting on the element, do)

N=

2nr 2lrl —

3 r=

2nr

dr

or,

d 11

w dr

dr

(2)

2 n / do) = N --3-

As in the previous problem N is constant when conditions are steady r

0)

Integrating,

2 n 1 ri fd w =

f—

N

dr3r

Rt

or,

23t/lw=

Putting

r

N

[

R2, CO = w2 , [

2yr/yi (02= -2-

1

11

(3)

we get 1

- 11 R1 R2

(4)

From (3) and (4), 1

1

(b) From Eq. (4),

D2 D2 N 1 %1 '2 N1 = —1 = 4 n ri (o2 2 2 R R 1.334 (a) Let dV be the dr then,

lfowing per second through the cylindrical shell of thickness

r2 dV = — (2 71 r dr) vo(1 — v?i). 2 n vo (r —

r3 dr R2

R

(

V= 2nvo

r3

r — --2-) dr = 2 n v

R o

R2

31 ...

° —4 —2 R

v°

Fr (b) Let, dE be the kinetic energy, within the above cylindrical shell. Then 1 1 dT = — (dm) v= — (27t r 1 d rp) v2 2 2 1 • 2 r2 2r

R

R

R

(r T= nip v20

2r3 — — +R4

nR2plvo =

6

Here frictional force is the shearing force on the tube, exerted

dv — dt. r2 dv dr And at

r=

r

— dv

2 vo R2-

AR = ' dr

2 vo

R

166

Then, viscous force is given by, F -

- Ti (2.7t RI) (

dv dr r . B

2v,,

= -27cRi/(-

= 47cTivo/ R

(d) Taking a cylindrical shell of thickness F= -

dr

and radius r viscous force, dv — dr '

Ti (231 rl)

Let Ap be the pressure difference, then net force on the element = Ap it r2 + 2 7( 11 r —

dr

dr

But, since the flow is steady, Fn et= 0 r dv - 2 1 t 1 i r ( - 2 v o . 71

- 2nhir— or,

dr

Ap-

= 4 .ri vo /A?

n r2

nr2

2

1.335 The loss of pressure head in travelling a distance I is seen from the middle section to be h2 - h1= 10 cm. Since h2 - h1 = h1 in our problem and h3 - h2 = 15 cm = 5 + h2 - h1 , we see that a pressure head of 5 cm remains incompensated and must be converted into kinetic energy, the liquid flowing out. Thus i,2 -P- - = pg A h where A h = h3 - h2 Thus

v = 15T h a 1 m/s

1.336 We know that, Reynold's number

(Re)

is defined as, Re.. p v

Iii,

where v is the velocity

1 is the characteristic length and Ti the coefficient of viscosity. In the case of circular cross section the chracteristic length is the diameter of cross-section d, and v is taken as average velocity of flow of liquid. Now, Ret (Reynold's number at x1 from the pipe end) -

Pdi v1 where v1 is the velocity m1

at distance x1 p d2 V2

.. Re ,

and similarly,

Re,

1 v1 - d V2 Re, d2

SO 1

From equation of continuity, Al v1= A2 v2 2 or, it ri vi = it r22 v2 or di vi ri .. d2 v2 r2 - ax 2

div 1 au

d2 V2 Thus

r2

r1

r

:

e

0 roe - axt Re,

—

e- a

Ar (as x2 - x1= Ax)

a ex - e

e. -

,

Re,

1.337 We know that Reynold's number for turbulent flow is greater than that on laminar flow: 2 p2 v2r2 pvd . 2 pi vi ri and (RA :. Now, (Re)iTI 1 11

167 But, (ROt a (R e)1 pi viri SO

v2

12

21

P2 r2

d

v 1.3311 We have R =

5 it m/s on putting the values.

at

mu

and v is given by

1

4 it

67rrirv=

(P =

r2 (p - po) g 3 density of lead, Po = density of glycerine.) v a.

2 11- (1) P°) g r2

4 and

=

dv m— dt

mg - 6 n

13°) g d2

=

Po) g

Po d3

5.2 nun on putting the values.

ry

dv 63trir —+ m v = g dr

or or

+kv. g,k.. Ltil

dv dt ict (IV

or

e la

Or

2 (13

18

1/3 d = [9 ri2/pc, (p - Po) g ]

1339

UTI (P

1

1

Thus

1

ve

m kt

v- ge

dT fice

kt

or -c/- e

dt

g

g

kt v - ge

- kt (where C is const.)

.. - e il + C or vCe k -k +

Since

kt

v. 0 for t= 0,

0= 1+C

The steady state velocity is fc. v differs from

by n where

1 t = - ln n

or

k

4n Thus

n

e-

1=

-

3

r3 P

4r2p

kWe have neglected buoyancy in olive oil.

--

d2p

6 n II r 18 1

18 r)

168

1.8

RELATIVISTIC MECHANICS

1.340 From the formula for length contraction

(10-10117 So, 1.341

2 V 1 - -2- =

(1 - 1)2

C

)- fi lo

or v - c Itri (2 - ri)

(a) In the frame in which the triangle is at rest the space coordinates of the vertices are

NIT

a — , - -0 , all measured at the same time t. In the moving 2' 2 frame the corresponding coordinates at time t' are

/3a (000), a( — , + -, 0 a

A : (ve, 0, 0), B :( g-v3-.N 2

r1 --.p2+ ve 1( 0) and C :: (lf- V31/1 - 132 + Vti , - g-, 2'

2

2

, 0

The perimeter P is then in

P= a + 2a (14 (1 -132) + 14)

= a (1 +114-737 )

(b) The coordinates in the first frame are shown at time t. The coordinates in the moving frame are,

8

B

L. oafs \ 2 ) —2- ,Cy

0)0,0) C ca,o,o A : (vt', 0,0), B : (11117112 + vti , a

_s_o), c:

(a VITT + ve , 0 , 0)

The perimeter P is then 1/2

P= a11 -72- +1[1-132 +3]

x 2- a ( 71

71 +"4 - 1-32- ) here 13=

1.342 In the rest frame, the coordinates of the ends of the rod in terms of A : (0,0,0) B : (l0 cos00 , /0 sin00 , 0) at time t. In the laboratory frame the coordinates at time t' are A : (ve, 0, 0), B :(10cos00\117-1-32 + ve ,10 sin00 , 0)

V c

proper length /0

)

169

Therefore we can write, 1 1 cos 00 = 10 cos00 117 2- and

Hence

q . (12)

cost 0 + (1 -

/ sin 0 = 10 sin00

132 1-12

) sine

0

- 132 sin2 0 -132

or,

A

1.343 In the frame K in which the cone is at rest the coordinates of A are (0,0,0) and of B are (h, h tan 0, 0). In the frame K', which is moving with velocity v along the axis of the cone, the coordinates of A and B at time t' are

Vt _A2

x'B - x'A

and the lateral surface area is, S = n h'2 secO' tan0' h2 (1 ti2)

tan° N

tan2

/32

50 11

cost

0

:

Here S0 - n h2 sec° tan° is the lateral surface area in the rest frame and h'= h111771 , 13- vie. 1.344 Because of time dilation, a moving clock reads less time. We write,

t- At= rili17 (32, 2

Thus,

or,

2At (At + 1t t v

c

=

1

.112

At- 2- A—f) t t

1.345 In the frame K the length 1 of the rod is related to the time of flight At by 1= v At In the reference frame fixed to the rod (frame K')the proper length 10 of the rod is given by

But

170

Thus,

v

v At' =

At

V-1-17

and 1.346 The distance travelled in the laboratory frame of reference is vA t where v is the velocity of the particle. But by time dilation Ato At =

So

v

=

c

/ /1

( At 0 / 6■0 2

V 1

v 2 /c 2

Thus the distance traversed is cAtV1 - (Ato/At)2 1.347 (a) If to is the proper life time of the muon the life time in the moving frame is VT TO

V — y2/c2

o

and hence 1.. 111-v2/c2 1

Thus

V 1 — V2/C2

toV

(The words "from the muon's stand point" are not part of any standard terminology) 1.348 In the frame K in which the particles are at rest, their positions are A and B whose coordinates may be taken as, A : (0,0,0),B = (10,0,0) In the frame K' with respect to which K is moving with a velocity v the coordinates of A and B at time t' in the moving frame are A = (ve, 0,0)B =

(10111.-7

+ve,o,o), p =

Suppose B hits a stationary target in K after time t'B while A hits it after time tB + At. Then,

/A

B

v At

So,

1°

— y2/c2

1.349 In the reference frame fixed to the ruler the rod is moving with a velocity v and suffers Lorentz contraction. If 10 is the proper length of the rod, its measured length will be

171

In the reference frame fixed to the rod the ruler suffers Lorentz contraction and we must have A x 2 NirT3' - /0 thus l0 =

and

1 — ir

Ax,

---

or V= c

AX2

rAirA—x-2

1— 6`x2

1350 The coordinates of the ends of the rods in the frame fixed to the left rod are shown. The points B and D coincides when 10= c1 — vt0 or to =

Cl — 10

The points A and E coincide when ci + /0

0= cl + 10 V1 7 — vti , tt

V

From this

1351 In K0, the rest frame of the particles, the events corresponding to the decay of the particles are, A : (0,0,0,0) and

(0,10, 0,0) = B

In the reference frame K, the corresponding coordintes are by Lorentz transformation

Now by Lorentz Fitzgerald contraction formula. Thus the time lag of the decay time of B is v/0 At

vI

vl

B decays later (B is the forward particle in the direction of motion) 1.352

(a) In the reference frame K with respect to which the rod is moving with velocity v, the coordinates of A and B are A: t, xA + v (t — tA ), 0, 0 B : t, xB+ v (t — ta),O, 0

172

Thus 1= xA - x B - v (tA- tB) = XA

XB - V OA

tB)

B

1353 At the instant the picture is taken the coordintes of A, B, are

A

, B' in the rest frame of A B

A: (0, 0, 0, 0) B: (0, 10, 0, 0) B': (0, 0, 0, 0) A' : (01 - /0117

A

, 0, 0)

In this frame the coordinates of B' at other times are B' : (t, vt, 0, 0). So B' is opposite to B at time t (B) —

10

. In the frame in which B' , A' is at rest the time corresponding this is by

Lorentz tranformation.

1

t° (B' ) =

to vio v ( c

2

V

v

Similarly in the rest frame of A, B, to coordinates of A at other times are A' :(t, - /0 V17722 + vt, 0, 0 c /0 A' is opposite t.,A at time t (A) .= The corresponding time in the frame in which A' , B' are at rest is /0 — t yt(A).. v

1

( t

1.354 By Lorentz transformation t' = 1- -7

vx

c

173

So

,

at time

1

vx

t= 0, t c2

Ifx > 0 t' < 0, if

x < 0,

> 0 and we get the diagram given below "in terms of the K-clock".

K: K: The situation in terms of the

1355

Suppose

K'

clock is reversed.

x (t) is the locus of points in the frame

K

at which the readings of the clocks of

both reference system are permanently identical, then by Lorentz transformation 1

(t

Vx (t))

t' =

= t c2

1/1 - V2/c2

C2( So differentiating x (t) =

1-

A nr72:. 1-

x (t)

-111 - tan h2

0)

=c

cosh0-1

1

cosh0

Vc o s h 0 - 1

c

e = ctanh— 2

c s in h 0

h

cos h0

sin

tan h0

1356

V

(1 - VF7)

= tan h0 , 0 s 0 < oo , Then

Let

-(tan

=

cosh0+1

s

v

0 is a monotonically increasing function of 0)

We can take the coordinates of the two events to be A: (0, 0, 0, 0)

For

B

B: (At,

a, 0, 0)

191 to be the effect and A to be cause we must have At >

In the moving frame the coordinates of A and A : (0, 0, 0, 0),

B :[

, y (a -

y (At -

B

become 1

V At) , 0, 01 where y

J.

Since 2 (At,

ai:

2

ai27)

112 1 At

-

(a - V At)2 C2

a' we must have At' >

=

2 (&)

>0

(Vc

2

2

174

1357

(a) The four-dimensional interval between A and B (assuming Ay = Az = 0) is :

52 - 32 = 16 units Therefore the time interval between these two events in the reference frame in which the events occurred at the same place is c (t'B - t'A ) or

Ct

= 4m

-8

t'B - t'A = 4=

i

7 6

x 10 S 3 (b) The four dimensional interval between A and C is (assuming Ay Az = 0) 32 _ 52 -16

•

B

5 4

c

4

3 2

So the distance between the two events in the frame

0

in which they are simultaneous is 4 units = 4m.

1

23

4

6' 6

1358 By the velocity addition formula vx

-V

v Nil -V2/c2 = y v V 1- x C2

V V vx ' Y

vx' =

1 - ----5-

C-

and

1359

v' =

= x

il(vx - v )2 +

y

(1 - y2/c2 )

1— vx 2V C

(a) By definition the velocity of apporach is dxl

v

ch = — -

aPPma in the reference frame K .

dt

dx2

v1

dt

- (- v2) = vi + v2

(b) The relative velocity is obtained by the transformation law

V =

1,1 -(-V2)

v1+ v2

1/1 (- V2)

Vi V2

1

1+ C2

1360 The velocity of one of the rods in the reference frame fixed to the other

V=

2v

v+ v -7r

= 1+13

11`

1+-2-

The length of the moving rod in this frame is

V lo

4 1/2/c2

1

(14.r2)2

= -1+ p2

di+ di+1 2

v " v2 2

1.361 The approach velocity is defined by

VaPPrcach = dt

dt

in the laboratory frame. So Vappmach - v vi + v2

rod is

C

175

On the other hand, the relative velocity can be obtained by using the velocity addition formula and has the components 1/2 SO

[ - v1 , V2 V 1 -—

__2

2

Vi

2

2 21 C V• = VI+ V2 C 1.362 The components of the velocity of the unstable particle in the frame K are

(V, V

1-- -,0 c

so the velocity relative to K is Vv2 +vr2_V

r 2 T72 v 2 C

The life time in this frame dilates to V2

vr2

v,2 V2

4

AtO / V 1

c

C2 C2 and the distance traversed is Vv2 + v/2 '64

v-'

(vr2 v2 ) c2

V 1 - V2/ c2 1/1 - V12 / C2

1.363 In the frame K the components of the velocity of the particle are v cos 0 - V Vix

V V COS 0 2 C

1

v'y =

v sin 011 -1.772 VV 1 - -T cos 0 v'

Hence,

ta

2-

n 0' -

v'Y

-

v sin 0 1/(1 - V2 )/c2 v cos 0 - V

1.364 In K' the coordinates of A and B are A: (t', 0, - v' t', 0); B : (t',1,- v' t', 0) After performing Lorentz transformation to the frame K we get

A:t=

B:t=y(t

,V1

yVe\x= y(1+Vt') y=

t' y= z=0 „

By translating t

t'

z=0 V/ t - -T, we can write

the coordinates of B as B : t = y t'

176

V1 y= - v' (t, - --I) , z= 0 c Thus

Ax = /1F 11 , Ay = V

yi c Hence

tan 0' -

O

2

VI V

C

2 \ F-1717 1 - -7_ -

1.365

t

I + dt

-4. v

-ii. --.). v + w dt

c

In K the velocities at time t and t + dt are respectively v and v + wdt along x - axis which --0 -0 is parallel to the vector V. In the frame K' moving with velocity V with respect to K, the velocities are respectively, v + w dt - V v V and V vV 1- , 1 - (v + wdt) --f

c"-

c

The latter velocity is written as

i v- V w dt + V v 1-v— 1-vV C2

+

C2

v- V v- V wV „. ,+ —1- "" = 4 V (1 -Y1 c I - V-C

2

C

2

wdt (A -

V2 2) c 2

(1 _ V;)

Also by Lorentz transformation

de =

1 - vV/c2 dt - V / c2 dx = dt Vi_v2/c2 111 - V2/c2

Thus the acceleration in the K' frame is

, dv' w = --= de

w 3

1-

V

712)3/2

(b) In the K frame the velocities of the particle at the time t and t + di are repectively (0, v, 0) and (0, v + wdt, 0) where V is along x-axis. In the K frame the velocities are (- V, v1/1 - V2/c2 , 0) and

(- V, (v + wdt)11171

, 0) respectively

177

Thus the acceleration =

wdt V(1 — V2/c2)

V2 • w 1 — — along the y—axis. 2

dt We have used dt' =

— V2/c2

1.366 In the instantaneous rest frame v = V and (from 1.365a)

3/2

(1 dv

So,

w' di 3/2

(

V2 1— C2

WI is

constant by assumption. Thus integration gives

w' t V

1° (

t ) 2

2 C

( + (WI 12 1 —, w 1.367 The boost time to in the reference frame fixed to the rocket is related to the time t elapsed

Integrating once again x

on the earth by .1/2

0 =

1 — -c72 0

1

dt f

dt

2

1+ (-"Lt-)

0

For

46 1,

1

1

mo V2 1-11T Arrii 1.369 We define the density p in the frame K in such a way that p dx dy dz is the rest mass dm0 of the element. That is p dr dy dz = Po dxo dyo dzo , where po is the proper density thco , dyo , dzo are the dimensions of the element in the rest frame K. Now

dy= dyo , dz= dzo, dx= ciao

/

1

2„ _ C2

178

if the frame K is moving with velocity, v relative to the frame Ko. Thus Po P= c

or

1- -4- i) . c irl (T#- '1l) a + To' 1 +1

v= C

1.370 We have Pri v

V

m0

or,

"` P

2.--

1

M

un

2

O+ c

-1- -712 2 c

Or

c

v=

Or

C

P

S

2

1ip2 + mo, c2

V 1 4. (mo 7 P

So

c — v .• c

2

1 (mo c

1— 1+ (-m 95)

lx 100 % • 7_. - -) x 100 % 2 p

P

1.371 By definition of TI, m0

v

v

= I) mo v

141-7 1— —

2

or 1— i = c

1 2

1

C2

1 — 2= .L.V1771 n 11 1372 The work done is equal to change in kinetic energy which is different in the two cases Classically i.e. in nonrelativistic mechanics, the change in kinetic energy is ?nco c2 ((0.8)2 - (0.6)2) s,— 0.28 = 0.14 mo c2 c 2 2 m° 2 Relativistically it is, . 2 2 2 mo c mo c m0 c mo c2 2 = m c (1.666 — 1.250) = 0-6 — 0.8 ° 1/1 — (0.8)2 V11 — (0.6)2 = 0416 mo c2 = 042 mo c2 or

v.c

179

1373

mo C

2 ' 2 m0

2

C

2

Or

g-

:

c v —..

or

V

1 —

or 2

1 =

2

V-3i.e.

v .•

4

c V3c 2

2 1374

2

Relativistically

T

1 2 m1 mo C (V177

PL lig

So

Thus

—Pro 'I

[

?

2 I(

3

2T

Mo

3 T2 I

-

.

-

\

2

Mo C4

4 2 - Mo C

T

m0 C 2 f3rei — Pet

2 SO

3 4

i•

a

mo c

Pct

T Hence if

2T )

1 3 Fr (1 — 4 mo c)

1/2

{27-.-But Classically, f3d=

3 (

/fto Cc2

no C

2T _ 2 C

3 4 +80 2

02 \ 2 2 -di (Prd) .I.

2T

1 2

—1 =

2


K or O RTV > V bR If a, b do not vary much with temperature, then the effect at high temperature is clearly determined by b and its effect is repulsive so compressibility is less.

'-7 +

' Ko {...

4


1 if P< Vind21 kT

Now

for 02 of 0 is 0.7 Pa.

VI d2 /

(b) The corresponding n is obtained by dividing by kT and is 1.84 x 1020 per m3 = 1.8414per c.c. and the corresponding mean distance is vs. 10-2 (0.184)1/3 x 105 1

1

2.228 (a) v

-

1-8 x 10- m = 0.18ttm.

k/

d2n

•74 x 1010 s-1(see 2.223)

(b) Total number of collisions is v a 1.0 x 1029 s cm-3 2n

Note, the factor

1

• When two molecules collide we must not count it twice.

1

2.229 (a) X -

V2- d2 n d is a constant and n is a constant for an isochoric process so k is constant for an isochoric process.

v= 1 (b)

=

, A,

kT

-

v2 d2 P

a T for an isobaric process.

v

for an isobaric process. T VT 2.230 (a) In an isochoric process X, is constant and a

v a 1ff a pV a Vp- a kT must decrease n times in an isothermal process and v must increase = _ rv2 nd2p n times because is constant in an isothermal process. (b)

259

2.231 (a)

1 1 V ka;"N/V"il

Thus

T1/2 V

ka V and v a

But in an adiabatic process (y - i here) TVY-1.. constant so 7V2/5411 constant or

T1/2 a V- 1/5 Thus v a V- 6/5

(b) X, a 1: P Y

_ vi

But p(-;7) ... constant or Taapp P

or Tap1-1/Y

A, a p-1ly ..1 p-5/7

Thus

1 +— 2 y

v-

X

a1'. ap VT

1/2

ai

L+.1

p 2Y = p 6/7

(c) A. a V But

7'V" - constant or V a T-572

Thus

A. a T-5/2 T1/2 3 v a - ,i a T

2.232 In the polytropic process of index n pVn .. constant, TV"' .. constant and pl-n rig constant (a) )s. a V T1/2 va l- V (b) A. a —T,

1-a

-n+1

2 V-1-

2

'V

, 1 T"ap"1 or Tap ' n

kap-1/11

SO

i

p 1-- +—1 2 v= 2n la p a vi,r ap x

n+1

2/1

n

(c) kaT,paTn-1 4 1

n

1 A

kar

1

" • ...........

n

1

T

K"

m T1-n

260 2.233 (a) The number of collisions between the molecules in a unit volume is 1 - nv 2

1 2 —7--7td2 n a

VT' 2 V

112

This remains constant in the poly process pV-3 I. constant Using (2.122) the molar specific heat for the polytropic process pVa = constant, is

1

C= R

)

Y-1 a - 1 Thus

1

C = R —y

11 1 + 74)

5+1

11

1(

as

R

1

It can also be written as (b) In this case

V

— R (1 + 2 i) where

_ constant and so pV = constant

C = R(-y1"f+

so

5

4

R (13+

= R

It can also be written as — (i + 1) 2 2.234 We can assume that all molecules, incident on the hole, leak out. Then, - dN

- d (nV).

S dt

4n

dt -n 4v/S =

dn- - n

Or

Integrating

. Hence =

n. no e

dt — y8RT M

2.235 If the temperature of the compartment 2 is times more than that of compartment 1, it 1 must contain — times less number of molecules since pressure must be the same when 1 the big hole is open. If M = mass of the gas in 1 than the mass of the gas in 2 must be So immediately after the big hole is closed. n

o

M mV

n

o

M

2

MI711

where m — mass of each molecule and 7,°, , n2 are concentrations in 1 and 2. After the big hole is closed the pressures will differ and concentration will become n1 and n2 where n +n 1

(1 2

m 171

+11)

On the other hand n1 = n2 i.e. ni = WI n2

261

Thus

n2(1 + rri)

÷n:li (1+

f(1+i1)

n2. n)

So

1 +Nrri 2.236 We know

T Thus

1

3 X, p

d2 m a if

changing a times implies T changing a2times. On

the other hand 1

1 Virf i

3

3

—k=

Thus D changing 13 times means

73/2

p

kT

It"' iiitd2p changing 13 times

a3 So p must change — times Pn 2.237 Da

NriccVNIT , nocif

(a) D will increase n times n will remain constant if T is constant (b) D a

T3/2

a

(P ,13i2

p

=

p1/2 T7312

p

a VT/. Thus D will increase n 312 times, ri will increase n1/2 times, if p is constant 2.238 D a VIrf , rl a if In an adiabatic process TVY-1= constant, or T cc 1 Now V is decreased — times. Thus

282 So 1 remains constant in the isothermal process

pV = constant, n = 1, here (c) Heat conductivity K 311

1 Cv

and Cv is a constant for the ideal gas Thus

2.240

71

n = 1 here also,

1 VcrcT m s

m

2

1/2 c/2

3

.,{7ncT 7t3

1 1/4 ( 2 ) /2 (m k or d = 3r1 3.( 37)

(

2.241

K=

d2 1/4

1/2 6) 2 x 10 3 x 18.9

2 3 x 18.9)

(4 x 8.31 x 273 x 10') 3 7E X 36 x 10"

4 4 x 834 x 273)1/ .,

1/2 = 10-1° (

1

(

0.178

m

x .36

1 — X, p cv 3 1

8 kT

3

mn Vind2n M Cv) - . Now Cv is the same for all monoatomicCv is the specific

( heat capacity which is ---"

1

mn

Cv

M

gases such as He and A. Thus Ka

K He=

or

1

1,41 a' dA2

8.7 =

KA

Viii"

547/dH

dA

CIH2 .

rcr

1.658 Pi F7 c/H.

VT.6

2.242 In this case N1

or

N,

2 R AR RA

= 4nriw

2 7t 1 CO R3 I"

4 7c w or

AR

To decrease N1, n times 1 must be decreased n times. NOW 1 does not depend on pressure until the pressure is so low that the mean free path equals, say, 1 AR. Then the mean free 2 path is fixed and ri decreases with pressure. The mean free path cquals I AR when 2 1 = AR (no = concentration) d2 no

263

Ili k T 2 a d AR The sought pressure is n times less Nr2- k T 10- 23 ..• 0.71 Pa P2 - 70.7 x nd nAR 10 - 1-8 X 10 - 3 The answer is qualitative and depends on the choice 1 AR for the mean free path. 2 2.243 We neglect the moment of inertia of the gas in a shell. Then the moment of friction forces on a unit length of the cylinder must be a constant as a fUnction of r. Corresponding pressure is po a.

N1 (1 _ 1) 3 2 n r Ti —dw a. N1 or (o (r) = dr 4 71 Ti r2 r2 ' 1 ' N1 1 _1 N1 (1 1 and oi . or ri . 4 it ri 1.21 r22 2- r 2) — 4 n 0) r1 2 2.244 We consider two adjoining layers. The angular velocity gradient . is cos Ti. So the moment of So,

the frictional force is a

71 a4 co N= f r•27trdr-nr—°). —1-h 2h 0 2.245 In the ultrararefied gas we must determine rl by taking X, .2 1 11 22 -3-

1 - h. Then 2

8kT 1 , . _ liiE . 1 V72M --h —WE XinX kT 3 7t RT "P 1 1/7171 N = po a4 p 2T7-,

SO,

2.246 Take an infinitesimal section of length dr and apply Poiseuilles equation to this. Then dV dt

From the formula

- 7t a4 pi 8T1 az

pV = RT • iitii RT _, pdV al — UM M dm It a4 M pit - Ti dx . . L 8 RT dt

Or

42. This equation impliei that if the flow is isothermal then p dr must be a constant and so 1 p22 _ p21 1 equals I

Thus,

21

in magnitude. 2 7C a4 M IP22 14 -Pi 16-1-0 T 1

264 2.247 Let T= temperature of the interface. Then heat flowing from left = heat flowing into right in equilibrium.

Thus, K1

T- T2 T1- T = K2 1 1 122

(xi Ti K2 T2+ 1 12 1 or T=

J.

K2)

, + 12

2.248 We have Ti-T K1

11

T- T2 =K2

IC

•

12

Ti. - T2 K , , 41 + 42

or using the previous result 2 K1 T1 K T2 , + K1 T1 - T2 /2 41 =K T1li 11 + 12 K 1

1

K1

1 1+ 1, Or

1

K2 —, (T1 - T2) Ki 12 T1 - T2 K2 =K + li K1 4. __ 11 12 11 12

11 + 12 or K do 1 1 ti gl -. + — Ki -K2

2.249 By definition the heat flux (per unit area) is dT d Q= -K—dx= -a —dr ln T= constant = + a x Integrating

ln .T=

In T1/T2 /

T2

7 ln —

1nT

T+1

1

where T1= temperature at the end x. 0 za So

T= Ti(-77-2)

/ a In T1 T2

and Q=

2.250 Suppose the chunks have temperatures T1, T2 at time t and T1- dri , T2 + dT2 at dt+t. Then

Thus

Hence

Ci c/Ti = C2 dT2 -

dAT= -

K

K S 7 (T1 - .7.2) dt

SII . 1 1 [&- + -a- AT dt where AT = Ti - T2 1 2

AT = (AT)0e-th where 1 =

t

KS

1

1

/ Ci + C2)

time

266

2.251

• Q =

K ar al

-A

ax

if ar

ax

aT 3

2 =- A 3

/2 (

ax '

A

=

constant )

3/2 3/2 ( T1 - T2 ) = 3A 1 2

T34 = constant - =xi( Ti3/2- T23/2)

Thus

1 at x= 0

T= T

or using

3/2 3 T3/2=

Tr + '"I ( T3/ 2 /2 - T3/2 ) 1

Or

( T

/2

1

=

)

+ X ( ( T2

Ti

1

T

)

-1

-L

1 \FITT 2.252 K En

-3-

am

vi

1 mn 2 . R3/2 i T3/2 7td2n Al 3n 3/2 d 2 1.617 NA

Then from the previous problem

(T1 + T2)

27°C = 300K. T = 2330 x 10 5III = 2.3.3mm » 5.0mm = 1

2.253 At this pressure and average temparature =

A=

1 KTV2 nd 2

P

The gas is ultrathin and we write X.=

Then

q=

1

where

K= iX

1 - here 2/ T2 - Ti.

dT dx MP R

K — mi K

1 1

p I

21x RTxy-lxM= 6T(y-1)

p q - 6T (y -1)(T2- TO

and

where

1

2

-

—

T2 +T1

here.

3t e have used T2 - T1 « M W Mn .

266 2.254 In equilibrium 2rr K dT = —A = constant. So T = B — dr

A In r LII K

But T= T1 when r= R1 and T= T2. when r=R2. T2 —

r In —

From this we find T = T1+ In Ri 2.255 In equilibrium 47t r

2

dT K— dr

— A = constant A l1 B+ — ztioc r

Using T T1 when r R1 and T T2 when r R2,

T2 — T T+

(1

1-1 R2 Ri

1

r

R1

2.256 The heat flux vector is — K grad T and its divergence equals w. Thus v2

T. K

or

_a _ r ar

aTr ar

in cylindrical coordinates.

K

0) T — B+Alnr-- r

or

2x

Since T is finite at r = 0, A =

0. Also T = To at r = R

SO Thus

B= T

°

+—R 4K

2

T= To+ --AW- (R2—r2) 44K

r here is the distance from the axis of wire (axial radius). 2.257 Here again v 2 7. .

w K

So in spherical polar coordinates, 1 a ( 2 aT ar r ar

or

— or r K

T= B —

A

w 3 2 aT r +A =— 3K a? ' w

r OK so finally Again

r

2

and B = To+ — w A =

2

6K R

T. T +w (R2_r2)

0

°

6K

PART THREE

ELECTRODYNAMICS

3.1

CONSTANT ELECTRIC FIELD IN VACUUM

2 3.1

2

Fd (for electorns) = :---q--2-and F r=

4 ncor Fei Thus

Fir

-"I r e

(for electrons) =

4neoym2

(1.602 x 10 -19 C)2

(

1

n) X 6.67

=4

X 1042

x 10 -11 m3 / (kg • s2) x (911 x 10 -31 kg)2

9 x 10' q2

Fd

Similarly

-1 Fr (for proton) = - 4ncoym =

(1.602 x 10 -19 C)2 (

1 9) X 6.67

x 10

-

11

= 1 x 1036

m3 /(kg • s2) x

(1.672 x 10 -27 kg)2

9 x 10 For Fd = Fr 42

v m2

'-----

4neor2=

r2

n

or

Zi = r4 7JT07

V 6 - 6 7 x 10 -11 m3 (kg — s2) =

9 x 109

0.86 x 10 -1° C/kg

3.2 Total number of atoms in the sphere of mass 1 gm =

1 63.54

6.023 x 1023

x 6.023 x 1023

-1 x 1.6 x 10 9 x 29 63.54 Now the charge on the sphere-,z. Total nuclear charge — Total electronic charge

So the total nuclear charge A. =

288

6.023 x 10 -

-19 23 x 1-6 x 10

63-54

x

29 x 1 ... 100

4.298 x 102 C

Hence force of interaction between these two spheres, 1

[4.398 x 10212

4 n eo

12

F. 3.3

N= 9 x 109 x 104 x 19.348 N=

1-14 x 1015N

Let the balls be deviated by an angle 0, from the vertical when separtion between them equals x. Applying Newton's second law of motion for any one of the sphere, we get, T cos 0 . mg and

(1)

T sin 0 - Fe

(2)

From the Eqs. (1) and (2) F tan 0=

e

(3)

mg

as x « /

(4)

> Fe

Fe.

Ty_

q

or

2

ing

mg z

47tE0X

2=

21

2 ,t eo mg x3 Thus

4

Differentiating Eqn.

2

-

(5)

I

(5) with respect to time da

2 it eo mg

4 s . 1 dx —dt . v

According to the problem

2 eh 3 xdt

= aAri (approach velocity is

1/2 (2 it E0 mg

3)

dq

3 x Eo mg x2 a

x

SO,

1

Hence

dt g.

3 -—a ' dt 2

1

Nrx-

2 Jt eo mg 1

3.4 Let us choose coordinate axes as shown in the figure and fix having

three charges, qi. , q2 and q3 --, --a. --,. position vectors ri, r2 and r3

respectively. Now, for the equilibrium of q3 -30

--0

+ q2 q3 ( r2 - r3 ) I r2 - r3 I 3

•

qi q3 ( ri. - r3 ) + --0 --a. 3 - 0 I ri - r31

dr ) dt '

269 q2

or,

I 2° -

r31 2 I. —10

--lb

—is

—II.

r2 - r3 ri - r3 1 r7- r3'I = - I ?.-- 731

because or,

qi. I 737 - 73'] 2

Nq; (IT- 3)=Nriii ( 3- 2) — —0

=

Vq2 r--■+ V Ii tii. .r2

—P.

or,

r3 = Nrii7+ V-4;

Also for the equilibrium of q1, • •-ilP

q3(

..41,

r

.10.

3 - ri ) + q2( r2 - ri ) = 117:- t7n3 1772- 77-113

or,

q3 -

-q2 ,

1 2-

r1

--l• 21

I

-1.2

r1 - r3 I

Substituting the value of r3 we get, - ql q2 q3 ( Nrq-1- + Nri; )2

3.5 When the charge q0 is placed at the centre of the ring, the wire get stretched and the extra tension, produced in the wire, will balance the electric force due to the charge q0. Let the tension produced in the wire, after placing the charge q0, be T. From Newton's second law in projection form F, = mw,.

dFe

Td 0 --j. - 1° (--2---r d 0) = (dm) 0, 4 n E0 r2 2 nr or,

T=

q qo 82t2E0r2

3.6 Sought field strength -. E-

1

q

4 n E0 I r-. 712

= 4.5 kV/m on putting the values. 3.7 Let us fix the coordinate system by taking the point of intersection of the diagonals as the origin and let k be directed normally, emerging from the plane of figure. Hence the sought field strength :

270

q + xk E— 4 n Eo (12 + x2)312 +

4

—q

l(—i)+xk

1E Eo

(12 + x2)3/2

--9,

—q 1j +xk (12 4nEo + x2)3/2

q l(-1)+xic* 4 rc Eli) (12 + x2)3/2

ql

Thus E —

n go (12 + x2 )3/2 3.8 From the symmetry of the problem the sought field. ±

+

where the projection of field strength along x — axis due to an elemental charge is

11f +

dq cos 0 _ q R cos 0 d 0 Ex

x

4 n £0 R 2

4-

di

E = f d Ex

9

0

-->

X"

>4.

t +

4 7t2 E0 R3

do ?

E

It/2

Hence E.'

q 4 Tt

cos 0 d 0

2

2 £0 R

q

++

2n2£0R2

N.,/2

3.9 From the symmetry of the condition, it is clear that, the field along the normal will be zero i.e.

E.= 0 and E = Ei dq

Now

dEi —

But

q

a

d =

2

cos 0

47t Eo (R 2 1- 1 ' )

1

dx and cos 0=

2nR

2 (R

Hence 2zR

dx

E= f dEi= f 0

4 n E0 (R

2 + / 2 )3/2

2 TER

1 ql or E — 47c e 0 (/ 2 +R 2 )3/2 and for 1» R, the ring behaves like a point charge, reducing the field to the value, E

1 q 4neo 12

2 +l

1/2 )

271

For

Emax,

So,(12 + R2

dE we should have

)3/2 _

7ll- =

0

1 (12 4. R 2 )1/2 2/ .

0

12+R2-312=

or

0

2

R

3.10

q

E

Thus /-= --,—_- and V2

max

6 13 - ne oR 2

The electric potential at a distance x from the given ring is given by,

q

q

cp (x) =

4 n Eo x

4 7t Co (R2 + X2)1/2

Hence, the field strength along x-axis (which is the net field strength in 'our case), dcp = - —= dx

q

4 it co

1

q

4 n go x2

qx 4 n Eo

(R

R2)3/2 3 { (1 + -T x X

2 + X2 )3/2

—1

x2 (R2 + x2 )3/2

q X3 4 7C Eo

1

x Neglecting the higher power of

2

(R

3 R4

—— 2x2

—— + 8 x

2

•••

I

3/2 + x2)

R/x, as x » R. 2 3qR E— 8

Note : Instead of cp (x), we may write 3.11

3 R2

n eo x4.

E (x)

directly using 3.9

From the solution of 3.9, the electric field strength due to ring at a point on its axis (say x-axis) at distance

x

from the centre of the ring is given by :

E

And from symmetry

at every point on the axis is directed along the x -axis (Fig.). Let

us consider an element (dx) on thread which carries the charge

(X. dx). The electric force

experienced by the element in the field of ring. kqx dx dF =

cbc) E (x) 4 It Eo (R2 + X2)3/2

Thus the sought interaction 00

kqxdx F = 6 / 4 n eo

(R

2

2.3/22)3/

+ x2

0 On integrating we get,

F—

4 TE eo

R

272

3.12

(a) The given charge distribution is shown in Fig. The symmetry of this distribution implies that vector Eat the point 0 is directed to the right, and its magnitude is equal to the sum of the projection onto the direction of E of vectors difrom elementary charges dq. The projection of vector dZonto vector E is dE cos cp — where dq= kRdq)=

1 1(1 cos fp, 4 n E0 R 2

R cos cp d cp.

Integrating (1) over qi between 0 and 2 7c we find the magnitude of the vector E: 2,

E—

, f 4 7c Eo R 0

2

cos

,/ yu

=

4 E0

It should be noted that this integral is evaluated in the most simple way if we take into account that = 1/2. Then 2x

cos2q dq

0. Let us start with Ex. The contribution to Ex from the charge element of the segment dx is dEx -

X dr in s a 4 eo r2 1

(1)

Let us reduce this expression to the form convenient for integration. In our case, dx = r da/cos a, r = y/cosa. Then k dEx =.r 4 n soy sin a d a. Integrating this expression over a between 0 and n/2, we find Ex= X/4 7r eo y.

In order to find the projection ;

it is

sufficient to recall that dE differs from dEx in that sin a in by cos a.

(1) is simply replaced

This gives dE = (X. cos a d a )/4

y and E = X./4 it e0 y.

We have obtained an interesting result Ex = ; independently of y,

: --3.

i.e.

E is oriented at the angle of 45° to the rod. The modulus of E is E. VEx2 +Ey2

3.15

kli/4neoy.

(a) Using the solution of 3.14, the net electric field strength at the point 0 due to straight parts of the thread equals zero. For the curved part (arc) let us derive a general expression i.e. let us calculate the field strength at the centre of arc of radius R and linear charge density k and which subtends angle 00 at the centre. From the symmetry the sought field strength will be directed along the bisector of the angle 00 and is given by

dl

+ 00/2 E_

f

(R dO) 4 IT e0 R 2

cos 0 -

00 sin — 2 it E0 R 2

- 0V2

In our problem 00 = 7t../2, thus the field strength due to the turned part at the point 12which is also the sought result. - 4 3r e R (b) Using the solution of 3.14 4 IT EoR

(a), net field strength at

0 due to stright parts equals

and is directed vertically down. Now using the solution of 3.15 2 7t E0 R

275 (a), field strength due to the given curved part

(semi-circle) at the point 0 becomes

and is directed vertically upward. Hence the sought net field strengh becomes 2 7t eo R zero.

3.16 Given charge distribution on the surface a =

r iiss shown in the figure. Symmetry of this distribution implies that the sought E at the centre 0 of the sphere is opposite to

dq= cr(2nrsin0)rd0= cs•t12nr2sined0= 2nar3sin0cos0d0 Again from symmetry, field strength due to any ring element dE is also opposite to a i.e. dE f Hence

a-,

de4 n Eo

0. 2

dq r cos 0 -a 0 r2 cos2 0)3/2 a---arsing the result of 3.9)k`) sin A+

(2 n a r3 sin 0 cos 0 d 0) r cos 0 (4 n eo r3 = 2

r 0 cos2 d a sin

a

0eo

Thus de= (- illrfsinOcos20d0 2 E0 0 Integrating, we get z =

-

Wr2

ar — =2E03 3Eo

3.17 We start from two charged spherical balls cad} of radius R with equal and opposite charge + densities + p and - p. The centre of the balls are at and- -a respectively so th 2 e2 –0 a a a equation of their surfaces are r - R or r - -cosON R and r + -cosOeg R, considering 2 2 2 a to be small. The distance between the two surfaces in the radial direction at angle 0 is acosO f and does not depend on the azimuthal angle. It is seen from the diagram that the surface of the sphere has in effect a surface density a = ao cos() when cr0 = pa. Inside any uniformly charged spherical ball, the field is radial and has the magnitude given by Gauss's theorm 47cr

4n = ar p/e0 Prr

Or

3 e.

In vector notation, using the fact the V must be measured from the centre of the ball, we get, for the present case

13 77*()--L1-(77.+ i) 3E0

2

3E0

2)

2 76 rw

= -pa/3E0 -

When kis the unit vector along the polar axis from which 0 is measured. 3.18 Let us consider an elemental spherical shell of thickness dr. Thus surface charge density of the shell a = p dr.= (54:r) dr. Thus using the solution of 3.16, field strength due to this sperical shell ar dE = - — 3 eo

dr

Hence the sought field strength R

a it E= - a fr dr = 3 Eo

6 eo

3.19 From the solution of 3.14 field strength at a perpendicular distance r < R from its left end k A (er) t(r) (-i )+ 4ne r 4nE r0 Here er is a unit vector along radial direction. Let us consider an elemental surface, dS = dy dz = dz (r d 0) a

figure. Thus

flux of E (r) over the element dS is given by E • dS = [

ci

=

A

(-i )+ ( d•dr(rd0)i 43tEr o 47te or er

4 n codr d0 (as e-7. I I- 4 °) R

The sought flux, (13 = -

2x

k f dr f d 0= 4 7r eo 0

If we have taken ciS t t ( -

A

R 2 Eo

0

then 41) were

X,R — 2 E0

Hence Eo 3.20 Let us consider an elemental surface area as shown in the figure. Then flux of the vector E through the elemental area, d (I) = 2q

di*= E dS = 2 Eo cos cp c/.5 (as Etf dS) I

4 E0 (12 + r2 ) (/ 2 + r2 )- 2

2q1 r dr de ( d 0) dr = 4 n eo (r2 + / 2 )312

277

where E0 -

magnitude

q2 2 2 +r ) 4 Jt go (/

of

field strength due to any point charge at the point of location of considered elemental area. 2:t

rdr

Thus (13 -11 1 - f 4 it go

f d0

0.2 + 12 )3/2 0

0

R

r dr

2 q/x 2n f

(r2 +

4 n go

i

/ [1 -

12)3/2 - EO

V/2- -;—.R 2 +

o

It can also be solved by considering a ring element 3.21

or by using solid angle.

Let us consider a ring element of radius x and thickness dx, as shown in the figure. Now, fl ux over the considered element,

dam= E • dS = E r dS cos 0 But EE

E

pr from Gauss's theorem,

r= —

3E 0

dS=

and

Thus

ro 2

x

, cos 0=

2

dcl) =

x dr

r -

_r° 2 nx dr to

3 E0 Hence sought flux R217r1 2

ap

ro

c1) -

2np

P ro

fx dr 3 go

3.22

ro (R2 - ro2)

3 Eo

0

The field at

P

2

due to the threads at A and

2-

ro)

3 Eo

B

are both of magnitude

2 2 a go (x

and directed along

AP

and

BP.

2

ned

2+

OP

12/4)

with

x

2 A, cos 0

E-

The resultant is along

+

1/2.1/2 =

n

Bo (.X2 + 12/4)

2 En X "

—- 2 4x

2 Vx

x 7, 71 .4_

2

Eo

7c -

2 v` r-

+1

A This is maximum when x = 1/2 and then

E=E

—max =

E0

1 I

2/1

1/2

278

3.23 Take a section of the cylinder perpendicular to its axis through the point where the electric field is to be calculated. (All points on the axis are equivalent.) Consider an element S with azimuthal angle q. The length of the element is R‘dq) , R being the radius of cross section of the cylinder. The element itself is a section of an infinite strip. The electric field at 0 due to this strip is R d5, Go cos cp (R dcp) along SO 2 n EoR This can be resolved into cro cos p d cp cos cp along OX towards 0 2 7c Eo

sin p along YO

On integration the component along YO vanishes. What remains is

23,

f

ao cos' cp dy_ cro

2 co along XO i.e. along the direction cp 2 n eo 0 3.24 Since the field is axisymmetric (as the field

=

7c.

Oc \\%\c'M%\l AcSttttilAmiten1), we conclude that the flux through the sphere of radius R is equal to the flux through the lateral surface of a cylinder having the same radius and the height 2R, as arranged in the figure.

Now,

But

Thus 3.25

E • dr- E. rS

(I)

Er =R

(I) = 11-S = -11 2nR•2R= 47caR

(a) Let us consider a sphere of radius r < R then charge, inclosed by the considered sphere,

qindosed= f 4nr2drp= f 4

0

7(

r2 po (1 - ti)dr

0

Now, applying Gauss' theorem, 2

Er 4 7t r =

or,

qinclosed

go

, (where Er is the projection of electric field along the radial line.)

(1)

279

And for a point. outside the sphere r > R. R

r =f 4nr2dr pc, 1 — — (as there is no charge outside the ball) q R o lm'cosed Again from Gauss' theorem, R

4

It

r2 dr po 1

_R r

Er 4 n r2 = go 0 Po [R3 R41 poR3 Er= r2 go 3 — 4R — 12 r2 e o

Or,

(b) As•magnitude of electric field decreases with increasing r for r > R, field will be maximum for r R, we can write. r

E 4Ttr2 — q inclosed— Eo

q go

1 f a +— — 4 7t r2 dr E0 r R

On integrating we get, q — 2 n a R2) 4 n a r2 E 4 n r2 — ( + 2 e0 Ea The intensity E does not depend on r when the experession in the parentheses is equal to zero. Hence

q= 2 n a R2 and E—

a 2 eo

3.27 Let us consider a spherical layer of radius r and thickness dr, having its centre coinciding with the centre of the system. Then using Gauss' theorem, for this surface, r Er 4 n r2 =

qi nclaved Eo

—

f pdV

eo 0

280

After integration 3

e-ar

Er 4 n r2 = 3 Eo a

3

or,

Er -

-C" 3Ea

Now when a r

3

r «

1, Err

X

3 "0 PO

3 And when a r » 1, Er.

3 e0 a r

2

3.28 Using Gauss theorem we can easily show that the electric field strength within a uniformly. charged sphere is E = 3 Eo The cavity, in our problem, may be considered as the superposition of two balls, one with the charge density p and the other with - p. Let P be a point inside the cavity such that its position vector with respect to the centre of cavity be 77:" and with respect to the centre of the ball r Then from the principle of superposition, field inside the cavity, at an arbitrary point P,

I. I+ +I --P— ( - -P--3EOa 3 Eo Note : Obtained expression for E shows that it is valid regardless of the ratio between the radii of the sphere and the distance between their centres. 3.29 Let us consider a cylinderical Gaussian surface of radius r and height It inside an infinitely long charged cylinder with charge density p. Now from Gauss theorem : Er 2 it r h -

qinclosed Eo

(where Er is the field inside the cylinder at a distance r from its axis.) or, Er 2 71 rh = P r2 h or Er = 0

r 2p Eo

Now, using the method of 3.28 field at a point inside the cavity, is E_

+

=

- 7°) = 2E0

'

2E-0

P,

281 3.30 The arrangement of the rings are as shown in the figure. Now, potential at the point 1, = potential at 1 due to the ring 1 + potential at 1 due to the ring 2. -q + 4 3T Eo (R2 + C12)112 4 n Eo R Similarly, the potential at point 2, q CP2 4 n- etloR 4 1t Eo (R2 + a2)1/2 Hence, the sought potential difference, -q A cP2 = = 2(—A— + 4 n eo R 4 n Eo (R2 + C12)1/2

3.31 We know from Gauss theorem that the electric field due to an infmietly long straight wire, at a perpendicular distance r from it equals, Er = 2 n . So, the work done is 0r 2 x f Er dr =

dr

2 n Eo r 1 (where x is perpendicular distance from the thread by which point 1 is removed from it.) Hence A X12 - 2 n eo in TI 3.32 Let us consider a ring element as shown in the figure. Then the charge, carried by the element, dq = (2 nfl sin 0)R d 0 a, Hence, the potential due to the considered element at the centre of the hemisphere, 41_ 2 naRsin 0 d0 a R sin d 0 1 d" 4 It Co R 4 1t E0 2E0 So potential due to the whole hemisphere x./2

oR cp = R ° sin 0dd 00 == Eo 0 2 Eo Now from the symmetry of the problem, net electric field of the hemisphere is directed towards the negative y-axis. We have 0 = dE - 4 n1eod Rq cos 2 2 co sin 0 cos 0 dO x./2 x./2 Thus E=E ' . a f sin 0 cos 0 d 0 = Y 2E0 0

, along YO 4 eo f sin 2 0 d 0 -

4 E0

282

3.33 Let us consider an elementary ring of thickness dy and radius y as shown in the figure. Then potential at a point P, at distance 1 from the centre of the disc, is

d

=

a2nydy 2 + 12 )1/2 4 eo(y

Hence potential due to the whole disc, f

a 2 xydy 431 so (y2 + 12 )1/2

(P3E

al

+ ( R/1) 2

2e0

From symmetry E Ei= cr 2 eo

when /

0, ('14

21 2V/72-47/2

aR 2e

E=

a 2 eo

dl cr 2 E0

1 + (R//)2

and when / »R ,

o

3.34 By definition, the potential in the case of a surface charge distribution is defined by integral =

1

2c--d3. In order to simplify integration, we shall choose the area element dS 4 7r e0 r in the form of a part of the ring of radius r and width dr in (Fig.). Then dS = 20 r dr, r = 21? cos 0 and dr = — 21? sin 0 d O. After substituting these expressions into integral cP=

1

a dS,

4ne0

r

we obtain the expression for p at the point 0:

0 = —°

R

f 0 sin 0 d 0.

n Ea We integrate by parts, denoting 0 = u and sin 0 d 0 = dv : fOsin Ode

—Ocos 0

+fcos 0d0= —Ocos 0 +sin 0 which gives -1 after substituting the limits of integration. As a result, we obtain = aRht e0.

283 =a r p --0. - -II,

3.35 In accordance with the problem

Thus from the equation : i= - g(49

E_ 3.36

-aix (ar x) i + — (ay y) j + -7,- (a, z) k l a, a °.z

[ax i + ay j + z k ] =

- a+

(a) Given, (4) = a (x2 - y2) -. --. ---. --. So, E= -Vcp..-2a(xi-yj ) The sought shape of field lines is as shown in the figure (a) of answersheet assuming a > 0: (b) Since (43) = axy So,

E= -VT.. -ayi-axj

Plot as shown in the figure (b) of answersheet.

3.37 Given, So,

Hence

p

= a (x2 + y2) + bz2

:E= - 1.4 = - [2 axr+ 2 ayr+ 2 bz ic] I Z1 = 2 Va2 (x2 + y2) + b2 2

Shape of the equipotential surface : Put

--9.

7 .

.-1.

p= xt+yi or

p 2=

X2

2 +y

Then the equipotential surface has the equation a p2 + b z2 - constant = p If a > 0, b > 0 then q) > 0 and the equation of the equipotential surface is -P-- + -1-2 cp/ a p/b

sim

1

which is an ellipse in p , z coordinates. In three dimensions the surface is an ellipsoid of revolution with semi: axislcp 7 a , 7a , iq7/72. If a > 0, b < 0 then p can be 2 0. If IT > 0 then the equation is p2

z2

cp/ a cp/Ibl

1

This is a single cavity hyperboloid of revolution about z axis. If (4) = ape - 1bl z2 = 0 or

z = ± 11;67 p

is the equation of a right circular cone. If cp < 0 then the equation can be written as lbj z2 - a p2 = 1(491 z2

_

p2

1

Or Id /1b1

14)1 /a

This is a two cavity hyperboloid of revolution about z-axis.

0 then

284 3.38 From Gauss' theorem intensity at a point, inside the sphere at a distance r from the centre 1

is given by, Er= p and outside it, is given by Er = 3 EE o •

q .2

71 Eo o r

(a) Potential at the centre of the sphere, R

cc

00

R2 dr + f dr = + q 4 Tr e ro 3e 2 4 It eo 3 Eo

f

E

0

3q

q q + 8 n eo R 4 n eo R

as

P ( as

8 21 eoR 4 7t R (b) Now, potential at any point, inside the sphere, at a distance r from it s centre. R

CO

49 (r) f —P— r dr f —q-- dr 4 it Eo 3 Eo r r

2 = cp [1 — LH

2 On integration

[1 — 1

(r) = 8 neoR

3 R2

3R2

3.39 Let two charges +q and —q be separated by a distance 1. Then electric potential at a point at distance r » 1 from this dipole, —q

+q

cp (r) =

+

4 n eo

But

4 n eo r_

1 cos

r-—r+=

(r_ — r+

q

0 and

4 It Eo

r+ r_

r2

r + r-

From Eqs. (1) and (2), —Jo

cp (r)

—310

p • r

q 1 cos 0

p cos 0

4 n eo r2

4 n eo r2 q)

4 It e r0

±

where p is magnitude of electric moment vector. Now, Er.

and E0 —

So

E=

acp

2p cos 0 4 it eo r3

ar

p sin

ra0 Er 2

0

4 n eo 02

3 V4 cost 0 + sine 0

E

4neor 3.40 From the results, obtained in the previous problem, Er =

2p

p sin

cos 0

4 n eo

3 and E0

0

4 n eo

From the given figure, it is clear that, ti

E, = Er cos 0 — Eosin 0

(3 cos2 0 — 1) 4ne0r

285

3p sin 0 cos 0

El= Er sin 0 +Eo cos 0 -

and When

E l p, lEl. El

4iteor3

and Ez =

0 1

So

3 cost 0 =

1

and cos 0 =

V3

Thus EZp-I.at the points located on the lateral surface of the cone, having its axis, coinciding with the direction of z-axis and semi vertex angle 0 = cos -11 / 3.41 Let us assume that the dipole is at the centre of the one equipotential surface which is spherical (Fig.). On an equipotential surface the net electric field strength along the tangent of it becomes zero. Thus p

- E0 sin 0 +

= 0 or - E0 sin 0 +

4

sin 0

-0

n eo

1/3

Hence

r

P

4 it co Eo

Alternate : Potential at the point, near the dipole is given by, p • r -

(13 -

z

constant,

• r +

4n cor (

P

- E0 cos @ +

4 it co r-

Const

For q' to be constant,

p 4 co

3

E0 =

0 or,

3- E 0

4 it co r

r

1/3

Thus

r-

3.42 Let P be a point, at distace

Thus

E at 13

r

»1

2 it co

2 r

+ r 1 cos 0

4 X 2 Hence

(1

n E0 r2

2

r2 + —1 2 - r 4

I rcos 0)

r3

/ E

and at an angle to 0 the vector /

r-I12

l2 +

P

4 it E0E0

2 It co .

+l/2 2 it e0 {

(

IE I

2 it co r

2

r» 1

cos 01

▪

286

▪

Also,

(I)

Ink7+1721—

2ngo

g

4n o

In

/72]

2 go

r2 + r 1 cos 0 +

/4

/ cos 0

r2—r/cos0+ /`/4

,r»1

2 go r

3.43 The potential can be calculated by superposition. Choose the plane of the upper ring as x = 1/2 and that of the lower ring as x = —1/2. Then

(1

9 l" q

q

2 112

4ngo[R2+x -1x]

q 4 n go (R2 + X2)112

(1 +

2 (R

It go[R + x2 + /x)1 ,,,, 4 7t E0 2 +

/2

4

lx 2

, +f)

1—

(B

rZ

„ 2 (R"+

For The electric field is E = ql 4 n go

(R2 4. x2)3Z2 4- 2

3 (R2

q1 2

+ x )5/2 4 n go

For

x 2x =

ql (2x2 — R2) 4 t go (R2 + X2)5/2

» R, E

ql 3. The plot is as given in the book. 2 n go x 3.44 The field of a pair of oppositely charged sheets with holes can by superposition be reduced to that of a pair of uniform opposite charged sheets and discs with opposite charges. Now the charged sheets do not contribute any field outside them. Thus using the result of the previous problem

R

I (—o)12nrdrx

(I)

=

4 n ea (r2

0 =

_

R

2

X2)3/2

2 +X

a x/ f dy

y3/2

4 co x Ex. —

o xl

2

_ -E.. ax

of

2 El)

2 eo \fR"7272

2

x

a /R2

R2 + 4 3/2

2 €0 (R2 .1. x2)3/2•

1

VR2 2 - ( +x

The plot is as shown in the answersheet.

▪ 287

3.45 For x > 0 we can use the result as given above and write ±

/ xj x 1j 12 eo ( (R2 + 2) 2

for the solution that vanishes at a. There is a discontinuity in potential for I solution for negative x is obtained by a - a. Thus alx (I) 2)1/2 + constant 2 eo (R + x Hence ignoring the jump al R2 E €0 (R2 + x2)3/2 ax = 2 for large Irl

(pm

and E

3 (where p = nR2 a 1)

4neox2

22reolxl

-)

Ee.

3.46 Here E

r

0 and F

r= 2 eo r ' = p aai (a) Fialong the thread. E does not change as the point of observation is moved along the thread. F= (b) p along F W*.=

kP 2 j:ze 2 it eo r

(c) iralong

—

F=p

a

,

On using — e= o) ar

nreor-

a

X,

ra0 2

7(

e o

r er

--0 P a er pX 2 rt eo r2 a0 = 2 J't eo r2

am

2 n eo r2.

3.47 Force on a dipole of moment p is given by, F=

(I) 29

In our problem, field, due to a dipole at a distance 1, where a dipole is placed,

ri

up ........... ■E■ ■

2 II eo /3

Hence, the force of interaction, 2 F-

3 2 eo r

2.1x10-16N

3.48 -dq= E dr= a (y dx + x d y) = a d (xy) = On integrating, p - a xy + C 3.49 - d or,

= E dr.= [2 a xy r+ 2 (x2 - y2):1] • [dx +dy1 ] d (19 xi: 2 a x y

+ a (x2 - y2) dy = ad (x2 y) - ay2 dy

xi = 0.

The

288

On integrating, we get, 2 = ay(Y-- x2) +c 3 3.50 Given, again --0

– d cp.= E•(Ft = a (y

7110

(ayi+(ax+bz)j+byk)•(cfri+dyj+cbck)

+ ax dy) + b (z dy + y dz) . ad (x y) + bd (yz)

On integrating, = - (a xy + b yz) + C 3.51 Field intensity along x-axis.

E

- --L = 3 a x2 ar Then using Gauss's theorem in differential from aEx

p (x)

ax

go

so, p (x) = 6a Eo x.

3.52 In the space between the plates we have the Poisson equation 32 PO ti)

a x2

E0

or,

cp= - P° X2 +Ax+B 2 co where Po is the constant space charge density between the plates. We can choose

cp (0) = 0 so B = 0

Po d2 Then

(d) = A

Now

E

2 E0

= Ad -

cp or, A -

x - A = 0 for x

ax

d

Po d + 2 E0 0

0 cp

A

Po d

= d + — 2 co

if

2E0A•zp then

Po =

Also

Ef

d2

Po d (d)

go 3.53 Field intensity is along radial line and is

Er=

-2ar

ar

From the Gauss' theorem, 4 3t T2 Er 'II

f

Eo, where dq is the charge contained between the sphere of radii r

Hence Differentiating (2) p = - 6 co a

and r + dr.

289

3.2 CONDUCTORS AND DIELECTRICS IN AN ELECTRIC FIELD 3.54 When the ball is charged, for the equilibrium of ball, electric force on it must counter balance the excess spring force, exerted, on the ball due to the extension in the spring. Thus F d = Fs pr 2

The force on the charge

Or,

2 = K X, (

4 it co (21) q might be considered as arised from attraction by the electrical image) or,q = 41 Vit eo K X, sought charge on the sphere. 3.55 By definition, the work of this force done upon an elementry displacement dx (Fig.) is given by q2

dA = Fx dx =

dx,

4 it eo (202

where the expression for the force is obtained with the help of the image method. Integrating this equation over x between 1 and 00, we find CO

3.56

dx

f

q2

A-

16 it eo

q2

x2 -

16 it eo /

(a) Using the concept of electrical image, on each charge,

it is clear that the magnitude of the force acting

2

q2

111= Nri

4 it eo / 2

4 it eo (Vi

02

q2 Nri -

1)

8 It E0 (b) Also, from the figure, magnitude of electrical field strength at P

1 E= 2

1-

1,— ) 5 V5 TC co /-

3.57 Using the concept of electrical image, it is easily seen that the force on the charge q is,

F-

(_ 02 4 it eo (202 + 4 it eo (2 V-2- 02

(2 V - 1) q2 2

32 it eo /

„ (It

is attractive)

290 3.58 Using the concept of electrical image, force on the dipole g

--a

F

p

a E , where E is field at the location of

.rdue to (-I") or, IF 1=

al

IP 327tE0/4

al 4=

as,

p 3 4 It to (21)

3

3.59 To find the surface charge density, we must know the electric field at the point P (Fig.) which is at a distance r from the point 0 . Using the image mirror method, the field at P, q

E= 2 E cos a = 2

1

4 it eo x

2

X

q1 2 It go (1 2 + r2 )3/2

Now from Gauss' theorem the surface charge density on conductor is connected with the electric field near its surface (in vaccum) through the relation a - e0 En, where En is the projection of E onto the outward normal titwith respect to the conductor). As our field strength it a

3.60

E

216

4. I17 so ql

2 7t(1

+ r2)3/2

(a) The force F1 on unit length of the thread is given by

E

=E

i

where 1 is the field at the thread due to image charge :

-x

E 1= 27(E0 (20 1 k2 Thus

47teo /

minus singn means that the force is one of attraction. (b) There is an image thread with charge density- X behind the conducting plane. We calculate the electric field on the conductor. It is

E(x)=En (4

it €0 (X2 + 12)

ou considering the thread and its image. Thus

291 3.61

(a) At 0 , 00

En (0) =

2

43TE0x2s. 231E01 So

a (0) A. to

En= 2711

00

dy 4 ?ACE()

3/2

on putting y = x

2

+ r2,

12+r2

A 2 n eo Hence a (r) =

E0

l 2 + r2

A

En=

231 V12

r2

3.62 It can be easily seen that in accordance with the image method, a charge —q must be located on a similar ring but on the other side of the conducting plane. (Fig.) at the same perpendicular distance. From the solution of 3.9 net electric field at 0, E=2

ql 4 it eo (R2 4. 12)"

( -4

where is

outward normal with respect to the conducting plane. Now

En. Eo

Hence

— ql

a= 2 it

/2 + 12)3

where minus sign indicates that the induced carge is opposite in sign to that of charge q> O. 3.63 Potential q is the same for all the points of the sphere. Thus we calculate its value at the centre 0 of the sphere. Thus we can calculate its value at the centre 0 of the sphere, because only for this point, it can be calculated in the most simple way. EP 2°

1 / + szp' 4 7t eo /

(1)

292 where the first term is the potential of the charge q, while the second is the potential due to the charges induced on the surface of the sphere. But since all induced charges are at the same distance equal to the radius of the circle from the point C and the total induced charge is equal to zero, p' = 0, as well. Thus equation (1) is reduced to the form, 1 q cP 4 it co 1 3.64 As the sphere has conducting layers, charge —q is induced on the inner surface of the sphere q and consequently charge + q is induced on the outer layer as the sphere as a whole is uncharged. 2=

Hence, the potential at 0 is given by, q (— 4) + cP0 — 4 7t E0 r 4 it E

q 4 it e R

0 R1

0

2

It should be noticed that the potential can be found in such a simple way only at 0, since all the induced charges are at the same distance from this point, and their distribution, (which is unknown to us), does not play any role. 3.65 Potential at the inside sphere, ql

q2

(Pa— 4ne0a+4itE0b

b

= 0 for q2 = — —a ql

Obviously

(1)

r a b,

When ql

q2

r, using Eq. (1).

(1-

(Pr — 4 n eo r

+

4 n eo r

4 n Eo

a

And when r s b ql CPr -

3.66

4 n eo r

(a) As the metallic plates

+

q2

4 Tt eo b

ql

(1

1

4 n eo r a

1 and 4 are isolated and conncted by means of a conductor,

= cp4. Plates 2 and 3 have the same amount of positive and negative charges and due to induction, plates 1 and 4 are respectively negatively and positively charged and in addition to it all the four plates are located a small but at equal distance d relative to each

293

other, the magnitude of electric field strength between

1 - 2 and 3 - 4 are both equal in

magnitude and direction (say E ). Let E' be the field strength between the plates 2 and 3, -4. -4. which is directed form 2 to 3. Hence E' t 1 E (Fig.). According to the problem E' d = Aq) In addition to 0931 - 9)4 = or,

31

I:1)2 -

(1)

(1)3

(9)1 - 9)2) + (c1)2 - 9)3) + (9)3 - 9314

0 - - Ed + Ay-Ed

or,

.. Ai 2d

AY - 2Ed or E

Hence

E . —' . (2) 2 2d (b) Since E a a, we can state that according to equation (2) for part (a) the charge on 1 / 3 rd of it lies on the upper side and the plqte 2 is divided into two parts; such that 2 / 3 rd on its lower face.

Thus charge density of upper face of plate 2 or of plate 3a 1 or plate 4 and lower face of 9) = eo E . A and charge density of lower face of 2 or upper face of 3 2d ' E' (3 ' Eo ' Eo d4TEO

Hence the net charge density of plate 2 or 3 becomes a + a =

3 Eo A 9) which is obvious 2a'

from the argument. 3.67 The problem of point charge between two conducting planes is more easily tackled (if we want only the total charge induced on the planes) if we replace the point charge by a unifonnly charged plane sheet. outward electric field on the two Let a be the charge density on this sheet and E1, E

sides of this sheet. Q Then

Ei +

E 2 -= CO

The conducting planes will be assumed to be grounded. Then E1 x = (1- x). E

2

a ,, ,

Hence

E1- - -

ki

, -

X) ,

a

r, E.2

2 -.

X 1

1 E O

0

This means that the induced charge density on the plane conductors are a

a

a1= - 7 (1 - -0, 02 m - T x

2

294 3.68 Near the conductor E = E,,.. E0 This field can be written as the sum of two parts E1 and E2. E1 is the electric field due to an infinitesimal area dS. Very near it Ei . ±

a 2 Eo

43 The remaining part contributes E2 = — on 2 eo both sides. In calculating the force •on the element dS we drop E1 (because it is a self-force.) Thus 02 dF a dS 2e0= 2E0

S

T

3.69 The total force on the hemisphere is /4/2 2 a F= i — cos 0 • 2nRsin0Rd0 0 2 E0 x/2 2

R2 02 f cos 0 Sin 0 d 0 2 eo 0 2 2 2 7C 1t2 1 q 2 £0 x 2 x 4nR2) 32ne0R =

LA____)

3.70 We know that the force acting on the area element dS --0 1 --* d F= — a E dS (1) 2 It follows from smmetry considerations that the resultant force F is directed along the z-axis, and hence it can be represented as the sum (integral) of the projection of elementary forces (1) onto the z-axis : dFz= dF cos 0

(2)

For simplicity let us consider an element area dS = 2 n R sin 0 R d 0(Fig.). Now considering that E = a/eo. Equation (2) takes the from 2 Tt cr2 R dFz = sm 0 cos 0 d 0 Eo MI

-

2 2 n clo R go

3 COS 0 d cos 0

of a conductor is,

295

Integrating this expression over the half sphere (i.e. with respect to cos 0 between 1 and 0), TC On F = Fz = R

we obtain

2 2

4 eo noP 3.71 The total polarization is P = (e - 1) eo E. This must equals where nois the concerntation of water molecules. Thus nop N= = 2.93 x 103 on putting the values (e 1) e0E

3.72 From the general formula -10 -11.11, -IP p p

2

1 3 — r E 4 rc eo r5 1 2ir E 4 Eco 13 where r= 1 and This will cause the induction of a dipole moment. a 1 2p P P 4 n Eo 13 x e° Thus the force, 13 2p a 1 2E. 3 13p2 13 al 4 7r Eo 13 4 7c2 4

7 t

3.73 The electric field E at distance x from the centre of the ring is, qx

E (x) - 4 7c eo (R2 + x2)3/2 The induced dipole moment is p = 13 Eo E =

q x

4 7c(R2+ X2)3/2

The force on this molecule is 2 8213 x (R — 2x2) 2 16 7E2 eo (R + x2)4

This vanishes for x =

V2

(apart from x = 0, x = co)

It is maximum when a x (R2 - A2 x 2) 4 ax (R2 + x2) (R2

or,

)

4 x2

(R24. x2)

8 x2_ (R2_ _ 2 2 = 0 R2

or, orrx =

2 x2) (R2 + x2

R4 —13 x2 R2 + 10x4 = 0 or, x2 = —20 (13 ± VI191 R

✓

13 ± Nifg (on either side), Plot of Fx (x) is as shown in the answersheet.

20-

296

3.74 Inside the ball

—• --* --1,. a Dr; ) = —A-- — = EEO E. 4 7t r3

Also

E-1

E0E+P. D or P—

E

f

Also,

i

q'=

•

cir

_E-1_q_

E

1 q T* 4 7t r3

r d c 2 ......... E

4,7( J

E

175 Ddiel =t EOE did m D conductor 1.

E—

-41'

D=

a or, Edid =

1 E

q

a E E0

Pis=

(E - 1) E0Edid

=

E—1

a

E

— Pal.

E-1

a

E

This is the surface density of bound charges. 3.76 From the solution of the previous problem q' in - charge on the interior surface of the conductor = — (E — 1)/E fo dS =

—e;1q

Since the dielectric as a whole is neutral there must be a total charge equal to E—1 q' outer = + q on the outer surface of the dielectric. E 3.77

(a) Positive extraneous charge is distributed uniformly over the internal surface layer. Let o0 be the surface density of the charge. Clearly,

E = 0, for r < a

For a < r E0E x

4 7t r2 = 4 it a2 oo by Gauss theorem. 2

or,

E—

GO

(a

ab

r

19_ E

So by integration from infinity where 00 a2 PC I`

E0 r

r>b

cp (cc) = 0,

297

a

R 2

By continuity of

B — 2133

at

so,

, _ _.2 '

r

*" tO r

nr2

pR3 14) =

3Eol.' R2

r>R

++ CC r 11, then using — xp = 0 in the closed circuit, (Fig.)

rs. + g - s2

or,

C2

0 1

(g2 gl) Cl C2

qua

(Ci + C2)

D

-sH q

C

C2

313 Hence the P.D. accross the left and right plates of capacitors, _ q _ & U C2 el + e2 (P1 Cl and similarly -q

(P2- C2

(g1-g2)C1

Ci + C2

3.119 Taking benefit of tire foregoing problem, the amount of charge on each capacitor I - il Ci C2 I ql ' Ci + C2 3.120 Make the charge distribution, as shown in the figure. In the circuit, 12561. -A9= 0 yields - 'VI+ 612 A -4 C2-42 3 C3 C4 41 + 41 - = 0 or, 4 • 21} q1- e3 + e4 C4 C3 and in the circuit 13461, q2 q2 /.., + (_, - g = 0 or, 4,..2

Now =

q2 -

B Cl C2 + e2

5

-11+41

•

-cisI itqi

%,-,1 q2

q1

(PA - q)B 1" Zli: - c13

[ C2 Ci + C2 C3

c+4 I -c4

Cl

6

[

CC -CC2 32 1 4 (Cl + C ) (C3 + C4)

C3

a

16 It becomes zero, when (C2 C3 - C1 C4) = 0. or

Cl_ C2 = C4

3.121 Let, the charge q flows through the connecting wires, then at the state of equilibrium, charge distribution will be as shown in the Fig. In the closed circuit 12341, using - Aq9= 0 (Ci V- q) a a + -A-+ =I-- - 0 Cl C2 C3

14-s

or q -

V (1/C1 + 1/C2

+(C v-4)

+ 1/C3)

- 0 06 mC CI7(cv,i)

3.122 Initially, charge on the capacitor C1 or C2, Ci C2 q - C1+ C2 , as they are in series combination (Fig.-a)

I

c2

314

when the switch is closed, in the circuit CDEFC from - Ay = 0, /(Fig. b ) q2 s -

c2

=

or q2 = C2

u0

And in the closed loop BCFAB from - Ay = 0

-q1 C1

q2 F2

0

Swi

C

IIIC2 T'

1

qi

F (b)

cad From (1) and (2) q1=

2

-14 A

0

Now, charge flown through section 1 =

(q1 + q2) - 0 = C2

and charge flown through section 2 = -

-q

c

CI C2 +c,2 1

3.123 When the switch is open, (Fig-a) 2 C'i.C2 go - el e2

7

1 Cl

3 -go 2 -qo T C 2 +g o

Sv4

3

2

and when the switch C and is q2closed, = q1= 1 C2

Hence, the flow of charge, due to the shortening of switch, ugh the section 2 = - q2 - (q0) = t through section 1 = q1- qo = C1 h r and through the section 3 = q2 o

C - C_I 2 CI + C2

2 4

C2 [

p , C CI - C2 = 3 6 I X C I + C 2

(q2 - q1) -

0=

(

C2 - C1) = - 60 [tC

315 3.124 First of all, make the charge distribution, as shown in the figure.

In the loop 12341, using — Aq = 0

-, + i q2 z--1 1 q c 3

q1

- 0

(1)

Similarly, in the loop 61456, using — Acp = q2

Z

1441

I

6

/ -(1211+42.

0

q2 - q1 +

C2

-

C3

=

(2)

0

From Eqs. (1) and (2) we have 2 C2q2 - q1 =

C2 + C3

Hence,

5

C1

C1 —+1 C3

CPA - CPB =

q2 - q1 c3

NB

42 C2 -

1C 1

Ci + C2 + C3

3.125 In the loop ABDEA, using — AT = 0

_

+ 3

ql+ql+q2+ C3

. 0

C1

1

(1) E.—if

1

-0,1+q2) ilqi-H2 D

CI

&I

Similarly in the loop ODEF, 0 ql

q2 C1

(2) F

112 -1.

1-i2+-er = 0 2

bz

Solving Eqs. (1) and (2), we get, C2 -

C2 -

1 C3 +

C

II 2

q21

12

qz

3C3

q1 4. q2 -

A

C3 C2 — +— +1 Ci

TO

2: (1)1

= -

h:iT h

3

Ci (q1 + q2)

Now, (Pi -

il l

5

C3

as ((o = 0)

Cl

i (c2 + c3) - c2 - c3 CI. + C2 + C3

2 (ci + c3) -1c11- c3 And using the symmetry, CP2 -

Ci+C2+C3

(Ci + C2) -

and

Ci -

C2

1:1D3 Ci + C2 + C3

The answers have wrong sign in the book. 3.126 Taking the advantage of symmetry of the problem charge distribution may be made, as

shown in the figure. In .the loop, 12561, -: Acp = 0

316

q2

or

C2

q2

ql

q1

C3

0

C1

5

6 or

q1

CI (C3 + C2)

q2

C2 (C1 + C3)

42

Cr

(1)

4

Cg

Now, capacitance of the network,

C3

T2-1, R1+ R2

C0

q1+ q2

(PA- 'PB

q2/C2 +

q1/C1

qg I -qg

1 (1 + q1/R2)

2 qi 111,

C2

(2)

C

( 1

—+ C2

q2 CI

From Eqs. (1) and (2) 2

C1 C2 + C3 (C1 + C2)

Co

C1 + C2 + 2 C3 qi 3.127

(a) Interaction energy of any two point charges q1 and q2 is given by is the separation between the charges.

+4.

+

4 n eo r

where r

+4. ■• • • • ■•■•

1 I

I

Hence, interaction energy of the system, 2

—4 Ua U —4 b 2

and

U, =

2 4n eo a

2

2

q

zineoa

47c eo a 2 q2 4n

E0

4ireo(v2q./— +2

— a

q2 r—

4 eo (V2 a) 2 q2

2 /27 q —

4 jt Eo WY

a)

4 TE eo a

I

317 3.128 As the chain is of infinite length any two charge of same sign will occur symmetrically to any other charge of opposite sign. So, interaction energy of each charge with all the others, -1+ + 2 3 4

U= —

4nEoa

up to 00 1

1

In (1 + x) = x - — x2 +1 x3 2

But

and putting

x

1

we get In 2 =

(1)

up to 00

1

1

1——+— 2 3

up to 00

+

(2)

From Eqs. (1) and (2), U=

- 2 q2ln 2

4 n eo a 3.129 Using electrical image method, interaction energy of the charge q with those induced on the plane. 2 q

U—

q2

4 n go (20

—

- 8nEO 1

3.130 Consider the interaction energy of one of the balls (say 1) and thin spherical shell of the other. This interaction energy can be written as f dcpq

q1

=1

p2

2

p2 (r)

4 n co R

2 n r sin 0 d 0 dr

—

0

2 r sin 0

d 0 dr

2 co (/ 2 + r2 + 21r COS 0)1/2

l+r

=

ql r dr • 2r 2 E0 l

2

p2 (r)

4 7t r2 dr p2 (r)

-. 4 ic co /

Hence finally integrating CO

ql q2

U. = int 4 It co 1

where, q2 = f 4

r2 p2 (r)

dr

0

3.131 Charge contained in the capacitor of capacitance C1 is q = C1 q and the energy, stored in it : U

9.2 1

= 1 C QD2 1

Now, when the capacitors are connected in parallel, equivalent capacitance of the system, C = C1 + C2 and hence, energy stored in the system :

318 C1 U 1 2

2

2 (40

(C1+ C2)

, as charge remains conserved during the process.

So, increment in the energy, 2 C1 cp2 ( AU= 2 (

1

1+ 2

1

—Ci

r, 2 - C2 `-' 1 IP

)

)

2 (Ci + C2)

= — 0-03 mJ

3.132 The charge on the condensers in position 1 are as shown. Here q qo q + qo ru Co — C + Co and

C Co Hence,

qo- Co + 2 C

+qo Co —qo

11--qo C +q+clo

—20-*71 ) 1:1,

ii-KoG.10 C —(1.-clo

C

+40 1

2

C +q

2 C

IO

(b) After the switch is thrown to position 2, the charges change as shown in (Fig-b). A charge qo has flown in the right loop through the two condensers and a charge q0 through the cell, Because of the symmetry of the problem there is no change in the energy stored in the condensers. Thus H (Heat

produced) = Energy delivered by the cell

= Aq

= q()

2 CCo — Co+ 2C

3.133 Initially, the charge on the right plate of the capacitor,

q = C (l;i -

switched to the position, 2. charge on the same plate of capacitor ; q

So,

Aq=

- C

1

q' = C

Now, from energy conservation, AU + Heat liberated = A ceo, where AU is the electrical energy.

2) and

finally, when

319

2

C l i-

2

-

C

is responsible for redistribution of the charge. So,

as only the cell with e.m.f. C 1 1 2— Hence heat liberated

,2)2+ Heat liberated = A q

2C

2+

Heat liberated

C

1 — C2 2

3.134 Self energy of each shell is given by 1-1, where q is the potential of the shell, created only by the charge q, on it. Hence, self energy of the shells 1 and 2 are : 2 2 W1. q1 a nd W q2 2 = 8 7t eo R2 87ts0 R1 The interaction energy between the charged shells equals charge q of one shell, multiplied by the potential q, created by other shell, at the point of location of charge q. q2 q1 q2 W So, 12 - '1 4ne R0 2 4n E R O 2 Hence, total enegy of the system, 12

Wi+ W2 + Wi2

qiq21 4 n eo { 2 R1 2 R2 R2 3.135 Electric fields inside and outside the sphere with the help of Gauss theorem : r> R) 2 (r s R), E2 A 1 E, ( > q r 4 e oR Sought self energy of the ball 1

qz

[ qi

CO

U- W1+ W2 co

R

ICO Elc'l2

n2

E0 E

22

1)

4 a r2 dr - —2--(1 + 2 2 , Wi 8 neoR 5 0 R U= 3q2 ana 2- 1 Hence, 1 1 3.136 (a) By the expression 4 n r2 dr, for a spherical layer. — E e E2 2 ° E 2 dV = 2e E° To find the electrostatic energy inside the dielectric layer, we have to lategrate the upper expression in the limit [a, b] =

4nr2 dr + f

4

iii

n

E 0 5

R

320 3.137 As the field is conservative total work done by the field force, 1 2 q ((pi - Cp2)

A fd = tit- Uf= - 1 q2 1 - 1 2 4 n E0 R1 R2

, q2 1 - 1 8 n Eo R1 R2

3.138 Initially, energy of the system, Cii = W1 + W12 where, W1 is the self energy and W12 is the mutual energy. So,

, 1 q2 qq0 ui = + 4 a coRi 2 4 a eoRi

and on expansion, energy of the system,

2 4 a eo R2

4 a eo R2

Now, work done by the field force, A equals the decrement in the electrical energy, i.e.

q (qo

A = (U, - Uf) -

+ q/2) ( 1

4 a 60

1

Ri - R2

Alternate : The work of electric forces is equal to the decrease in electric energy of the system, A= Lli - Uf In order to find the difference UL - Up we note that upon expansion of the shell, the electric field and hence the energy localized in it, changed only in the hatched spherical layer consequently (Fig.). R2

R 1

where E1 and E2 are the field intensities (in the hatched region at a distance r from the centre of the system) before and after the expansion of the shell. By using Gauss' theorem, we find

1

E -

g + go

I- 4 n E0 r2

and E2 -

1

As a result of integration, we obtain A-

q (qo + q/2) ( 1

471E0

-

go

4 a go r2

1

R1 R2

3.139 Energy of the1 charged sphere of radius r, from the equation 4 1q q 2 cPq42 n 8e eo r n o If the radius of the then work done is shell changes by q2/8n eor2 ,Thus unit ( r2a)2 a2 sought area 4 force per , e 4 n r2 (8 2 --A____

r

U= —

dr

a r2 Fu dr = - dU =

7C

Fu -

321

r2 x8n eo r2 4 Initially, there will be induced charges of magnitude and on the inner and outer surface of the spherical layer respectively Hence, the total electrical energy of the system is the sum of spherical a and their ofself shells,having n mutual energies radii d energies including the point charge 24 neo 24n 4n eo 4 neo 4neo eo r__ _I.1

7C E 0 rl

TE

2 so

3.140

-q

+q

a

b,

q.

U

1

q2

1 (-02

+

b

Or

_q q

a

+

qq

a

2

ui = ____q___

+

+

- qq

b

b

8 eo n Finally, is at infinityhence,Uf = 0 charge Now, y the agent= incrementin the energy work done b (a)Sought alent e work performed against the work is to th electric field created by equiv one plate, and to er plate away. Therefore the holding at bringthe required work, rest oth where b a

q

= U1- U = —q—2 [1

3.141

1

11

8 n eo a b

E-

2a create d by ti one ys plate t o at the fh locatio e t n of h other. ei A nf it ee ln ds i age nt =q

E (x 2x1 ),

0

2

So, A (x2 x1)

agent = q ,1

L'

eo 2 eos (x2 - x1) A ext =AU(as field is potential) 2 2 eoS 2 eoS When voltage is kept const., the forceacingon each plate of capacitor will depend on the distance betweentheplates So, elementary work done byagent, in a its displacement over a distance relative to the other, But, Fx = -

Alternate :

—

2 q— X

2 — --q—

2 X — —q--

(x2 —xi)X1)

E0S

2

(b)

dx,

0(x)

2eo

x2

dA = - Fx cbc

S a(x) and a(x) =

EoV

Hence, A= fdA 1 e° —s =5 v2 2x

dx

e 2 x1 x2 o S V 2 [ 1 1 1 -

x

322 Alternate : From energy Conservation, Ur— Ui = A eell + A agent 1

or

coS

2 1 eoS 2 V — -- V 2 x1

"2 x2

eoS

Mg

— e0S x 1

x2

( as Aceo = (q1— qi) V = (Cr Ci) V2 )

EoS V2 [ 1 So

3.142

A agent '

2

g V2 + A a ent

1

xi x2

I

(a) When metal plate of thickness rid is inserted inside the capacitor, capacitance of the EOS

system becomes Co - d (1 _

1)

eo S V' Now, initially, charge on the capacitor, q0 = Co V= d (1 — TO 805 d Finally, capacitance of the capacitor, C = As the source is disconnected, charge on the plates will remain same during the process. Now, from energy conservation, Uf — Ui = A

agent (as cell does no work)

q02

1 q°2

1

or,

I c — i Co - A agent

Hence

1[

60 S V

2

1

{

A agent — 2 d (1 — TO

a—

(1 — i) C

1 c v21-1 2— wi 2 (1 — 1)

1.5 mJ

(b) Initially, capacitance of the system is given by, Ce Cn

'1'

(this is the capacitance of two capacitors in series)

ri (1 — E) + E

So, charge on the plate, q0 = Co V Capacitance of the capacitor, after the

glass plate has been removed equals C

From

energy conservation, A agent = Uf — Ui 1q2 { 1 -

i q°

1.

C' . — Co

1 C V2 E Ti (e — 1)

-

0 .8 mj

2 [E _ 1 (E — 1) J2

3.143 When the capactior which is immersed in water is connected to a constant voltage source, it gets charged. Suppose ao is the free charge density on the condenser plates. Because water is a dielectric, bound charges also appear in it. Let

a' be the surface density of

bound charges. (Because of homogeneity of the medium and uniformity of the field when

we ignore edge effects no volume density of bound charges exists.) The electric field due ao a' to free charges only —; that due to bound charges is s — and the total electric field is 80 eo ao —

. Recalling that the sign

of bound charges is opposite of the free charges, we have

Eeo

323

a ao a' -— or, a' - ( e - 1 ) a0 o = EEEE )( E0 E0 E Because of the field that exists due to the free charges (not the total field; the field due to the bound charges must be excluded for this purpose as they only give rise to self energy effects), there is a force attracting the bound charges to the near by plates. This force is 2 1 , 0, ( e 1) 00 per unit area. -C 2eE0 e0 r 2 1 The factor - needs an explanation. Normally the force on a test charge is qE in an 2 electric field E. But if the charge itself is produced by the electric filed then the force must be constructed bit by bit and is

F =j. q(E' )dE' if

1 This factor of -, is well known. For example the energy of a dipole of moment irin an -4 electric field E0 is -p • E0 while the energy per unit volume of a linear dielectric in an 1 electric field is - • E0 where P is the polarization vector (i.e. dipole moment per unit 2 volume). Now the force per unit area manifests ifself as excess pressure of the liquid.

V ao -=— d EE0

Noting that

Ap -

We get

E0E(6 - 1)V2 2d2

Substitution, using E = 81 for water, gives Ap = 7.17 k Pa = 0.07 atm. 3.144 One way of doing this problem will be exactly as in the previous case so let us try an alternative method based on energy. Suppose the liquid rises by a distance h. Then let us calculate the extra energy of the liquid as a sum of polarization energy and the ordinary

gravitiational energy. The latter is 1

2 h • pg • Sh = - pgSh2

2

If a is the free charge surface density on the plate, the bound charge density is, from the previous problem, E-1 a' a E This is also the volume density of induced dipole moment i.e. Polarization. Then the energy is, as before 1 2

-1 - a,E0 = 2

,a ac o

" (e - 1) (32 2cog

324 and the total polarization energy is

-S(a+h)

(e — 1) 02 2e0e 4 1 ". " •• ■•, •

The actual height to which the liquid rises is determined from the formula

dU .U' (h).0 dh E — 1 )Cr2

This gives

h

2e0 e pg

3.145 We know that energy of a capacitor,U = aU —

Hence, from F =

ax

x

q Const.

• 2

2C

we have, Fx

C2

(1)

Now, since d «R, the capacitance of the given capacitor can be calculated by the formula of a parallel plate capacitor. Therefore, if the dielectric is introduced upto a depth x and the length of the capacitor is 1, we have, 2 7t eo eRx R eo (/ — x) C— (2) From (1) and (2), we get,

Fz = eo(e — 1) R dV

2

3.146 When the capacitor is kept at a constant potential difference V, the work performed by the moment of electrostatic forces between the plates when the inner moveable plate is rotated by an angle dp equals the increase in the potential energy of the system. This comes about because when charges are made, charges flow from the battery to keep the potential constant and the amount of the work done by these charges is twice in magnitude and opposite in sign to the change in the energy of the capacitor Thus

N

au ir,2 ac V

Z

ay 2 atp Now the capacitor can be thought of as made up two parts (with and without the dielectric) in parallel. eoe (It —

eoR2cp Thus C

)R2

+ 2d

2d

1cp. 2 Differentiation then gives as the area of a sector of angle pis —R 2 E0R2v2

Nz — — 4d The negative sign of Nz indicates that the moment of the force is acting clockwise (i.e. trying to suck in the. dielectric).

325

3.4

ELECTRIC CURRENT

3.147 The convection current is I

dt here, dq = k dr, where A, is the linear charge density. But, from the Gauss' theorem, electric field at the surface of the cylinder, E—

(1)

2 n co a

Hence, substituting the value of A. and subsequently of dq in Eqs. (1), we get 2E n co adx I=

dt

dx , v dt 3.148 Since d

R

Rx

(R., + 2R) R if

Rx = Rx + 2R + R Rx2 + 2R Rx - 2 R 2 = 0

or,

On solving and rejecting the negative root of the quadratic equation, we have, Rx = R (VT - 1) 3.152 Let Ro be the resistance of the network,

A

A• R2

13

RI

•

1

Ro R2 then, Ro =

R

or R2-R0RI-RIR2 = 0 0 + R2

On solving we get,

R0

/?0

B

328

3.153 Suppose that the voltage V is applied between the points A and B then

V= IR = I0R0, where R is resistance of whole the grid, I, the current through the grid and 4, the current through the segment AB. Now from symmetry, //4 is the part of the current, flowing through all the four wire segments, meeting at the point A and similarly the amount of current flowing through the wires, meeting at B is also //4. Thus a current //2 flows through the conductor AB, i.e.

R0

Hence, R= T 3.154 Let us mentally isolate a thin cylindrical layer of inner and outer radii r and r + dr respectively. As lines of current at all the points of this layer are perpendicular to it, such a

le

7

, .-a•o• / . •C'N — 0

layer can be treated as a cylindrical conductor of thickness dr and cross-sectional area 2 It rl. So, we have,

;I

%\

;

li

III 0 I I I I1 I, ‘I

dr

dr dR = P .S -) P It rl

i

t

i

■12--

and integrating between the limits, we get,

--1--

I 1

if- ............ - -4— ,

+

11.

I

---=---1---- ==--- - ----'\-Jz)

R . --P-- in 2/Id a 3.155 Let us mentally isolate a thin spherical layer of inner and outer radii r and r + dr. Lines of current at all the points of the this layer are perpendicular to it and therefore such a layer can be treated as a spherical conductor of thickness dr and cross sectional area 4 n r2. So

dR = p

dr

2

4nr And integrating (1) between the limits [a, b], we get, P 1 1 R= 4 n[a b Now, for b —* 00, we have

R

-— P4 n a

p [1 1 3.156 In our system, resistance of the medium R = (1-— — 4n 61' where p is the resistivity of the medium The current 4 n la b

(1)

329

f

or,

dam_

At Cp[1 4n

bi

At4nab Cp(b-a)

in

or,

a

Hence, resistivity of the medium, 47t At ab C (b - a) ln 3.157 Let us mentally impart the charge +q and -q to the balls respectively. The electric field strength at the surface of a ball will be determined only by its own charge and the charge can be considered to be uniformly distributed over the surface, because the other ball is at infinite distance. Magnitude of the field strength is given by,

E= 4 7t eo a2 So, current density j =

I=

1 q and electric current P 4 n eo a2

f

iS -

q

,

43t eo a-

4 a2 eo

But, potential difference between the balls, cp+ - Cp_

2 4 It eo a

Hence, the sought resistance, - ozp_

2q/4 n Eo a

p

R= I

f i lP E0

2 na

3.158 (a) The potential in the unshaded region beyond the conductor as the potential of the given charge and its image and has the form (19 =A

1-1 r1 r2

where r1, r2 are to distances of the point from the charge and its image. The potential has been taken to be zero on the conducting plane and on the ball cp a. A (

1 a-2

=V

330

So A as Va. In this calculation the conditions a « 1 is used to ignore the variation of (pi over the ball. The electric field at P can be calculated similarly. The charge on the ball is Q and

Va

Ep

42tEoVa „

coo =

2a1V

1 Then j = —E = 2a1V § normal to the plane. pr (b) The total current flowing into the conducting plane is

I = f 27txdxj = f 0

2a1V

21txdx

2)3/2 P (n2 + 1

o

(On putting y = x2 + 12 ) 00

I=

2

7C

p

V

Hence 3.159

alV f dy J y3/2

47taV

p 2na

(a) The wires themselves will be assumed to be perfect conductors so the resistance is entirely due to the medium. If the wire is of length L, the resistance R of the medium is 1 a — because different sections of the wire are

connected in parallel (by the medium) rather than in series. Thus if R1 is the resistance per unit length of the wire then R= R1/L. Unit of R1 is ohm-meter. The potential at a point P is by symmetry and superposition (for / » a) A T1 A r2 cp si — ln — In 2 a —2 —a A T1 am 2 In r2 Then qh =

V

=

A

a

In

7

(for the potential of 1)

p

331

I — V/In — a V 2 In 1/a In ri/r2

A

Or,

and

s`

We then calculate the field at a point P which is equidistant from 1 & 2 and at a distance r from both : Then

E

x 2 sin°

21n117 /a (r1 ) VI

1

21n 1/a r2 and (b) Near either wire and Then

J

aE

1 p

V 1 21n1/a r2

V 1 21n1/a a 1 V J E p 21n1/a V 7r I= V= L — J 2 aL R E

Which gives

R1 = -e In I/a —q to the plates of the capacitor.

3.160 Let us mentally impart the charges +q and Then capacitance of the network,

ceof En& C

(1) 9)

Now, electric current, --3. i= f j•dS= faEndS as j T T E.

(2)

Hence, using (1) in (2), we get, 1.5 ti A

iV a=

--

EEO

P Ego 3.161 Let us mentally impart charges +q and —q to the conductors. As the medium is poorly conducting, the surfaces of the conductors are equipotential and the field configuration is same as in the absence of the medium. Let us surround, for example, the positively charged conductor, by a closed surface S, just containing the conductor, — (1) asitti then, R .rd f a En clS.

f

and So,

geof ER clS

C = g-Ego RC = — a

p EE0

332 3.162 The dielectric ends in a conductor. It is given that on one side (the dielectric side) the

electric displacement D is as shown. Within the conductor, at any point A, there can be no normal component of electric field. For if there were such a field, a current will flow towards depositing charge there which in turn will set up countering electric field causing the normal component to vanish. Then by Gauss theorem, we easily derive a = Dn= D cos a where a is the surface charge density at A. The tangential component is determined from the circulation theorem

fI. ch7

0

It must be continuous across the surface of the conductor. Thus, inside the conductor there is a tangential electric field of magnitude,

(Conductor)

D sin a at A . eoe This implies a current, by Ohm's law, of D sin a .1

eoe P 3.163 The resistance of a layer of the medium, of thickness dx and at a distance x from the first plate of the capacitor is given by, 1 dx dR = — (1) a (x) S Now, since a varies linearly with the distance from the plate. It may be represented as, a = al +

(02 —

al)

x , at a distance x from any one of the plate.

d From Eq. (1) 1

dR = CI1 +

dx

(02 — al) d

x

S

d or,

01

+

(

R=

1 dx 3.- 02 — al)

I

d

d In (a2 x S — al)

02 al

v S V (a2 — al) — — 511A R a2 d In — al 3.164 By charge conservation, current j, leaving the medium (1) must enter the medium (2). Thus Hence,

1=

j1 cos al = j2 cos a2 Another relation follows from Elt = E 21

333 which is a consequence off E d r 0 1 . 1 Thus — ji sin al = —)2 sin a2 ai 02 tan al

tan oi2

al

02

tan al

al

tan a2

_

(2)

Or,

or,

(1)

J

02 3.165 The electric field in conductor 1 is

P11

x. LI 1

1.

R P2

and that in 2 is E2 on

2

R Applying Gauss' theorem to a small cylindrical pill-box at the boundary. Pi/

R

P21 c/..5 + --- dS

cl*5 Eo

nR

1 Thus,

a = E0 (P2- Pi)

and charge at the boundary= e0 (p2 - pi)/ 3.166

We have,E1 dl + E2 d2 = V and by current conservation 1

E 11

151 Thus,

P2

E2

0

+V

Pi V

Ei =

di

PI dl + P2 d2 E2

d2

P2 V

P1 dl + P2 d2 At the boundary between the two dielectrics, a= D2 - Di = E0 e2 E2 - eo el Ei

Eo V P1 di+ P2d2 (E

2 P2 - el Pl)

dx

3.167 By current conservation E (x) = E (x) + d E (x) = dE (x) p

(x)

p (x) + d p (x)

This has the solution,

E (x)

C p (x) =

/ p (x) A

d p (x)

0

4

E

Etclk

I

334

Hence charge induced in the slice per unit area

da= E

°A

I

/ — [{E(x)+dE(x)Hp(x)+dp(x)}-E(x)p(x)]= E

°

— d[E(x)p(x)]

d Q = Eo Id [ E (x) p (x) I

Thus,

Hence total charge induced, is by integration, E Q = o /(E2 P2 - El Pl)

3.168 As in the previous problem E (x) = C p (x) = C (po + pi x) where

po + p1 d = 11 po or,

Pi -

(Ti - 1) Po

d

d

By integration V= 5 Cp(x)dx = C po d (1 + -2 o 2V Thus Cpo d (ri + 1) Thus volume density of charge present in the medium dQ

- E dE (x)/dx Sthc °

= 2E0 V

3.169

1 C 0 d (i + 1) 2 •°

(i - 1) po

2E0 V (i) - 1)

(a) Consider a cylinder of unit length and divide it into shells of radius r and thickness dr Different sections are in parallel. For a typical section,

d Integrating, or,

1 )._ 2n r dr _ 2 n r3 dr R1 (a/r2) a

(

R1 =

2

1

nR4

R1

2a

S2 27ta

a 2 7c2 , where S= it R

S

= E o ( Z axis is along the axiz of the conductor).

(b) Suppose the electric filed inside is Ez

This electric field cannot depend on r in steady conditions when other components of E are absent, otherwise one violates the circulation theorem -->

f

--).

E • dr = 0

The current through a section between radii ( r + dr, r) is

2nrdr

1 cc/r

Thus

I=

0

2 E = 2nr

3

Edr — oc

E - ItR4E cc 2cc

335

3.170 The formula is, q= C V0 (1 _ e-t/RC ) or,

V= -(= V (1_ e-t/R C ° t/RC

_V

or,

1_ e-t/RC

V0

Vo— V

or, eHence, t = RC In

Vo Vo — V

— Vo

V0

— R C In 10, if V = 0.9 Vo.

Thus t = 0.6 tiS. 3.171 The charge decays according to the foumula - t/R C

=q0 e Here, RC = mean life = Half-life/in 2 So, half life= T = R C In 2 q

But,

C=

Hulce,

p=

EEO ln

pd A 1 — 1.4 x 10 3E2•m

Eeo A d 2

R=

3.172 Suppose q is the charge at time t. Initially q=

at t = 0.

Then at time t, —iR— =0 But,i = —

F

dq dt (- sign because charge decreases)

- iR So

rig + C

dt

dt RC

R t i/R C =

or,

e

or,

e

t n/R C

'

q—

+A e- "VR rl

i -- - 1, from q= Cat t= 0

A= Cl; (

t/R

Hence,

q= C dq dt

Finally,

11 (1— 1) e- t TIM C

3.173 Let r = internal resistance of the battery. We shall take the resistance of the ammeter to be = 0 and that of voltmeter to be G. Initially,V=

,I— r G

336 G V 2° r + G

So,

After the voltmeter is shunted V

r (Voltmeter) RG r+ R G

—

=

11

and

(Ammeter) r G+

RG r+ R+G

From (2) and (3) we have V r +G

(4)

R

From (1) and (4)

Then (1) gives the

G — = r + G — rir or G 11 required reading V +1

3.174 Assume the current flow, as shown. Then potentials are as shown. Thus, -IR1+ p1=cp 1

1 -IR2-

2

I.

Or,

R1 +R2 P2=

And So,

eigrIRM

p1— /

(Pi — (P2 =

Rr

R1+

+

+

R2

2 R2

—1

R1

2

2

6 R2 3

9-I 7 -61-1R R1)/(R1 + R2) = — 4 V 2 3.175 Let, us consider the current i, flowing through the circuit, as shown in the figure. Applying loop rule for the circuit, — A p = 0 = — ( 1 R2

+

—2 or,

iRi + iR2 + iR = 0

g- R1

i (Ri + R211- R)=

or,

i= R + Ri + R2

Now, if

itP2 = —

+ iRi =

0

R

2 R1 = and 2R1 = R2 + R + R R + R + R2i or, R R1 — R2, which is not possible as R2 > Ri or,

Thus,

CP2 — CP3

-1R2 337

2

g

Or,

mi

R + Rl+ R2

So,

R..

R2 - R1,

which isRthe required resistance. 2

l‘r

Na R

3.176

(a) Current, i=

= a, as N

R

(b) cpA — cps = pz

N

R

- n i R . -

n a R

g=

- n a R =

aR

(

given)

0

3.177 As the capacitor is fully charged, no current fl ows through it. So, current 6.14 i=

(as

2> U

R1+R2

And hence, PA

— cps =

t =

1

1-

- 452 + R

1

1

2 + i R2

1

b-, ,

11 701 B

D

+R2 R2

3.178 Let us make the current distribution, as shown in the figure. g

Current i =

(using loop rule)

RI R2 R + R1+R2

So, current through the resistor R1, R2 it -

Ri R2

RI + R2

R + R i + R 2

2

—

1.2 A

RRI+RR2+RiR2

and similary, current through the resistor g

if R +

R1

R2,

1

= =

RiR2 R1+R2 RRI+RiR2+RR2m R i + R 2

x Ro R

3.179 Total resistance= I-

I

i x

RO +

xR0 R + I

x R R 0

I - x 1

D

R°+

IR + xR0

1 - x xR

zg '1 0 [

c

1

+ IR + x Ro

1

0

.8 A

338 xR

Then V= V

1(

°1R+xR0 For

I

_x+ 1

xR

1

VoRx I {1R +Rox(1-7)}

xR0 + 1R )

R»RolVsi Vo

3.180 Let us connect a load of resistancc K between the points A and B (Fig.) From the loop rule, A q' 1 — 11 R1

iR = and

(1)

— (i ii) R2

iR =

Or

= 0, we obtain

i (R + R2) =

+ i1 R2 (2)

Solving Eqs. (1) and (2), we get 1R1

1 ==

+

R 1R2

2R2 +R2

R+

1 RI g2 R2 R1 + R2

where

Ri+ R2

R + Ro

(3)

R

R1 R2 and Ro

R1 + R2

Thus one can replace the given arrangement of the cells by a single cell having the emf and internal resistance Ro. 3.181 Make the current distribution, as shown in the diagram. Now, in the loop 12341, applying — A fp = 0 iR+iiRi+

1= 0

R2

5

(1)

4

and in the loop 23562, iR—

+ (i—

R2 = 0

(2)

Solving (1) and (2), we obtain current through 2

the resistance R, R1- 1 R2) +RR2++ RR 1R2

=

3

i

— 0.02 A

and it is directed from left to the right 3.182 At first indicate the currents in the branches using charge conservation (which also includes the point rule). In the loops 1 BA 61 and B34AB from the loop rule, — A cp = 0, we get, respectively 2"4" i R3 +

i1)R2 3 — (i — ii)

+

Ri = 0

(1)

=0

(2)

6t

On solving Eqs (1) and (2), we obtain —

( 31 —

2) R3 + R2 (.1 +

R1 R2+ R2 R3+ R3 R1 — cpB = R a 0.9 V Thus 1A — it 1

sto 0.06 A

5

339 3.183 Indicate the currents in all the branches using charge conservation as shown in the figure. Applying loop rule, - A cp = 0 in the loops 1A781, 1B681 and B456B, respectively, we get R2

k) = (io - i1) R1 • i3R3 + ii R2 =0

(1) (2) and

21

6

5-

8

0 (3) go Solving Eqs. (1), (2) and (3), we get the sought current _ (R2 + R3) + ,3R3 -13) R (11 (R2+ R3) + R 2R3 3.184 Indicate the currents in all the branches using charge conservation as shown in the figure. Applying the loop rule (- 6, cp = 0) in the loops 12341 and 15781, we get - i + i3 Ri - (4 — i3) R2 = 0 (1) 7 R3 8 --.\./../ ..... "--and (i1- i3) R2 — 2 + it R3 = 0 (2) A118 — i3) R —

(i1

— i3 R3 =

Solving Eqs. (1) and (2), we get i

13

3

(R2

+

R

Hence, the sought p.d. (PA — (PB = —

s,

1

+ R3) + '2 R2

R1R2 + R2R3

I,

ii v

3R1

3

67 — i3 Ri

2 R3 (RI + R2) —

i R1 (R2

+ R3) —

-1V

R1 R2 + R2R3 + R3R1

3.185 Let us distribute the currents in the paths as shown in the figure. Now, cpi - 4)2 = iRi + it R2 (1) '

and (p1 - cp3 = iRi + (i- i1) R3

(2)

Simplifying Eqs. (1) and (2) we get R3 i

=

((Pi — cP2)

+ R2 (T1 — (p3)

R1 R2+ R2R3 + R3 R1

=

0.2 A

3.186 Current is as shown. From Kirchhoff's Second law

_A/WV`

iiki = i2R3 ,

iiRi + ( ii - i3 )R2 = V , i2R3 + ( i3 + i2 ) R4 m V

R, C

2-i3 .-

?3

A

22 R3) D

R2 R4

12+13

--o B

340

Eliminating i2

2 Hence

i3 {

4 R ( R1

+

R2 )

+ - - - - -

D --( R 3 + R 4 )

/1..3

-11[(Ri+

R2 ) - 1 7: 1

(

R3 + R 4 )

/13

R3 (

or,

RI

+

i3 - R3R4 ( RI +

R2

) - R i ( R3 + R 4 )

R 2 ) + R iR 2 ( R 3 + R 4 )

On substitution we get i3.. 1.0 A from C to D 3.187 From the symmetry of the problem, current flow is indicated, as shown in the figure. (1) Now, TA — q),3 .. ii r+(i—ii)R In the loop 12561, from — A q) . 0 (i — OR + (i —24) r — iir = 0 (R + r) i

(2) I11 3r + R Equivalent resistance between the terminals A and B using (1) and (2), Or,

Ro. (PA—cPB i

[R+r (r—R)+Rl 3r+R i 2g

i

2g

r

r

r(3R+r) 3r+R

3.1118 Let, at any moment of time, charge on the plates be +q and

—q respectively, then voltage

3

6

341

From (1) and (2),

41 dt R =

2q 1-- C

dq . dt t _ 21 R

or,

(5)

On integrating the expression (5) between suitable limits, q

t

dq

1

f dt R0

-

t _ 21

`'

-C

g

C

CV= i 1

Thus

C

- —In

or,

= t—

R

g (1 _e-2t/Rc)

3.189 (a) As current i is linear function of time, and at t =

0 and At, it equals io

respectively, it may be represented as, i= io 1At

At

io At

q= f i dt = 5 io(1 o o

Thus

Aoft

2

; .__ 2q io At

So,

2/(1 _ t ) At At

i.

Hence, The heat generated.

At At

H= f i2 R dt = o

3iti t ) 12 R dt = 4q2 R 3 [ At - rt

o (b) Obviously the current through the coil is given by t/At

1= to (12) io At Then charge

q=

f idt = f io 2 -"I dt - In 2 0

0

; _ q In 2

So,

‘

At

And hence, heat generated in the circuit in the time interval t [0, co],

co

and zero

142

1.190 The equivalent circuit may be drawn as in the figure. Resistance of the network = Ro + (R/3) Let, us assume that e.m.f. of the cell is current i

then

Ro + (R/3)

Now, thermal power, generated in the circuit P = i2 R/3 =

(R/3) (R0 + (R/3) )2

P For P to be maximum, d—dR . 0, which yields R. 3 Ro

&,R0

3.191 We assume current conservation but not Kirchhoff's second law. Then thermal power dissipated is P(ii) il2R1

R1

2 R2

( )

= i12 Ri + R2) - 2i4R2 + i2R2 2 R2 = [R1 + R21 1+ ] [i R1 R2

2 R1R2 4-1 R 1+ R2

-11

The resistances being positive we see that the power dissipated is minimum when R2 i -i 1 R1+ R2

This corresponds to usual distribution of currents over resistance joined is parallel. 3.192 Let, internal resistance of the cell be r, then V = - ir (1) where i is the current in the circuit. We know that thermal power generated in the battery. Q = i2r

(2)

Putting r from (1) in (2), we obtain,

Q=

-

= 0.6W

In a battery work is done by electric forces (whose origin lies in the chemical processes going on inside the cell). The work so done is stored and used in the electric circuit outside. Its magnitude just equals the power used in the electric circuit. We can say that net power developed by the electric forces is P = - IV = - 2 • OW

Minus sign means that this is generated not consumed.

343 3.193 As far as motor is concerned the power delivered is dissipated and can.be represented by a load, Ro . Thus V I.

R + Ro V2

and

P . 12 Ro. _

Ro

i/

(R0 +R)2

This is maximum when Ro = R and the current I is then V I= —2R The maximum power delivered is V2 -P 4R 2 The power input is R + Ro and its value when P is maximum is The efficiency then is 12- = 50% 3.194 If the wire diameter decreases by 6 then by the information given __ v2 P . Power input = — = heat lost through the surface, H. R Now, H c< (1 - 6) like the surface area and R cc (1 - 8)— 2 So,

But

V2 (1 - 6)2 . A (1 - 6) or, V2 (1 - 6) = constant Ro V «1 +1 so (1 + 1)2 (1 - 6) = Const = 1 6 = 2 i = 2%

Thus 3.195 The equation of heat balance is V2

R

Put So,

C

T

- k (T - To). C

d

R

T - To= V2 +k . R or, d ka

2

dt

k -,-C

V

V2 CR

/u/c

Of,

-di (e c) - crif e y

V2

ek" = kR e

Of,

ac

+A

where A is a constant. Clearly = 0 at

t= 0,

so A =

-

v2 and hence, kR

7 (i - e - c l , 0 + —kR V2 T- i

")

v2 2R

344 3.196 Let,

PA

— cpB = cp

Now, thermal power generated in the resistance Rx, =

R

if2 R.. _ [

P

2

R + CPR2 Rx R2 +Rx + 1 R 2 Rx

Rx, •

For P to be independent of Rx,

R

dP— 0, which yeilds dRx Ri. R2 .

R

12 Q x = R1+R2 = 3.197 Indicate the currents in the circuit as shown in the figure. Appying loop rule in the closed loop 12561, — Acp = 0 we get iiR — i'f iRi= 0

(1)

and in the loop 23452, (i — i1) R2 + 2 — it R. 0

(2)

Solving (1) and (2), we get,

62

M24-g2Ri i = 1 RR1+R1R2+RR2

So, thermal power, generated in the resistance R,

{ P= ill R =

I.

2 R2 + RR1+R1R2+RR2 dP

12R fR2

R1

For P to be maximum, ---fi — 0, which fields

RI R2 R..

Hence,

PMaX —

1-1,

1

3

R1+ R2

( 1 R2 + g2 R1)2 4 Ri R2 (Ri + R2)

3.198 Let, there are x number of cells, connected in series in each of the n parallel groups

R Now, for any one of the loop, consisting of x cells and the resistor R, from loop rule

i iR + —xr — .x. ..

0

n

N_ So,i —

—

x xr R+— n

Nr' R+ n2

using (1)

345

Heat generated in the resistor R, 2

Q= i R -

( 2

/

)

■ / " 7:

2 R

(2)

n R +NR dQ and for Q to be maximum, tht = 0, which yields TF —=3 R 3.199 When switch 1 is closed, maximum charge accumulated on the capacitor, n-

qmax — C

,

(1)

and when switch 2 is closed, at any arbitrary instant of time, (Ri +R2) (—

= q/C, dt because capacitor is discharging. q t 1 1 or, f . qdq -(R1+ R2) C fodt q„. Integrating, we get

-t

—t In 2— = gmax (Ri + R2) C Differentiating with respect to time,

4, . qmax e (R, +Rd c

or,

(2)

-t 1 i (t) =

dt

=

a

i

e (R1 "2)C (

)

(R1 + R2) C

-t C t or

e (R1 +R2)C

i (t) (Ri + RO

C

Negative sign is ignored, as we are not interested in the direction of the current. thus,

i (t) — (Ri+ R2)

e

(R +R )C 1

2

(3)

When the switch (Sw) is at the position 1, the charge (maximum) accumalated on the capacitor is, q-C When the Sw is thrown to position 2, the capacitor starts discharging and as a result the electric energy stored in the capacitor totally turns into heat energy tho' the resistors R1 and R2 (during a very long interval of time). Thus from the energy conservation, the total heat liberated tho' the resistors. H= Ili = fc-,- =121 Ce

346 During the process of discharging of the capacitor, the current tho' the resistors R1 and is the same at all the moments of time, thus H1 oc R1 and So,

H R1 R)

H1-

H2 cc R2

( as H

+ H2)

21 RiC+RiR2 Hence

R2

e

=

3.200 When the plate is absent the capacity of the condenser is C=

Eo S d

When it is present, the capacity is S C' = d(1 —r))

1-Ti

(a) The energy increment is clearly. AU— 1 CV2 — —1 C'V2 C 11 2 2 2 (1 —

V2

(b) The charge on the plate is qi.

CV initially and q1. CV finally. 1 —1-1

CVTI CV2Ti A charge 1 — has flown through the battery charging it and withdrawing units 1— of energy from the system into the battery. The energy of2 the capacitor has decreased by 1 CV must be the work done by the external 2 1— agent in withdrawing the plate. This ensures conservation of energy.

just half of this. The remaining half i.e.

3.201 Initially, capacitance of the system = C e. 1 2 — (C E) V 2 1 2 and finally, energy of the capacitor : Uf= CV So, initial energy of the system : Ul =

Hence capacitance energy increment, it

AU=

2

C V2

—2

(C e) V2 = — CV2 (e — 1) = — 0-5 mJ 2

From energy conservation A U— Amu + Aagen

t

(as there is no heat liberation) But Acell = (C1— C,)1/2 =

(C — Ce)V2'

347

Hence Anent

= A U— Awn

1 = —C (1 — e) V2 = 2

mJ

3.202 If Co is the initial capacitance of the condenser before water rises in it then 1 Ul = —2 C0 V2 , where Co

eo 21n R

(R is the mean radius and 1 is the length of the capacitor plates.) Suppose the liquid rises to a height h in it. Then the capacitance of the condenser is C—

EE0h27GR

E023tR

e (1— h) 2TER d

d

d

(1 + (e — 1) h)

and energy of the capacitor and the liquid (including both gravitational and electrosatic contributions) is E o 27ER d (1 + (E — 1)h)V2 pg (2 R hd) 1 2 If the capacitor were not connected to a battery this energy would have to be minimized. But the capacitor is connected to the battery and, in effect, the potential energy of the whole system has to be minimized. Suppose we increase h by oh. Then the energy of the capacitor and the liquid increases by

2

E° 2nR Oh

2d

(E —1) V2

p g (2.7tRd)h)

and that of the cell diminishes by the quantity Aceo which is the product of charge flown and V E0 (2itR) Oh

d

(e — 1 ) V2

In equilibrium, the two must balance; so pgdh —

E0 (E - 1) V2

2d E0 (E

Hence

h=

1)V2

2pgd2 3.203 (a) Let us mentally islolate a thin spherical layer with inner and outer radii r and r + dr respectively. Lines of current at all the points of this layer are perpendicular to it and therefore such a layer can be treated as a spherical conductor of thickness dr and cross sectional area 4 it r2. Now, we know that resistance, dr dR = S (r) Integrating expression (1) between the limits,

dr 4 n r

(1)

348 b

R

fdR=fP drr2

Or,

4n

4n,

a

b

a

4 Tt eo e Capacitance of the network,C = [1

1

a b and

q= C

also,

—

where q is the charge q) at any arbitrary moment]

-

R as capacitor is discharging. , From Eqs. (2), (3), (4) and (5) we get,

q-

dt

4 it eo

411 [1 [ dt p a

E

1 bi or

41t

,

q

PEE°

r

f —amd Integrating

dt

1 r dt

q

dt

PEe

P 8Eo

go

-t q

Hence

e

P

E°6

(b) From energy conservation heat generated, during the spreading of the charge,

2

H=

1 q0

— Uf [because ken

1 = I 4 a Ee a 3.204

1 b

0=

go

2

8 1t E08

0]

b—a ab

(a) Let, at any moment of time, charge on the plates be (qo — q) then current through the resistor, i = —

d (go — dt

' because the capacitor is discharging.

-00-4)

i= dt

Or,

--q)

Now, applying loop rule in the circuit, —

or,

or,

go —

—o

C go

dt

dq

1

R

qo — q

—

RC

dt

dt

349

-▪ At t= 0, q= 0 and at t= i, q= q So,

Ingoqo

q

RC

/ q = qo (1- e -T R C

Thus

--r

= 0.18 mC

(b) Amount of heat generated = decrement in capacitance energy 2 q0 1

4f

1

2C

2

q0 (1 —

e

- s/R C)]

[

3.205 Let, at any moment of time, charge flown be q then current i =

dq

dt Applying loop rule in the circuit, -

( CVo SIR-

- q) a

•=1-0 C

C

dt dq

or,

0, we get :

R

1

CV0-2q

RC

C170 - 2q So, In CV0

dt

t - 2 R C for Ostst

i=o8 dt

C Vo or,

q=

1-eR2cj)

2

dq

Hence,

=

C Vc,

2

- 21/R C

=

dt

2 RC

=

Vo

2t/R C -

Now, heat liberaled, • 00

co Q=

-41

f

i2Rdt=

V 2 o

2R

1

f e RC dt= — CV2 4 °

3.206 In a rotating frame, to first order in co, the main effect is a coriolis force 2 m v x This unbalanced force will cause electrons to react by setting up a magnetic field B so that the magnetic force e Tic B balances the coriolis force. Thus

-

e B = co 2m

or, B = -

2m --• — e

The flux associated with this is J= Nnr2B= Nnr2 —2m w

350

where N=

1 is the number of turns of the ring. If co changes (and there is time fo

2r rn the electron to rearrange) then B also changes and so does O. An emf will be induced and a current will flow. This is I= N r2 -2-n1 co/R The total charge flowing through the ballastic galvanometer, as the ring is stopped, is

q= N n r2 I -2Ln co/R e m

So,

2Nitr2co qR

lwr qR

3.207 Let, no be the total number of electorns then, total momentum of electorns, p= nomevd no e Now,

I= p

Sx Vd

(1) ne

- v-S xV d=

T

Vd

Here Sx = Cross sectional area, p = electron charge density, V = volume of sample From (1) and (2) p=

11= 0.40 tiNs

3.208 By definition nevd= j (where vd is the drift velocity, n is number density of electrons.) Then

1

nel

Vd

J

So distance actually travelled
t

nel < v >

( = mean velocity of thermal motion of an electron) 3.209 Let, n be the volume density of electrons, then from I- p ;vd, I= neSx1 I =

So,

t=

n e Sx1 I = 3 vs.

(b) Sum of electric forces = Knv)er1=Inslepli, where p is resistivity of the material. = nSlep—= nelpI.

IAN

(2)

351 3.210

From Gauss theorem field strength at a surface of a cylindrical shape equals, where k is the linear charge density. 2 it eo r'

(1)

I or,

1= k v

or, A. -

I =

, using (1)

v

1277 me

I Hence

Nrne

E =

- 32 V/m 2iteor

2eV

(b) For the point, inside the solid charged cylinder, applying Gauss' theorem,

or,

2nrhEe- nr2h

q 2 eo n R 1

q/1

Ar

E=

So,from

, 2iteoR`

E '

r=

2iteoR 2

- -d-R th .

R 14)2

Or,

Hence,

3.211

Between the plates

49 -a x4/3 or,

acp

1/3

= a x -Ix ax 3 d2cp 4 —2- in v ax ax

2/3 . - p/ eo

4 eo a or, PLet the charge on the electron be - e,

9

x

- 2/3

352 1 - mv2 - e 2

then

= Const. = 0,

as the electron is initially emitted with neligible energy. 2_ts

V2=

So,

,

v=

4e a ° V 9

j= -pv=

(j is measued from the anode to cathode, so the - ve sign.)

3.212 E_ = So by the definition of the mobility

v and

V

+ V u0d,v

+

j=

u0 - d eV ) 7

(n+ uo + n_

(The negative ions move towards the anode and the positive ion towards the cathode and the total current is the sum of the currents due to them.) On the other hand, in equilibrium

= n_

So,

I (uo+ + uo"

n + = n_ = Id

eV

)d

- 2-3 x 108 cm-3

e V S (u+0 + uo ) 3.213 Velocity = mobility x field Vo or, v= u 7- sin co t, which is positive for 0 s w t So, maximum displacement in one direction is x

=f u—

Vo smoatdt I

2 u Vo Iw

0 2u Vo At co = coo x = 1, so,

=1 1 co 2 CO

Thus

U 2 170

3.214 When the current is saturated, all the ions, produced, reach the plate. = .!1'111- = 6 x le cm3 eV (Both positive ions and negative ions are counted here)

Then,

n.

dn The equation of balance is, -tii= ni - rn2

353

The first term on the right is the production rate and the second term is the recombination rate which by the usual statistical arguments is proportional to n2 (= no of positive ions x no. of -ye ion). In equilibrium, dn n — = ll dt n

so,

as

Nit

71:.-= 6 x 10

7

CM- 3

3.215 Initially n = no = ViT77

Since we can assume that the long exposure to the ionizer has caused equilibrium to be set up. Afer the ionizer is switched off,

dt

d

21

''

rn 2

t dn r dt = - — o rt = 2 ' r, n

Or

But

1 —

n

+ constant

1- 1

n = no at t = 0, so, rt =

n no The concentration will decrease by a factor Ti when 1 1 r t„ = no/Ti no u

Ti - 1 no

to - T1 -1 = 13 ms 1/7771.-, 3.216 Ions produced will cause charge to decay. Clearly, evi — V = decrease of charge = ni e A dt = V 71 11 C d or,

eo 11 V t- . 2 = 4.6 days nie d Note, that n1, here, is the number of ion pairs produced. or

3.217 If v = number of electrons moving to the anode at distance x, then

7dv x -

a v or v= v

o

e

11 X

Assuming saturation, I= e vo e a d 3.218 Since the electrons are produced uniformly through the volume, the total current attaining

saturation is clearly, d

I= f 0

1)

Thus,

354

3.5 CONSTANT MAGNETIC FIELD. MAGNETICS 3.219 (a) From the Biot - Sayan law, S

dB—

11/4 dl x F7so 4n r3 I

4

(R d 0) R R3

(as dl

From the symmetry 2

110 i f dB"if 47c 0

and

0

110 2R

6.3

f dl xR* 1144

T

R dl 714 dl

23tR2ir

Here n is a unit vector perpendicular to the plane containing the current loop (Fig.) and in the direction of x 2 nR2i

B= 4 31 (x2 +R .2)3/2 n

Thus we get 3.220 As L AOB =

2

3 t—

, OC or perpendicular distance of any segment from centre equals

R cos —It . Now magnetic induction at 0, due to the right current carrying element AB

µo

i

4n

3t R cos —

2 sm — n

(From Biot— Savart's law, the magnetic field at 0 due to any section such asAB is perpendicular to the plane of the figure and has the magnitude.)

355

x x

a 2 R c o s - sec OdO n

f 1-1. B - f —° i 43t

dx — cos() = r2

Po i 43t

ROi

n COS u

2sin; 47t

2

l it 2 R cos— sec 0 n

x M

1

al

a Rcos;

As there are n number of sides and magnetic induction vectors, due to each side at 0, are equal in magnitude and direction. So, go

Bo =

=

ni

n 2 sin - • n 4n 7t n R cos n

ni — tan 2 ir R n

and for n --* 00

Ro i jtall7t Bo- -i- R- .Lt ItIn

Ro i =

3.221

i ii

B

We know that magnetic induction due to a a perpendicular distance from it is given by :

tto — - (sin 01+ sin 02 ), 4 3t r

Bwhere

straight current carrying wire at any point, at

r is the perpendicular distance of the wire from the point, considered, and 01 is the

angle between the line, joining the upper point of straight wire to the considered point and the perpendicular drawn to the wire and 02 that from the lower point of the str aight wire.

Here,

go and

i

B2 = B4 =

( sin 1 + sin 2 1

4 n (d/2)

cos

1 2

2 Hence,

the

magnitude

of

total

magnetic

induction at 0, = B B0

4i as

cos I 2

43t d/2

sin 2

in1 +cos/ s

-

4

1+ B 2 +B3 +B4

2

4 Ro i stp d sin

0.10 mT

3 1

356 3.222 Magnetic induction due to the arc segment at 0, go i — (2 It — 2 cp) 4 It R and magnetic induction due to the line segment at 0, B.

go

o

i

[2 sin cp] Blin = e 4 it R cos cp So, total magnetic induction at 0, Bo=

B arc +

ne= Bli 231 Ri

[it —

„

2

+ tan p]= 28 it T

3.223 (a) From the Biot-Savart law, (d1 x PO i 4n r3 So, magnetic field induction due to the segment 1 at 0,

dB =

B1= also and

4 7C a

(2 it — q)

B2 = B4 = 0, as dl T t B = PO cp 3 4 it b

Hence, Bo = Bi + B2 + B3 + B4 i{ 2it—cp+1 4tt a b (b) Here, B1=

Poi 3 it

4 n—aa

, B = 0, 2

sm 45°, B3 -47Cb go

B4 = 4 nb sm 45°, and

B5 = 0

So, Bo. B1+B2+B3+B4+B5 i 3n ila i Poi = tto— — — + 0 + -— — sin 45° + 0 sin 45° 4 it a 2 4xb +47Cb = 110t.{ 3n — +— 4 it 2a b

-.SS!!

• s/

357 3.224 The thin walled tube with a longitudinal slit can be considered equivalent to a full tube and a strip carrying the same current density in the opposite direction. Inside the tube, the former does not contribute so the total magnetic field is simply that due to the strip. It is B P

o (//2 n R) h 2n r

1101 h 2 4n Rr

where r is the distance of the field point from the strip. 3.225 First of all let us find out the direction of vector Bat point 0. For this purpose, we divide the entire conductor into elementary fragments with current di. It is obvious that the sum of any two symmetric fragments gives a resultant along 'shown in the figure and consequently, vector B will also be directed as shown So,

I B1= f dB sin fp

(1)

R di sin q) It

=

f 0

Po . , is in q) d q), as di= 2 n` R

Hence 3.226

i

—dcp

n

B = go i/n2 R

(a) From symmetry Bo= Bi +B2 + B3

3 ft 311

110 +

i

4 nR

n

It +

-— ILO

i

4R 4T

(b) From symmetry Bo = Bi +B2 + B3 Po i

i

Po i

r

3 3n

(c) From symmetry Bo= Bi +B2 + B3 tio i

Po

11° 11/4 I4ni4-471R " 4n/= 4nR 3.227

+ n)

K. 4+;32 or, I go I =

= 2.0 T, (using 3.221) 4nl�

358 3.228

(a)

4 R So,

tto i

1B Ism

2

---iTt171

4

4 so

0'30 IAT

R

(b) K-4+4+4 110 =

k

-

)

i

i

4-3-171i71 (-

tr

)

Pv i [ k +(n+ 1) 4itR

So,

II3: = —411°it

+ (It + 1)2 = 0.34µT

(c) Here using the law of parallel resistances + i2 = i and

i2

So,

1 - = 1:2 3'

4 3 1. -;1-

3. z2 - -4- z, and it

Hence

Thus ro=

'to

4 it R

4nR

4n

2 R

:4I1° (n/2) i2

R

lio i --a. (j+k)+0 4 31 R i µ0 =4 R "1 T 3.229 (a) We apply circulation theorem as shown. The current is vertically upwards in the plane and the magnetic field is horizontal and parallel to the plane. Thus,

110 f

B

d ri .

2 B1 = }toil or,

R-

i (b) Each plane contributes tinv

2 2- between

the planes and outside the plane that cancel. Thus

B

1.10 i between the plane 0 outside.

359

X=0

3.230 We assume that the current flows perpendicular to the plane of the paper, by circulation theorem, 2B dl = go (2 x dl ) j or,

B- xj, I xl s d

Outside, 2 B dl go (241 dl ) j

vodj Ixlz d.

or,

3.231 It is easy to convince oneself that both in the regions. 1 and 2, there can only be a circuital magnetic field (i.e. the component B,). Any radial field in region 1 or any B, away from the current plane will imply a violation of Gauss' law of magnetostatics, Bcf, must obviously be symmetrical about the straight wire. Then in 1, Bq,27tra [to/ Bq,

Or,

In 2,

llo I 2 7t —r

B •271r = 0, or

B, = 0

2 4.4.0/R 2 3.232 On the axis,B

Bx, along the axis.

3/2 =

2 (R 2 + X2 )

0

00 00

fB

Thus,

dx dr. f

Exdr. NI R2

2

— 00

=

go I R

31/2

(R 2 j — 00

R sec2 0 d 0 3 3

f

2 2 R sec 0 - x./2

r

+ x2)3/2

, on

putting x = R tan 0

x/2

1 cos 0 d 0 = 11/4/ tio / — 2 - x/2 The physical interpretation of this result is that f Bxdx can be thought of as the circulation ao

of B over a closed loop by imaging that the two ends of the axis are connected, by a line at infinity (e.g. a semicircle of infinite radius). 3.233 By circulation theorem inside the conductor 2 B 2 7c r = tio jz it r or, B,- tio iz r/2 B

—•

1

—

2

°

jxr

Similarly outside the conductor,

360

iz X R 2

By 2 3t r =

. R2 14°Jz r

---0 1 R2 B. Li. 0)61. —

So,

2 ° 2.234 We can think of the given current which will be assumed uniform, as arising due to a negative current, flowing in the cavity, superimposed on the true current, everywhere including the cavity. Then from the previous problem, by superposition. -0,- 1

B

1

or, By

1

r2

-°

- 1 ) x (A P - B P ) = go j x / 2 1° -2

If 1 vanishes so that the cavity is concentric with the conductor, there is no magnetic field in the cavity. 3.235 By Circulation theorem, r

B •2irr= t.tof j(ri)x21tr' dr' or using By.. brainside the stream,

So by differentiation, (a + 1) bra = [to j (r) r Hence,

j (r) =

b ( a + 1)

_

r

3.236 On the surface of the solenoid there is a surface current density js=

nIeip A

e x ro Then,

B= -

4

31nIf Rdcpdz

ro

where 70a is the vector from the current element to the field point, which is the centre of the solenoid,

-v2

Thus,

B Bz

gon.1 4x

x2xR-

f -U2

dz (R 2 + 43/2

361 + tan

-I

/ — 2R

1

cos a d a

= ptoni

f

-1

tan

(on putting z= R tan a)

1 2B 2

1/2

= ponisina= gonI

go nil V1+ ( 1

1 (1/2) + R

=

3.237 We proceed exactly as in the previous problem. Then (a) the magnetic induction on the axis at a distance x from one end is clearly, nI B=

x2 7tR

dz

ER 2

2f

(z x)

213/2 -

2 1 µo n

IR

dz (z2 + R 2)3/2

2 x

0 n/2

1 =— 2

n I f cos0d0= 1Ron/ 1— 2 _1 -

Vrx27iR2

tan

0 menas that the field point is outside the solenoid. B then falls with x. x < 0 means that the field point gets more and more inside the solenoid. B then increases with (x) and eventually becomes constant, equal to I,to n I. The B — x graph is as given in the answer script. X>

Bo — 8B (b) We have,

1 =

xo R

or,

= 1 —71

[1

—

+ xo

xo

R2 + x02

=1—2

Since Ti is small (.1%), xo must be negative. Thus xo = and

is,

xo 11R 2 + 1 x0 12

1 — 2ri

— I xo

362

By comparision with the case of a solenoid and a hollow straight conductor, we see that field inside the coil = to 11: 1/1 — (h/2 Tt R)2 (Cf. B = go nI ). Outside, only the other term contributes, so I h Bq, x 2 7tr = [to h x 2 3tR x 23TR

or,

[to

B

=

2/

TC

q,

7

t

r

Note - Surface current density is defined as current flowing normally across a unit length over a surface. 3.239 Suppose a is the radius of cross section of the core. The winding has a pitch 21tR/N, so the surface current density is I --1. I ---3. J id e e + s 2aR/N 1 23ta 2 where 4 is a unit vector along the cross section of the core and il is a unit vector along its length. The magnetic field inside the cross section of the core is due to first term above, and is given by B4).2.7rR = go NI (NI is total current due to the above surface current (first term)) Thus,

B, = p.0NI/ 27r,R.

The magnetic field at the centre of the core can be obtained from the basic formula. -

-9

di34.. tto J., x r dS and is due to the second term. 4n r 3 o So,

or,

B = B z a=

Bz=

—• 110 ez 4 7t 27ta

f 1. Rthp x 27ta R3

1101 2R

The ratio of the two magnetic field, is =

N —it

3.240 We need the flux through the shaded area.

Now by Ampere's theorem; B

q)

2.7tr = go

I 2 -I • ;Er 3tR

ttO r B = — .1 --I 4) 2-it R The flux through the shaded region is, Or,

363 R

cpi - 5 1-dr.Bv(r) 0 R

= f dr - ill r =

2it R2

o

NI

4n

3.241 Using 3.237, the magnetic field is given by, B = 1 Ito nf 1-

2 1

,x -2 x +R 2

1

µ n I = B , here /30. 1. ni, At the end,B = -2 o -2 0 w t0 is

the field deep inside the solenoid. Thus, (I) = ito nIS = 00/2, where (1) = vo nIS 2 is the flux of the vector B through the cross section deep inside the solenid. 3.242

Bq, 27Er = 1.4.0N I go

or

NI

23tr

Bp 7 b

Then, cl) - 5 B

92

Ro

:4 .,-,- 2N l h In Ti, where 11= b/a

hdr,as rs b=

a

3.243 Magnetic moment of a current loop is given by pm = n i S ( where n is the number of turns and S, the cross sectional area.) In our problem, n =

o 1, S = aR 2 and B = g — -

2 RR So,

pm=

2BR

n.12 2 —

BR3 21(

go

110

3.244 Take an element of length rd 0 containing —

N nr

• rd0 turns. Its magnetic moment is

lld0--Id2/ It 4 normal to the plane of cross section. We resolve it along OA and OB. The moment along OA integrates to x

f

11c12/dOcos 0. 0 4

o while that along OB gives a

N d2I

pm= f

0

sin 0 d0 = 2

N d2/

364 3.245

( a )

Fr o

m

l oo p

at

B i o t - S av a r t ' s

its

c e nt r e

is

l a

w,

th e

g iv e n

ma g ne t i c

i

n duc t i o n

e

du

a

to

c i r c ul a r

c u r r e nt

c a r r y i ng

wi r e

by ,

B,. = -111

i

2r

Th e

pl a

f r o m

ne

a

sp i r a l

b.

to

i s

m ad e

Th e r e f o r e ,

up

t he

of

c o nc e nt r i c

c i r c ul a r

m ag n e t i c

to ta l

f

i

nd u c t i o

RO

Bo

oo p s ,

l

n

at

h a v i ng

th e

d i

f f e r e nt

ra dii,

v a r y i ng

c e nt r e ,

(1)

dN

2 r

Ro wh e r e

the

-

i

2r

i nt e r v a l

is

(r,

the

r

c o nt r i but i o

n

of

one

N

dN = ng

in

r a d i

us

r

dN

a nd

is t h e

nu mb e r

of

t u r n s

e q u a t i o

n

(1)

a

n d

b

a

dr

ng

i nt e gr a t i

t h e

r e s u l t

o ver

r

b e t we e n

a

a nd

b,

we

ob t a i n ,

b

Ni

B

Th e

m ag n e t i c

N

2r

° (b )

=

moment

f a

of

2

turn

o f

r a d i

u s r is

P=f ( a )

Let

us

ta

k e

a

hy pm t

r i ng

e l e m ent

i

j

n

In

b

a

dr =

(b - a) a

lioiN (b - a)

b

3.246

N

r 2

dr

b

pm = in r2 and of all turns, ni

_

a

3

(b3 - a2) (b - a)

a

o f

r a d i u s

r

a

nd

t h i c kne s s

dr,

then

ch a r g e

o

n

dq= a2nrdr (a a n d

So ,

c u r r e

nt ,

m ag n e t i c

i

due

to

th i s

nd u c t i o n

ele

at

ment ,

the

2

n

r

2

i r

dr) co

tii =

c e nt r e ,

=

du e

to

th i s

ele

CO dr r

a

ment

:

dB =

ILO 2

R f an d

h ence ,

fro

m

s y

mm e t r y

Ro

a

co

r

dr

:B=.1dBi.

tic, "

r

Magnetic moment of the element, considered, dpm = (di) nr2 = a co dr n r2 = on to r3 dr

H e nc e ,

the

so u

ght

m a g ne t i c

m o me nt ,

R

pm = f dp m= fanco r3 dr = a co n 4

"'.. CF °A 2

o

(b )

in

dr)

+

i . e .

Su b s t i t u t i

o f

t ur n

R4

di r

th e

r i n g

e l e ment . ,

365 3.247 As only the outer surface of the sphere is charged, consider the element as a ring, as shown in the figure. The equivalent current due to the ring element,

di = -2w7t (2 nr sin 0 rd 0) a

(1)

and magnetic induction due to this loop element at the centre of the sphere, 0,

dB .

di 2 n r sin 0 r sin 0 4n

I-to

4n

r3

di

sine 0

r

[Using 3.219 (b) ] Hence, the total magnetic induction due to the sphere at the centre, 0, 2t/2

0

N./2

B=

Hence,

f I-to

a

wr 3 sin 0 d 0 4n

2 —3 o cr (or

29 pT

0 3.248 The magnetic moment must clearly be along the axis of rotation. Consider a volume

q 3 dV' The rotation of the sphere causes this charge 4 n/3 R to revolve around the axis and constitute a current. element dV. It contains a charge

3q 4 R

2,., av x 3

2n

Its magnetic moment will be

3g

dV x 2n x nr2 sine 0

43TRSo the total magnetic moment is R

o o

f

r2 sin ode

x

2 sin2 0 r

2R 3

2

dr

3.q_x1(1 xR5 x4 2R3

2

5

1 3

314

2

—5 qR

The mechanical moment is 2

— tR co, So,P

5n

m M = 2m

3.249 Because of polarization a space charge is present within the cylinder. It's density is pp

— div P. — 2a

Since the cylinder as a whole is neutral a surface charge density ap must be present on the surface of the cylinder also. This has the magnitude (algebraically)

366

a x 2aR = 2 a nR 2

or,

cr = aR p

P

When the cylinder rotates, currents are set up which give rise to magnetic fields. The contribution of pp and op can be calculated separately and then added. For the surface charge the current is (for a particular element) oi orRx2andxx— = aR2w dx 2.7c

Its contribution to the magnetic field at the centre is goR2(aR2codx) 2 (x2 + R 2 )3/2 and the total magnetic field is

Bs =

f

2

(a R 2 oidx) 2 2 (x + R 2)3/2

lioR

go a R 2

40)

i

r

2

2.3/2 m

2

(x + R )

- 00

00

„4 Ro a iv 0)

dr

2 x --2-= 1.10 a R2(0 R

As for the volume charge density consider a circle of radius r, radial thickness dr and length dx. w The current is- 2cc x 2ardrdrx = - 2 ar dr co dr 2n The total magnetic field due to the volume charge distribution is R

co g0

B„= - f dr f dr2nrco 0

R

2

2 (x

- 00

2

r + r2)

co

-f a t.to w r3 dr f dx (x2 + r2)3/2

3/2 21

0

-00

R

= -f a go cor dr x 2 = - 1.10 a co R 0 3.250 Force of magnetic interaction, Where,

g

B

M

2

B= Bs + By= 0

e (17'x B )

lio e (vr1

47r

r3

If

2

Fma = r1) e [v---,x (v--,. x 0] t Llit—r3

So,

=

µ0 e3 zln r3

t 1 V1.1

And

171: X 17- ( 0

Fe e - geE-= e

I mi ag I

Hence,

so,

IFe lectic I

= - V2Ro E0 •It

X 71 =

1

4 n go

V2 (—) c

go e2 (_ v2 y1 4n r3 -4. er

1 713 = 1.00 x 10- 6

367 3.251

\ (a) The magnetic field at 0 is only due to the curved path, as for the line element, dl tt F.'

B =

Hence,

Po i

I-to i it (- k ) = 4nR

-2 PO 1

--*

Thus i: = iB (- .1.) -

-3"

— (- k ) 4R —1'

(- i )

4R

-2 not So, F. = 4R – 0.20 N/m -4. (b) In this part, magnetic induction Bat 0 will be effective only due to the two semi infinite segments of wire. Hence

B . 2• li° i sin -I! (4n =

(I) 2

I-to i(_

2

r)

)

itl

Thus force per unit length, 0 :. _ I-Lo ( r) n1 3.252 Each element of length dl experiences a force BI dl. This causes a tension T in the wire. For equilibrium,

T d a = BI dl, where da is the angle subtended by the element at the centre.

dl T = BI — = BI R da

Then,

The wire experiences a stress

BIR ic d2/4 This must equals the breaking stress crm for rupture. Thus,

a d2 a. B

=

4IR

3.253 The Ampere forces on the sides OP and 0' P' are directed along the same line, in opposite directions and have equal values, hence the net force as well as the net torque of these forces about the axis 00' is zero. The Ampere-force on the segment PP' and the corresponding moment of this force about the axis 00' is effective and is deflecting in nature.

368 In equilibrium (in the dotted position) the deflecting torque must be equal to the restoring torque, developed due to the weight of the shape. Let, the length of each side be the density of the material then,

1 and p be

1 ilB (1 cos 0) = (S / p)g — sin 0 + (S / p)g —1 sin 0 2 2 + (S 1p) gl sin 0 or,

112 B cos 0 = 2 S p g12 sin 0

Hence,

B - 2S

pg I

tan 0

3.254 We know that the torque acting on a magnetic dipole. —4.

_a,

—a.

N. pmxB A

A

But, /5: = i S n , where n iis the normal on the plane of the loop and is directed in the direction of advancement of a right handed screw, if we rotate the screw in the sense of current in the loop. On passing a current through the coil, this torque acting on the magnetic dipol, is counterbalanced by the moment of additional weight, about 0. Hence, the direction of current in the loop must be in the direction, shown in the figure. 13: x :9.= - /.>: Ainr Or,

N iSB- Amg1 B .-A1171 -1 = 0.4 T on putting the values. NiS

So, 3.255

(a) As is clear from the condition, Ampere's forces on the sides (2) and (4) are equal in magnitude but opposite in direction. Hence the net effective force on the frame is the resultant of the forces, experienced by the sides (1) and (3). Now, the Ampere force on (1), µo

iio

I 2

F . 1

21( (1 714

1

and that on (3),

3

to F3

i 1

So, the resultant force on the frame =F1- F3, (as they are opposite in nature.)

.„„e.._._a÷.___),, 4

369 2 1.10 iio

0.40µN. a (4 ri2 — 1) (b) Work done in turning the frame through some angle, A= f id (I) = i (43f — cl)i), where Of is the flux through the frame in final position, and (Di, that in the the initial position. — cl) and

I (Dfi -

Here,

— (Df

A1= 210 and A= i2(13

so,

A= 2i f Bds

Hence, a (71 + —1)

2

f

=21

go i a— r= 2.7t r

go

Uo a

3.256 There are excess surface charges on each wire

+1) In 2ri — 1

(irrespective of whether the current is

fl owing through them or not). Hence in addition to the magnetic force Fm, we must take into account the electric force Fe Suppose that an excess charge X, corresponds to a unit length of the wire, then electric force exerted per unit length of the wire by other wire can be found with the help of Gauss's theorem. Fe. ),E

k

1 _

21X, eo

2A2 4 nE0/'

(1)

where 1 is the distance between the axes of the wires. The magnetic force acting per unit

++++++++++++-F

length of the wire can be found with the help of the theorem on circulation of vector B [to 1.2

F.= — 4n /

where i is the current in the wire.

(2)

Now, from the relation, X C p, where C is the capacitance of the wires per unit lengths and is given in problem 3.108 and cp = iR i aEo t. R or ,, ne o R In i Dividing (2) by (1} and then substuting the value of — from (3), we get, Fm

(3)

110 (1nri)2 eo 312R 2

The resultant force of interaction vanishes when this ratio equals unity. This is possible when R. Ro, where

370

„ _ Vrio nin — 0.36 IQ /to — Eo it 3.257 Use 3.225 The magnetic field due to the conductor with semicircular cross section is µ0I B= n2R r2

aF

Then

go i

al — = BI = n2 R 3.258 We know that Ampere's force per unit length on a wire element in a magnetic field is given by. A

A

dFn= i (n x B) where n is the unit vector along the direction of current.

(1)

Now, let us take an element of the conductor i2, as shown in the figure. This wire element is in the magnetic field, produced by the current i1, which is directed normally into the sheet of the paper and its magnitude is given by, IBI=

I/ (2)

2 nr

.(--- b ---).

From Eqs. (1) and (2) /2

A

--lb

d Fn = — dr (n x B), (because the current through the element equals b A

12

— dr b

---3'

(IF = 11° '112 —dr ,towards left (as n .1! B ). n a b r'

So,

Hence the magnetic force on the conductor : a+b

tio 1112 —

"g..

F

n

=

2n

b

dr

f

— a

ks(ttowardowar ds left) r

a+b PO 1112 (towards left). = — —F. in 23t

a

Then according to the Newton's third law the magnitude of sought magnetic interaction force 110Ill = a b 3.259 By the circulation theorem B =

2

In

a +b

a

µo i,

where i = current per unit length flowing along the plane perpendicular to the paper. Currents flow in the opposite sense in the two planes and produce the given field B by superposition. 1 The field due to one of the plates is just —B. The force on the plate is, 2 1 2-B x i x Length x Breadth =

B2 — per unit area. 2t.tc,

(Recall the formula F = BB on a straight wire)

371

3.260 (a) The external field must be

B1 + B2 2

which when superposed with the internal field

Bl — B2 2 (of opposite sign on the two sides of the plate) must give actual field. Now B1— B2 1 2 2 11°1 B1— B2

Or,

2—B2 F B1 2 21.to

Thus,

82

Br

Bl — B2 (b) Here, the external field must be upward with an internal field, 2 on the left and downward on the right. Thus, B1+ B2 i= 1-to (c)

+ B2 2

'u

pward

2— 2 B1 B2 and F — 21,to

Our boundary condition following from

Gauss' law is, B1 cos 01 = B2 cos 02. Also,(B1 sin 01 + B2 sin 02) = µo i

where i

= current per unit length.

Br

The external field parallel to the plate must be

B1 sin 01 — B2 sin 02 2

(The perpendicular component B1 cos 01, does not matter since the corresponding force is tangential) B 2 s;,,,2 _ r,e 2 ,;n2 0 n °All Thus, F — 1 2 per unit area 2110 2 2per unit area. B1 B2 2 µo

The direction of the current in the plane conductor is perpendicular to the paper and beyond the drawing. 3.261 The Current density is —aL , where L is the length of the section. The difference in pressure produced must be, 1 Ap = —aL x B x (abL)/ab

IB — a

82

372

3.262 Let t — thickness of the wall of the cylinder. Then, J = 1/2 n R t along z axis. The magnetic field due to this at a distance r t t (R —i < r

J x n,

-3■

this gives rise to a surface current density in the z —direction of —

.11._. 2tR

The total molecular surface current is, = —

/' s

xi (27r,R) =

— x/.

231R

The two currents have opposite signs. 3.276

We can obtain the form of the curves, required here, by qualitative arguments.

f gal: 1,

From

we get

H (x » 0) . H (x « 0) . nI

Then

B (x » 0) . titio nI B (x _< 0) = t.t.onI

Also,

B (x < 0) = [to H (x < 0) J (x < 0) = 0 B

is continuous at x = 0,

sheet.

H

is not. These give the required curves as shown in the answer-

377 3.277 The lines of the B as well as H field are circles around the wire. Thus H17tr+H2nr= I or, Hi +H2 = Also

1..to µ1H1= µ2H2 [to = B1= B2 = B I

Thus

II = H2 —

[ii

112 nr 111 I

111

/41 + IA2 nr 111 1-12 I and

B = µoo 111+ 1-12. nr 3.278 The medium I is vacuum and contains a circular current carrying coil with current I. The medium II is a magnetic with permeability The boundary is the plane z= 0 and the coil is in the plane z = 1. To find the magnetic induction, we note that the effect of the magnetic medium can be written as due to an image coil in II as far as the medium I is concerned. On the other hand, the induction in II can be written as due to the coil in I, carrying a different current. It is sufficient to consider the far away fields and ensure that the boundary conditions are satisfied there. Now for actual coil in medium I,

SO,

B2

where

tloPm (2 cost 0, — sin2 0,) and B zin

1A0Pm (— 3 sin 0' cos 0') 47t

, I (n a2), a = radius of the coil. pm =

Similarly due to the image coil, Bz—

1A0Pin, 2 4n (2 cos 0' — sin2 0'), Bx =

7t

m

(3 sin 0' cos 0'), p' m= I' (Jr a2 )

As far as the medium II is concerned, we write similarly A, Oat (2 cost 0' — sin2 02 Bx— l-toP (— 3 sin 0' cos 02 p"m = Bz— tioP

r

a2 )

378

The boundary conditions are, p,,,+ p',„ = p",,, (from Bin = B2,,) - p,„+ p'„, = - 14p" „, (from 1111= H21) Thus,

/1 ' = --- ---p, + 1

1 r= m-=-1-I t.t, + 1 '

In the limit, when the coil is on the boundary, the magnetic field enverywhere can be obtained by taking the current to be -2E-I. Thus ,i.'= -----. /3., 11 + o [I, + 1 3.279 We use the fact that within an isolated uniformly magnetized ball, -4. 4%J H' = - J/3, B' = , where J is the magnetization vector. Then in a uniform magnetic 3 field with induction Bo, we have by superposition, -1.. -4. 2110./ -4. B. = Bo+--s—, H.

B

MT.

0 '..

T /I J 1 ..)

110

B„,+ 21.4.0Hi„ = 3 Bo

or, also,

B:= ti Ito irin

3 Bo -+ H. = and B. = 3 11B° 1.1. + 2 is 110 (Ix + 2) m 3.280 The coercive forcelf, is just the magnetic field within the cylinder. This is by circulation Thus,

theorem, 11,.=

N

1

I

= 6 kA/m

•d r= I, total current, considering a rectangular contour.)

(from f

1-1 3.281 We use, f 11.111=

0

Neglecting the fringing of the lines of force, we write this as H(td-b)+-11.b. 0 1-10 -Bb or,

Hi=

, - 101 A/m [to 'ra The sense of H is opposite to B 3.282 Here, f

Hence,

Bb 4:d r= NI or, H(27(R)+---- NI, so, H= [to 1-1.

=

B 2nRB lioNI-Bb tioH

NI lito -Bb 2 TER tto

- 3700

B µB H from the given graph. The 4 -H graph H starts out horizontally, and then rises steeply at about H= 0.04 k A/m before falling agian. It is easy to check that Ix. ft, 10,000.

3.283 One has to draw the graph of IA =

379

3.284 From the theorem on circulation of vector Bb Hnd+ — = NI

or,

tioN I

[to d

b

b

B

H = (1.51 — 0-987)H,

where B is in Tesla and H in kA/m. Besides, B and H are interrelated as in the Fig. 3.76 of the text. Thus we have to solve for B, H graphically by simultaneously drawing the two curves (the hysterisis curve and the straight line, given above) and find the point of intersection. It is at H 0.26 kA/m, B = F25 T Then, 4000. 1-to H 3.285 From the formula, F (p: • 1V3)r__.i_

pf (J. ,g)rdv,

Thus

(B-V

tt [to or since B is predominantly along the x—axis, x L aB Sgo f dB,2, Fx f Bx Sdx = lAo ax 211 x 0 3.286 The force in question is,

—3.

--0

„(p, • V ) B

F

since So,

B

-

x SB2 212 po

x SB2 2µo

BV dB ,

is essentiatlly in the x-direction. x V dB2 _ X Bo2 V d Fx sa. 2µo dr 2µo cbc

x Bo2 (e 2 ax2 )

This is maximum when its derivative vanishes i.e. 16 a2 x2 — 4 a — 0, or, xn, = The maximum force is,

—

4

ax

e

2110

V

1 4a

2

F max . 4a

1 {4;

So, 3.287 Fx

x

e

— 2µo

-1/2 X B0 V

x B02 V.I

Ve

(R0Fnia,, T) / VB02 = 3.6 x 10-4

Bx x131

dB

x V dB2

2µo dx

dx This force is attractive and an equal force must be applied for balance. The work done by applied forces is, x- L A f Fx dx _LK (_ B2 Ix. x VB2 go ° 2 go x 0 -

380

3.6

ELECTROMAGNETIC INDUCTION. MAXWELL'S EQUATIONS

3.288 Obviously, from Lenz's law, the induced current and hence the induced e.m.f. in the loop is anticlockwise. From Faraday's law of electromagnetic indcution,

0—> cV.).

dt d0= B•dS= —2Bxdy,

Here, and from

y = ax2, X =

a

2B -V-TA

Hence,

1 = By

a dt

774, — , using dt a

X

12WY

3.289 Let us assume, B is directed into the plane of the loop. Then the motional e.m.f.

and directed in the same of (i7x So,

i—

R+

Ri R2

•dr

— (v-*x

gin=

= vBl

B) (Fig.)

_L.

R1 R2

R2

ByI R+

Ri+ R2

As R1 and R2 are in parallel connections. 3.290 (a) As the metal disc rotates, any free electron also rotates with it with same angular velocity w, and that's why an electron must have an acceleration w2r directed towards the disc's centre, where r is separation of the electron from the centre of the disc. We know from Newton's second law that if a particle has some acceleration then there must be a net effecetive force on it in the direction of acceleration. We also know that a charged particle can be influenced by two fields electric and magnetic. In our problem magnetic field is absent hence we reach at the conslusion that there is an electric field near any electron and is directed opposite to the acceleration of the electron. If E be the electric field strength at a distance r from the centre of the disc, we have from Newton's second law. m = wn

m o.)2 r

eE = m r o.)2, or, E —

e

and the potential difference, a

cpcen— cprbn= f 0

a

--+ •dr =

2 f

M 03

e

r dr, as Et 4. di:*

381

(Pcen —(Prim ' A cp = m

Thus

e

c

o2

2 a — 3.0 nV 2

(b) When field ifis present, by definition, of motional e.m.f.

:

2

CP1 — (P2 = f

1-7X 13) • di.

1 Hence the sought potential difference, a

a

(PrimMr f — V Bdr= f — cor B d r ,

(Pcen

0 Thus

(Prim

(as v= wr)

0 419cen p=

1 — co

2 Ba =

20 mV

eB (In general w < — so we can neglect the effect discussed in (1) here). 3.291 By definition, 7E= C

So,

f

( 17X 11)

C • Cr it f

A

d ) •d

- (TX

f

A

- y B dr

0

But, v = wr, where r is the perpendicular distance of the point from A.

Hence,

f

—o)Brdr= —1(oBd2= —

A

This result can be generalized to a wire C

A

10mV

2 AC

of arbitary planar shape. We have

C

C

A

A C

(B•r w-B•co r)•dr

- - f A

1 =—— 2 d

being

AC

A

Bwd2,

and r being measured from A.

3.292 Flux at any moment of time, 143 IB • 1=

d S1= B

1

R2 2 is the sector angle, enclosed by the ield. where q f Now,

magnitude of induced e.m.f. is given by,

382 d'' gin =

BR2 dcp

- dt

2

BR2

dt

co,

2

where co is the angular velocity of the disc. But as it starts rotating from rest at t = 0 with an angular acceleration 13 its angular velocity w (t)= 13t. So, BR2 2 i(t. According to Lenz law the first half cycle current in the loop is in anticlockwise sense, and in subsequent half cycle it is in clockwise sense. L-

Thus in general,

= (- 1)

The plot gu, (t), where t„ =

nBR 2

ilt, where n in number of half revolutions. 2 1/2 n n/(3 is shown in the answer sheet

3.293 Field, due to the current carrying wire in the region, right to it, is directed into the plane of the paper and its magnitude is given by, [to i B = — – where r is the perpendicular distance from the wire. 2n r As B is same along the length of the rod thus motional e.m.f. 2

=

.

- f (nit) • dr

vB 1

1 and it is directed in the sense of (v B ) So, current (induced) in the loop, i,, t. in

1 go/ vi

R

2nRr

3.294 Field, due to the current carrying wire, at a perpendicular distance x from it is given by, I-to i B (x ) = — 27c x - (173x134.) • d r

Motional e.m.f is given by

There will be no induced e.m.f. in the segments (2) and (4) as, ill d / and magnitude of e.m.f. induced in 1 and 3, will be gi =

-"9 ) a andln=v ( 110 i )a 27t x 27t (a + x) '

v(

respectively, and their sense will be in the direction of ( ill< B ). So, e.m.f., induced in the network = i -

_

a v 1.1,0 i 1

1

[as 11 >

_

v a2 110 t

] •

27c

x a +x

2 7 c x (a + x)

383 3.295 As the rod rotates, an emf. d 1 2 1 — a 0.B=dt2 2

2 B w

(t) - - 1

is induced in it. The net current in the conductor is then

2

2

B

w

R

A magnetic force will then act on the conductor of magnitude direction will be normal to B and the rod and its torque will be

BI

per unit length. Its

a of

i

(t) - -21 a2 B w) dx B x R

0 Obviously both magnetic and mechanical torque acting on the C.M. of the rod must be equal but opposite in sense. Then for equilibrium at constant w 1 2 (t) - a Bw 2 R

or,

2 in

Ba

A mga s

2

2

cot

mg R

1 (2 B2 (.0 + 2 mg R sin w t) sin w t 2 aB 2 aB (The answer given in the book is incorrect dimensionally.) (t) =

1 a2 B w +

3.296 From Lenz's law, the current through the connector is directed form A to B. Here in = vBl between A and B where v is the velocity of the rod at any moment. For the rod, from Fx = mwx or,

mg

sin a -

i 1 B = mw

For steady state, acceleration of the rod must be equal to zero. Hence,

mg

But,

sin a =

i=

Lin

.

R

ilB

(1)

vBl R

in From (1) and (2,) v -mg s

aR B212

3.297 From Lenz's law, the current through the copper bar is directed from 1 to 2 or in other words, the induced crrrent in the circuit is in clockwise sense. Potential difference across the capacitor plates, -dI=

in

or,

q=C L

\VP

384 Hence, the induced current in the loop, dq -C dt

t=

dt

But the variation of magnetic flux through the loop is caused by the movement of the bar. So, the induced e.m.f. gin = B lv d dt

and, Hence,

dv - B1 — = Blw dt

i= C

CB lw

dt

Now, the forces acting on the bars are the weight and the Ampere's force, where = i 1B

Fa mp

(CB 1 w) B= C 12 B2 w.

From Newton's second law, for the rod, Fx = mwx mg sin a - C 12B2 w= mw

Or,

Hence

w=

g sin a

mg sin a

i2B2c

C12B2+m

1+

m

1.298 Flux of B, at an arbitrary moment of time t : na CDt = B•S= B 2 From Faraday's law, induced e.m.f.,

2

cos co t,

=d

(I)

dt 2

-fL- COS 0) t) 2

d

B

— —

7C

dt and induced current,

=

a2 (0 .

t.

Sill

2

R

2co sin co t. a

B

—

2R

Now thermal power, generated in the circuit, at the moment t = t : P(t) = gin

(13

—

x

a2 oo)2 1 2 R

2

sin 0)t

and mean thermal power generated, [B a a2 2

2 WI

T 1

fsin2 co t dt 0

P>=

1 2

2R (n(.0 a2

J

dt

2

Note : The claculation of which can also be checked by using motional emf is correct even though the conductor is not a closed semicircle , for the flux linked to the rectangular part containing the resistance R is not changing. The answer given in the book is off by a factor 1/4.

385 3.299 The flux through the coil changes sign. Initially it is BS per turn. Finally it is — BS per turn. Now if flux is cl. at an intermediate state then the current at that moment will be d (130 —N dt i= R So charge that flows during a sudden turning of the coil is q= f i dt = —Ls-I- polo — (— (1)]. 2 N BS /R R 1 qR Hence, B=— = 0.5 T on putting the values. 2 S 3.300 According to Ohm's law and Faraday's law of induction, the current io appearing in the frame, during its rotation, is determined by the formula, d cl) L d io io' —dt ' dt Hence, the required amount of electricity (charge) is, 1 q= f iodt = -R f (d (I) + L dio) = -R (Acic,+LAio) Since the frame has been stopped after rotation, the current in it vanishes, and hence A io = 0. I I It remains for us to find the increment of the a I lfux A cl) through the frame (A (13= (1:02 — 49. i Let us choose the normal iito the plane of the 1 frame, for instance, so that in the final position, :1 n is directed behind the plane of the figure (along B ). it- a ---)4 0' 0, while in the initial position, • Then it can be easily seen that in the final position, 43102 > ctli a •

;27E7.= —na2 !Joni or, Ev = — gonI a2 /2 r The meaning of minus sign can be deduced from Lenz's law. •

3.309 The e.m.f. induced in the turn is go n I n

d2

4

The resistance is —" p. go n/Sd

— 2 m A, where p is the resistivity of copper. 4p 3.310 The changing magnetic field will induce an e.m.f. in the ring, which is obviously equal, in the two parts by symmetry (the e.m.f. induced by electromagnetic induction does not depend on resistance). The current, that will flow due to this, will be different in the two parts. This will cause an acceleration of charge, leading to the setting up of an electric field E which has opposite sign in the two parts. Thus, So, the current is

—n aE= rI and

/,

1+ naE= '2 where is the total induced e.m.f. From this, 2

l= and

(fi + 1) rI,

— 1) d —

E= 2 a

2 7t a

+ 1

But by Faraday's law,l = n a2 b so,

—1

1 E= — 2

+1

3.311 Go to the rotating frame with an instantaneous angular velocity

53%4 In this frame, a

Coriolis force, 2 m v x (51*(0 acts which must be balanced by the magnetic force, ev --a• xB Thus,

-4"

(t)

w(t)= — 2m

(It is assumed that Zis small and varies slowly, so 0)2 and to. can be neglected.)

390 3.312 The solenoid has an inductance, L . 1.10 n2x b

1,

2

where n = number of turns of the solenoid per unit length. When the solenoid is connected to the source an e.m.f. is set up, which, because of the inductance and resistance, rises slowly, according to the equation, NJ-Li= V This has the well known solution, V t R/L 1 = - (1 - e).

R

Corresponding to this current, an e.m.f. is induced in the ring. Its magnetic field dF 2 B = 1.1,0 n I in the solenoid, produces a force per unit length, — = B i - [to n2 n a2 II/ r

dl

.11 ( 23, n a2 v 2 ( n2 -

r

e - tR/L (1 - e- t R/L ),

RL

acting on each segment of the ring. This force is zero initially and zero for large t. Its maximum value is for some finite t. The maximum value of

e

- t R/L

(1- e - :RA ) = 2

d F. dl

So

1_ 4

V

r

1 is

2

2

2

[to ir a

2

1_ e_ Lila

2

n2 4RL

4 •

2

[to a V 4rR1b2

3.313 The amount of heat generated in the loop during a small time interval dt,

dQ-

/R dt, but, l

d(1) = -— = 2 at - a -r,

dt

(2 at - a -02

d Q -

So,

R

dt

and hence, the amount of heat, generated in the loop during the time interval 0 to T

Q=

f

(

2 at - a T)2

R

1 a2 T3 dt - 3 R

0 3.314 Take an elementary ring of radius r and width dr. The e.m.f. induced in this elementary ring is x r 2 3. Now the conductance of this ring is.

d (1). R

h dr p 2 Itr

so d I -

h r dr ,, 2p P

Integrating we get the total current, b

I -f

hr dr 2p

13 -

h 13 (b2 - a2) 4p

T.

391

3.315 Given L =

110 n2 v . l_to n2 10 nR 2 ,

Thus,

n=

where R is the radius of the solenoid.

•FL---- 1

I-to Ion R •

So, length of the wire required is, ‘1 7aI T = 0.10 km.

1= n1021tR= 1-to 3.316 From the previous problem, we know that, /I =

vL 14 It

length of the wire needed=

, where 1= length of solenoid here.

I-to PPS r Now, R = — (where S = area of crossection of the wire. Also m = p S I') S' Thus, where po = resistivity of copper and p = its density. Equating,

Rm

LI

PPo

Ro/4 n

go m R L= — 4n ppo /

or,

3.317 The current at time t is given by, V 1 (t).. — (1 - e- tRiL ) R V The steady state value is, 10 = i i

1 (0 and or,

/0 t0

= r1= 1 — e- t

1 — R = In

°L 1

3.318 The time constant t

is given

'n

RA

- t.R/L or , e

or, t = °

= 1 - T1 L

R

1 In

1- ri

= 1.49 s

by

L T= —= R

L 10 ' Po s

where, p0 = resistivity, /0= length of the winding wire, S = cross section of the wire. p But m= 1 0 S So eliminating S,t —

L ,

—

mL

PO 10

in/ P lo

PPO 10

392 7111L

From problem 3.315 10 =

(note the interchange of l and 10 because of difference in notation here.) Thus,

i

=

m mL -0.7 ms, 4 - [to 4 n P Po / PP° —L 1

3.319 Between the cables, where a < r < b, the magnetic field H satisfies H 2 7t r = / or,

H9,=

2 r

8P

11, So

B`P

/

2 7t r b

f

The associated flux per unit length is,(13 =

2 it r

r= a

1-to I x 1 x 2 7t dr -

[to/ b In — a

(13

Hence, the inductance per unit length L

-t o

— IM — In Th Where

1= I=

2n

= — a

We get L1 = 0.26 IAH m 3.320 Within the solenoid,1/ -2 n r = NI or (1)

ii-cp = a+b

1-ttio f a dr and the flux, = N41)1, = N — N I 2 it r b

Finally, 3.321

L=

t-to

2

N a ln 1 +

Neglecting end effects the magnetic field B, between the plates, which is mainly parallel to the plates, is B = 120E. (For a derivation so 3.229 b) Thus, the associated flux per unit length of the plates is,

R0,xhxi=

a

NI r

B

CP

NI IA11° 2 r

So,L1 = inductance per unit length =

110

= 25 nH/m.

393 / (r > a). For the double line cable, with current, q) 2 r flowing in opposite directions, in the two conductors,

3.322 For a single current carrying wire,B

Rol

B

c

— between the cables, by superposition. The associated flux is, 3tr

d—a =

f tiol dr x 1 7r r

Rol

d In

=

In rl x /, per unit length

a

Hence,

lnrl

L1=

is the inductance per unit length. 3.323 In a superconductor there is no resistance, Hence, J.,

dI dt —

= +

d (13 dt

So integrating,

M n a2 B = —= L L

because

=

Also, the work done is, A = f I dt =

3.324 In a solenoid, the inductance L =

-

ctof = n a2 B, 43i= 0

014 ) f I dt — - iL I2 — dt 2

µµo n2 V =

Ro

N 2S

4

1 7c2 a

2

B

2

L

,

where S = area of cross section of the solenoid, 1= its length, V = SI, N = nl = total number of turns. When the length of the solenoid is increased, for example, by pulling it, its inductance will decrease. If the current remains unchanged, the flux, linked to the solenoid, will also decrease. An induced e.m.f. will then come- into play, which by Lenz's law will try to oppose the decrease of flux, for example, by increasing the current. In the superconducting state the flux will not change and so, I

I _ -/c• Hence,

/ - /0 '

= constant

or, I= I0 / 10 = 10 (1 + 11)

3.325 The flux linked to the ring can not change on transition to the superconduction state, for reasons, similar to that given above. Thus a current I must be induced in the ring, where,

aaB n a2B '4: I= 0 8a 8a L [to a In 7;,- - 2) µ0 In b -

- 2)

394

3326 We write the equation of the circuit as,

,. L di , nt + — — = q

11 dt ' for t a 0. The current at t = 0 just after inductance is changed, is

i = ri R4, so that the flux through the inductance is unchanged. We look for a solution of the above equation in the form i= A + Be- tic Substituting C= i

R, B= i- 1, A= ?i-

r

i . II (1 + (II _ 0 e- rill t/L )

Thus,

di

3.327 Clearly, L — = R (I - 0 = g - RI

So,

dt di 2L — = g dt

- Ri

This equation has the solution (as in 3.312) i= 1- (1 - e - tR/a)

R

3.328 The equations are, di1 die L1 ii. = L2 ii_

— 1? (4 +

=

i2)

d

Then,

—(L 1 il - LL2212)== 0

1)

or,

L1 it -L2 i2 = constant

But initially at t = 0, it = i2 = 0

27f12

so constant must be zero and at all times, Li ii_ = L e h In the final steady atate, current must obviously be

ill-if

= 4 Thus in steady state, R.

gL1 Q.

ii-22 R (1, 1+ L2) and i2 - R (Li +L2) Poi

3.329 Here, B=

2

5tr at a distance r from the wire. The flux through the frame is obtained as,

a+l

1.12 = f 1

mo I bdr -

2 ar

12 Thus, 1L

,12 =

/

—

pob — I in (1 +1 2n 1

—µ0bb l n__ 2 7t

a (1 + --) /

I

395 b

Rol 3.330 Here also, B =

2 ar

I f h dr xr c1"2 " 2 it r v

and

a

II IAD hN b In — 2n a 3.331 The direct calculation of the flux 402 is a rather complicated problem, since the configuration of the field itself is complicated. However, the application of the reciprocity theorem simplifies the solution of the problem. Indeed, let the same current i flow through loop Thus,

L 12=

2. Then the magnetic flux created by this current through loop 1 can be easily found. Magnetic induction at the centre of the loop, : B —

So, flux throug loop 1,

43 12

[to i 2b

2 tioi 3t a 2b

and from reciprocity theorem, 2• iko a I (1321 14312 = (I)21 2b 4312

So, L12 —

1

2 po n a /b

3.332 Let pm be the magnetic moment of the magnet M. Then the magnetic field due to this magnet is, 3 (rm " zin

r

Pm

r5

r3 *

The flux associated with this, when the magnet is along the axis at a distance x from the centre, is

4). f

5

where,c1,2 = a

and (1,1 =

3 µo Pm x2 f 2 it pd p 4n (x2 + p2)5/2 0

2

PO Pm X 2

(

1 x3

1 (x2 + a2)3/2 2

So,

PoPm a 1, — 2 2 (x + a2)3/2

3

( 1 3

1 —

396 d N d (I) lfows. The — is induced and a current -— — dt R dt total charge q, flowing, as the magnet is removed to infinity from x = 0 is,

When the flux changes, an e.m.f. - N

q= — cl) (x = 0) R NR

IA° P m 2a

2aqR

or, Pm_

N [to

3.333 If a current I flows in one of the coils, the magnetic field at the centre of the other coil is, 1.1,0 a21 po a2! B - 2 02 a2)3/2 = 213

ag 1» a.

The flux associated with the second coil is then approximately 4 1-10 71 a Hence, L12 -

µ0n a4 1/2 13

2 13

3.334 When the current in one of the loop is I1= at, an e.m.f. L12

d11 — - L12 a, is induced in dt

the other loop. Then if the current in the other loop is /2 we must have, d12 L 2 dt + RI2= L12 This familiar equation has the solution, -rR a L12 ( 1- e 2 which is the required current /2 3.335 Initially, after a steady current is set up, the current is flowing as shown. In steady condition i20 = R1.

•, t= 10

1.

•

R0 When the switch is disconnected, the current through R0 changes from i10 to the right, to i20 to the left. (The current in the inductance cannot change suddenly.). We then have the equation,

•

•

This equation has the solution t = /2 e—

t(R +RoYL

The heat dissipated in the coil is, cx)

Q= f 122 R dt = i2

20

Rfe-2t

t (R +120)/La

397 3.336

To find the magnetic field energy we recall that the flux varies linearly with current. Thus, when the flux is cl) for current i, we can write 0 = A i. The total energy inclosed in the field, when the current is I, is d(

W= f

i d t - f N — id t dt

I 2 ni f Ndcloi... f NAidt- -NAI 2 0 The characteristic factor

-

1 iNOI

appears in this way. 2

3.337

We apply circulation theorem, I1-2

nb- NI,

H= N 1/2 70.

or,

Thus the total energy,

W — 1BH -27th • n a2 . It2a2 b BH. Given N, I, b we know H, and can find out B from the B — H curve. Then W can be calculated.

3338

i l- d r= N I,

From fi

B 11-70+ — • b . N I , ( d

» b)

P.O NI

B= µµ 0 H.

Also,

Thus, H -

nd + lib'

Since B is continuous across the gap, B is given by,

N I B = It tio

both in the magnetic and the gap. nd + ith '

B2

Wgap

_

11 2

(a)

Wmagnetic

B

xSxb

.EL)

0 _

2

70 .

x S x otd 2 Kt°

(b)

The flux is

Nf

NI N giA,0 nd 0

ii'd S

+

_

SN

S'

Ro

2 rd ,

b+— R

140 SN2 L ...

So,

ad ' b+—

R Energy wise; total energy

tf, -2 =

wri

:2-1.- +b 22 µ0 ti

S-

1 2

,2 =

µ0N25 b +Lrd

li

i

1 2 - L I 2

398 The L, found in the one way, agrees with that, found in the other way. Note that, in calculating the flux, we do not consider the field in the gap, since it is not linked to the winding. But the total energy includes that of the gap. 3.339 When the cylinder with a linear charge density X rotates with a circular frequency co, a surface current density (charge / length x time) of 1.co t = — iis set up. 2it The direction of the surface current is normal to the plane of paper at Q and the contribution of this current to the magnetic field at P is --4. where e

S._ lio iceX 7) ds 4 it r3

is

Q

the

direction of the current. In magnitude, --).. —3. 'x' FT= r, since e is normal to r and the direction of dB is as shown. It's component, d B0 cancels out by cylindrical symmetry. The component that survives is, --1.

IB_LI= where we have used

1-to f id S Rai —471 ---1- cos 0 = —43t f do = po i , r

d S cos 0

= d Q and f d Q = 4 it, the total solid angle around any

r point. The magnetic field vanishes outside the cylinder by similar argument. The total energy per unit length of the cylinder is, 1 W1 =

1 1102 2 pc,

2 x

21=1-2-

2 it

it a2 =

t

IO a2 x2 0)2

Ba

3.340 WE = - 1 - eoE 2, for the electric field,

1 2 wB - 2 go B for the magnetic field. Thus,

1

2

2µo

B

=

2

1

e E 2 °

'

B

when

E=

‘cFto

3.341 The electric field at P is, EP P

ql 4 TE 80 (a2 + /2 )3l2

8 3 x 10 V/m

399

To get the magnetic field, note that the rotating ring constitutes a current i = q w/2 it, and the corresponding magnetic field at P is, µoa2i

=

WE Thus, Wm

2 (a2

EO PIO E B2

12 )3/2 • 2

ql x 2 e0 RO

)

4 7C eo

a

2

2.

2

1

1 E o tio a2co) Wm

2 4 .2 E0 ILO (k) a /I

or, WE

3342 The total energy of the magnetic field is,

dV =

(T3-

f B (-1113;11—j1"

f B B dV —12.1J • B dV. 2tio The second term can be interpreted as the energy of magnetization, and has the density 1 — — • B. 2J 3343

(a) In series, the current I flows through both coils, and the total e.m.f. induced. when the current changes is,

dI

dI

dt L'= 2L

dt

— 2L — = — L' or,

/ (b) In parallel, the current flowing through either coil is, — and the e.m.f. induced is 2

—L

1

dl

2 dt Equating this. to — L'

, we find L' = 2 L . J

2 3344 We use L1= µo n12 V, L2 = Ito n2 V = L12 µo n1 n2V =VEIT;

So, 3345 The interaction energy is 1

2

r

2

--a, 2

r 2µo f

J ri+B21 ay=

1B 11 dV 1

-f Bi •

--a.

B2 dV

--a.

1

r

2

ill21

dV

Here, if Bi is the magnetic field produced by the first of the current carrying loops, and B2, that of the second one, then the magnetic field due to both the loops will be B1 + B2.

4 0 0

3.346

We

can think

of the

aller

sm

coil

as

constituting

a m

agnet of dipole

moment,

pm- 7t a2/1 I t s d i r e c t i o n i s n o r ma l field,

due

to

t o t h e l o o p a n d ma k e s a n a n g l e 0

the bigger

loop.

This m agnetic

i nt e r a c t i o n e n e r g y h a s

the

3.347

2b

ma g ni t u d e ,

I W I =

Its

is,

Po 12

B2 = The

field

PI)2 Ibi 1" 2 it a2

COS 0

s i g n d e p e n d s o n t h e s e n s e o f t he c u r r e nt s .

(a)

There

the

is

a

radial

inner sphere,

outw

ard

conduction

current.

2 =

dQ

—

the

other hand

A

t the given

_. a5 ( id - at — ddt

Q 4

?) .

moment,

Ohm 's la w,j

=

q

— = P

47c

A

r

q

f

The

surface

He r e

also

'

integral

mu s t b e

It

4

'

-ye

r

, 4 n E0

•I d. dS =

A

q

—

h i=

and

r

oepr2

7* Then,

... i".*

A

E=

-:-. r

by

E

pr "

f dS cos 0 p

20

....s___

r2

EOE P 71'.

b e c a u s e j d , b e i n g o p p o s i t e o f j , i s i n wa r d . —111,

w e

73.

to,div

0,

see

that neglecting

Curl

H=

0,

edge

e ffects, jd =

71. —j.

3.349

Given

I = Im

sin wt. We

If it

is

zero,

B = —.$ B = 0.

constant

ins

sin wt = — jd.

S

D=

I.

7

V

/c

m

Bo

reduce

=

•

can

be

thought

of as

ap at

/m cos wt,

co S

Bo

equations

see that

j= —

or,

Ma x we l l ' s

=

solution of this equation is

e x t r a n e o u s m a g ne t i c f i e l d .

Thus

if ,,,,g.

A

general

dtr*

it r2

4 n eo E r2

and

be the instantaneous charge on

A

4 n r2

q (b)

Q

1

-

J—

or,

dt

On

Let

then,

j x 4 n r

3.348

w ith the direction of the m agnetic

so,

Em m

is Eo w S

the a

mp l i t u d e

of the

electric

field

and is

an

401

3.350

The electric field between the plates can be written as,

V.

V.

E = Re --- et' ',

— cos wt. d

instead of

This gives rise to a conduction current,

a jc= a E=

Re — d

i Vme "

and a displacement current,

OD — 1"

.1e1

V. . eo e

--el

at

The total current is,

a"

jr Q where,

tan a -

on taking the real part of the resultant.

80 8 (t) The corresponding magnetic field is obtained by using circulation theorem,

r = nr2jr V , or, H = H„, cos (wt + a), where, Hm = 712—da VG- + (E0E w)2 H.2 n

3351

Inside the solenoid, there is a magnetic field,

B=

tio n ins sin wt.

Since this varies in time there is an associated electric field. This is obtained by using,

E.(177.=

fB dt

For

r< R,

2

rE

- B. it

r> R, E = -

For

s

r2, or,

BR

dr E_ -

Br 2

2

2r The associated displacement current density is, aE

id=

60-67 a`

B r/2 •• 2 — Eo B R /2 r - eo

The answer, given in the book, is dimensionally incorrect without the factor

3.352

In the non-relativistic limit.

E*=

r 3

43tEor (a)

On a straight line coinciding with the charge path,

aE e° at

—

q 4n

[-V

3rr i r4

(

dr_ using,

E

O•

402

But in this case, r..

r

—v and v —

r=

(b) In this case,r = 0, as,

—I. v, so, Id =

4 it r3

71 t7..Thus, _. id '

3353

2 qr

We have, Ep us

qx 4 it co (a2 + X2)3/2

then

1d =

aE al at

aD — = Eo

at

--1; 4 Tt r

qv

4 n (a2 + x2)-5/2 (a2 — 2x2)

This is maximum, when x = xn, = 0, and minimum at some other value. The maximum displacement current density is

CI d) max 1` To check this we calculate

tiV 3 4na

a dI — ax

a id .... e._ rt._ 4x (a2 + x2) _ 5x ,a2 _ k ax 4 it 1 ‘ This vanishes for x =

0 and for x =

2

2x-- ) I

VT — a. The latter is easily shown to be a smaller 2

local minimum (negative maximum). 3354 We use Maxwell's equations in the form, 111 --Jr __), f E•dS, B•dr= co 1.10 at when the conduction current vanishes at the site. We know that, --1 A: f F. d .i.

_i_f 4 3t co =

dS r r2

V

I—f c/S2 = g 2 at (1 — cos 0), 4 n co 4 n co

where, 231(1 — cos 0) is the solid angle, formed by the disc like surface, at the charge. Thus,

f

ri•c/FL 23taB=2Roq•sin0.0

On the other hand,x = a cot 0 differentiating and using —dx = — v, dt • v 3z a

cosec2 0 0

lio q yr sin 0 Thus,

Ba.

43t?

403

µo4

(v 4 n r3

This can be written as, B -4. xr

and TH=

n 3 3.355 (a) If B = i.(t), then,

(The sense has to be checked independently.)

Curl /-

0 at ' So, gcannot vanish.

(b) Here also, curl E 0, so E cannot be uniform. (c) Suppose for instance, r= 5-7 (t) where W' is spatially and temporally fixed vector. Then -

aB

= curl E

0. Generally

dt ap 0 for in this case the left speaking this contradicts the other equation curl H= at hand side is time independent but RHS. depends on time. The only exception is when f (t) is linear function. Then a uniform field E can be time dependent. ap 3.356 From the equation Curl H --= j at

74'

We get on taking divergence of both sides

a — - div D= div j at

But div D = p and hence div j + 3.357 From

at = 0

V% r._

at we get on taking divergence a --• 0= - divB at

This is compatible with div B = 0 3.358 A rotating magnetic field can be represented by, Bx = Bo cos co t ; By = Bo Sill cot and Bz= Bz. aB -— Then curl, E at So, - (Curl 4 . _0) Bo sin cot = - coBy - (Curl Hence, where,

MM

w

BO

cos

Curl E = - ctrx B , (0 =

e3 CO

wt

MI

coBx and - (Curl A . 0

404

-▪

▪

-

3359 Consider a particle with charge e, moving with velocity 17,.in frame K. It experiences a force F = ev x In the frame K , moving with velocity 17: relative to K, the particle is at rest. This means that there must be an electric field E in lc, so that the particle experinces a force, F'= e E' F- ev00xB Thus,

E'-

3360 Within the plate, there will appear a

17x B (ii*x B) force, which will cause charges inside the

plate to drift, until a countervailing electric field is set up. This electric field is related to B, by E- e B, since v &B are mutually perpendicular, and E is perpendicular to both. The charge density ± a, on the force of the plate, producing this electric field, is given by

3.361 Choose Cir.!. f B along the z-axis, and choose i;.as the cylindrical polar radius vector of a reference point (perpendicular distance from the axis). This point has the velocity, v= wxr, and experiences a (T)xi B) force, which must be counterbalanced by an electric field,

- (01x 11 x

-

B

There must appear a space charge density, - 2 E0 co • B = -8pC/m3

p= eo div E

Since the cylinder, as a whole is electrically neutral, the surface of the cylinder must acquire a positive charge of surface density, 2 E 07.)* • B 0=

+

Tt a 2

2na

eo a (.3) • B

+ -2 pC/m2

3.362 In the reference frame K , moving with the particle, -3. E+

3

7/3/ a if- 17;x ://c2 = 0. Here, vo = velocity of K , relative to the 1 frame, in which the particle has velocity v. Clearly, i,-0).= t7.*From the second equation, B

vxE c2

q

vxr

E°11°x 4 n Eo r3

µo4 (v 47c

r3

405 3.363 Suppose, there is only electric field E, in K. Then in K', considering nonrelativistic velocity

17,* E' = E , B = -

i7x E 2

c

E' • B' = 0

So, In the relativistic case,

41 011 11

.w Ai' c2

- v2/c2

El

-

B'

El • B' = E'

Now,

II. B'II

+ E' B = 0, since 1 '1

E.,. •B,I. _4,_ • (TX Te)/(1 - v2/c2) = - 4 • (7x El ) / ii _ -v,2 A

3.364 In •K, B = b

y

)

A

i -xj 2 x2 + y

b = constant A

In

K',

A r K,E = a 2, r= (x 1+ y

,

by

X2 + r

The electric field is radial (r = x i + y j

:

2

).

)

K, "Tir1

In

A

= Tic73= by Y A

3365 In

.0

7.x E

a

rox

2 C r2

2

The magnetic lines are circular. 3366 In the non relativistic limit, we neglect v2/c2 and write, E__11 Elm El+vxll

E,11

r

_a, II= _j B'1 at -

ric2

These two equations can be combined to give, E'= E+v-10xB, B'= B-vxE/c` 3.367 Choose E M the direction of the z-axis, E =

(0, 0, E). The frame K' is moving with velocity

t7.= (v sin a, 0, v cos a), in the x - z plane. Then in the frame K',

Te r

E'

r eo l i r 1 =

-4°

B'

0

v

E/c` - v2/c2

The vector along T'is e-t

(sin a, 0, cos a) and the perpendicular vector in the x - z plane

is,

f = (- cos a, 0, sin a),

-

406 (a) Thus using

E=

E cos a Wa+ E sin a :f7

E

cos a and

Esina

E'„-

111-v2/c2 So

(b)

V1- (32 cos2 a

E'= E

B'H=

0,

tan a

and tan a' =

1 - 02

V T : 727

TS( Z/c2

-

-v

2 2/C f3 E sin a

B' -

cir=i7 -3. 3.368

Choose

Bin

then in the

the

z

(v sin a, 0, v cos a) in the x -

direction, and the velocity v

K' frame,

1 1

=

II =

0

3T'11-

_1. B

B - 1/2/c2

i_v2/c2 c13B We find similarly,

E' =

sin a

fI 7:131

132 B' = B 3.369

(a)

We see that,

V

E' •13; =

cost

a

tan a tan a' -

1-132

V17-(37

B, vx B)-( -

vx

,E)

v2

1- 2

EH • Bll+

But,

SO,

• 13.1 - (17),C

2

7)•(vxE)/c

2 1- V2iC -4. -4. -4. ri • .81 - (179x /3.1 )2 (17.x )/c2 =E . + l BII l V —2 AxB•CxD= A •CB •D -A •DB • C,

=

4 • 4 + 4..K 2 -rs 1

c2

z

plane,

407

= E2 — c2 B2

1 +

v2

17x E

vB )2 — c2

(Ei

ii

2

c2

1

1—

2

E2

= E —c + ll 2 B11 since, (v

3.370 In this case,

)

2

B2, c2

v2 Al

E•B=

0, as the fields are mutually perpendicular. Also, 2

Y- is — ye.

E2 — c2 B2 = — 20 x log Thus, we can find a frame, in which

/

2 2 B' = — v c B- — E- = B c 2 3.371

Suppose the charge

q

E' =

E2

1—

(

m

0, and

— 0.20

4 X 104

—

3 x 108x 2 x 10

c2 B2

_4

)2 = 0.15 mT

moves in the positive direction of the x—axis of the frame

us go over to the moving frame

K',

at whose origin the charge is at rest.

K. Let

We take the

x and x' axes of the two frames to be coincident, and the y & y' axes, to be parallel.

In

the

K'

frame,

E

—

q r—w

1

" Eo r13

and this has the following components,

qx' E ,

1

Ex=

4 it E0 7/3

Now let us go back to the frame coincide, we take

t=

K

1 4 it

Y

qy' 3

eo r,

*

At the moment, when the origins of the two frames

0. Then, v2

x=

rcos 0 = x'\11

,y=

rsin 0=

E

Also, From these equations,

r'2

= r2

y

=

Ex= Ex', Ey (1 —

v2

p2

/c 2

0) sing

1-132 1 _ /32 sin2 0)3/2

— 4 7C E0 r3

p23/2 () (

67 ..

q 4 3t eo r3 (1

.3/ j A )

x,

1—N2) R2 Sint 0)3/2

32

408

3.7

MOTION OF CHARGED PARTICLES IN ELECTRIC AND MAGNETIC FIELDS

3372

Let the electron leave the negative plate of the capacitor at time t =

A. E. 1.

EX= ..... x dx ,

As,

0

at

1

1'

and, therefore, the acceleration of the electron,

eE eat w=—=—

ml

m V

ml

0

But, from

eat

' dt

ml

t

1 ea

2

or, v — — - - t 2 m/

f dv - ell f tdt,

or,

dv

— am

Or

(1)

0

s = f v dt, 1 (6 m12)3

r 1 ea f 2 m/

1=

eat3

2 t dt =

6

ml

or,

t

ea

0 Putting the value of

t

in (1),

1 (9ale)5- .

2 1 ea 6m12 3 v ... 2 m/ 3.373

ea

16 km/s.

'

2 m

The electric field inside the capacitor varies with time as,

E = at. Hence, electric force on the proton,

F .. eat and subsequently, acceleration of the proton,

eat w= — m Now,

if t is

the time elapsed

during the motion of the proton between the plates, then

/ t = —, as no acceleration is effective in this direction. (Here v vit

is velocity along the length 11

of the plate.) dv1 —w

From kinematics,

dt V1

SO,

dv 1 = f w dt,

f

o (as

r

o

initially, the component of velocity in the direction, 1 to plates, was zero.)

409

r ea t2 ea 'v 1-= f m 2m = 2m v2

or

12

0

Now,

II

eal 2

VI

tan a =

V

2mv

II

1 eal2

-

2m

2e

as v =

3 '

11

2e T m

2 m

al 2 4

3

11

, from energy conservation. /7)

{.77/-

2 eV3

3.374 The equation of motion is, dv dv g_ ,,, -- = v — =dx m clt

kco - ax)

Integrating 1

I.

1

(E x - axe) = constant. 2 1/2 — m °- -2 But initially v = 0 when x = 0, so "constant" = 0 v2= 2m x - 1 ax2)m o 2

Thus,

2E0 Thus,v = 0, again for x = x. =

a

The corresponding acceleration is, (dv) dt

E0 Xm

= -q- (E0 - 2 E0) = m

m

3.375 From the law of relativistic conservation of energy 2 "10

- e Ex = mo c2.

1/1 — (1,2 / C2)

as the electron is at rest (v = 0 for x = 0) initially. Thus clearly

T= e Ex. 2

On the other hand,V1 - (v2 / c2) -

or,

or,

MO C

m0 c2 + e Ex

v V(m 0 c2 + e Ex )2 - m20 cal _ c mo c2 + eEx

ct= f cdt=

2 Ono C +

eEx) clic

NI (mo c2 + eEx)2 — m20 c4

410

f

1

dy

1 , 1,

—

so

2 eE

—

Vy _ mo2 c4

V

eE

Vno c2 +

eEx)2 - mo2 c4 + constant

The "constant" = 0, at t = 0, for x = 0, 1./ ., m 2o c4. . ct - — v (moc2 + eEx) 2

So,

E

Finally, using T = eE x,

3.376 As before, T = e E x Now in linear motion,

d mov dt V 1 - v2 / c2

mow Vi - v2/C

mo (1 _ v2 / c2)3/2 e E m2 c6 So, w=

„ =

(T + m 0c2)3 w = e E, 2 6 MO C

w

_3

eE

(T + moc)-

mow /c2)312 c2 W (1- V2

2 +

1+

m0

T mo c

2

3.377 The equations are,

d( dt I

movx - (v2 c2)

d ( dt

0 and

movy

vo

vx - constant

Hence,

1/1 - (4/ c2)

- v2/c2 Also, by energy conservation,

2

MO c

I1 - (v2 / c2)

I1 - (4/ c2)

vo E0 Dividing

Vx -

E E0 + eEy '

°

al

2

1 - (I? c2) C

7 11 - (v2 / c2) (e0+ e Ey) vy

MO

+eEy

e0+ e E y 2

MO

Also,

Thus,

2

MO

C2

E t + constant

"constant" = 0 as v = 0 at t = 0. Integrating again, e°

y+

1 E y2 -2 e

1 2 2 = - c E t + constant. 2

j

=eE

411

or,

ceEt =

or,

E0

Hence,

(eo

+

2

eEy)2 - Eo

eEy = 11E02 + c2

e2E2t2 C2 e Et

vo E0

=

also, v =

_2 tO

and

_ e2

2 E2 r 2

tan 0 =

vx

co2 +c

e E t 111 mo vo

Y e

V 2 222 E t

/ c2) .

3.378 From the figure, d

sin a =

dqB -

R

As radius of the

m

'

v arc R - m , where qB

v is the

velocity of the particle, when it enteres into the field. From initial condition of the problem, 1 qV = — mv2 2

dqB

Hence, sin a =

3.379

dB

2mV

FF11-:

m

and a = sin-

2qV

or, v = v

m

1 dB

= 30°,

VIZ

2 mV

on putting the values.

(a) For motion along a circle, the magnetic force acted on the particle, will provide the centripetal force, necessary for its circular motion. 2 my

i.e.

R

=

and the period of revolution,T =

evB

or,

eBR v-

2n = 2 ;IR w v

2 sc m eB

(b) Generally, dt = -0 But,

dt

-

d

mo

dt

I 11 1 - (V2/C2)

2-•

v • 174)

mo

mo

3/2 - (112 / C2)

(1

(V2 / C2))

For transverse motion, 1,: i-,= 0 so,

di* dt

NT. V i - (v2 1 c2)

mo - (v2 / c2)

v2

, here.

2 C

412

V2 MO V

Thus,

- Bev or,

r 1I I. - (v2 / c2) Ber

V_

or,

.,2

2

C Va,r- e 2

2

r- + mo c

It //to 2-7t V

2 nr T= v

Finally,,

2

2

2

B e r2 +mo c

- cBe

eB111-v2/c2

2

3.380 (a) As before,p = B qr. Ve2p2

(b) T =

+ m20 c4 = Vc2 B2 q2 r2 + m

c4 0

V

(c) w -

2

r

—

C

2

r [1 + (mo c I B qr)2 1

using the result for v from the previous problem. 3.381 From (3.279), T=

2n

E

(relativistic) ,,

c` eB Here,

moc2 /111-v2/c2 = E 23IT

Thus,

oT= c2 eB OT T 1= _____T , To mo c

Now, 3.382

T = eV =

1

—mv

, (T= K.E.) so, T = i mo c2

2

2 (The given potential difference is not large enough to cause significant deviations from the nonrelativistic formula). Thus,

So,

V=

VIeV m

vII =

VV — m

cos a, vi =

2 m v1

Now,

MV.L.

= Bev I or, r - Be '

r ,r

and Pitch

i

„, p ... v

i II

2nr 2nm = — v1 Be

2 7CM \Fi r ( ■ — — Be

m \

,, cos a .. L n

2M V — cos a

eB2

2eV m

sin a

413 3383 The charged particles will traverse a helical trajectory and will be focussed on the axis after traversing a number of turms. Thus

So,

1 vo

2 7cm n. — 4B1

n B

n +1 B2

2 am (n+ 1) 4B2 1 B2

- B1

Hence,

or,

or,

m

2 (B2 _ B1)2

3.384 Let us take the point A as the origin 0 and the axis of the solenoid as z-axis. At an arbitrary moment of time let us resolve the velocity of electron into its two rectangular components,v along the axis and )73 to the axis of solenoid. We know the magnetic force does no work, so the kinetic energy as well as the speed of the electron I vl I will remain constant in the x-y plane. Thus v—ig. can change only its direction as shown in the —3. Fig.. vii will remain constant as it is parallel to B. Thus at t = t vx = vl cos wt = v sin a cos wt, = v sin cot = v sin a sin co t and

vz

= v cos a ,

w

V,

here w = eB m

As at t = 0, we have x = y = z = 0, so the motion law of the electron is. z= v cos a t v sin a x— sin wt y=

v sin a (cos co t — 1) co

(The equation of the helix) z= 1, so t —

On the screen,

Then,

1 v cos a

) 2 v2 sin2 a ( )/ 1 — cos a 2 V cos a (0 leB co/ 2 v sin a . sm =2 sm a sin r= a) 2 y cos a 2 my cos a eB r 2 =X 2 + y

2

-

414

3.385 Choose the wire along the z—axis, and the initial direction of the electron, along the x—axis. Then the magnetic field in the x — z plane is along the y—axis and outside the wire it is,

I B = By= —2 7rx , (Bx = Bz=

0, if y =

0)

The motion must be confined to the x — z plane. Then the equations of motion are,

eV B

mvx = d(mvz)

- + evx By dt

Multiplying the first equation by vx and the second by vz and then adding, dvx v

dvz +v

x dt

=

0

x dt

or, dvx vx dx =

Then,

e — vo- vx

vxdvx

Role dx

or, 11--2

2 7T1ri X

7

Vo — vx

V

Integrating,

2 7TX

Role

72 2

vo — vx -

2

x

ion In —a

on using,vx = vo , if x = a (i.e. initially). vx = 0, when x = x„„

Now,

SO,

Xm = a e 0

Role

v

/b , where

b -2

3.386 Inside the capacitor, the electric field follows a 1law, and so the potential can be written as r V ln

r1a

-V

9)

1

E lnbla

'

ln blar

Here r is the distance from the axis of the capacitor. my2 — r

Also,

qV

1

l n b l a r r

2

or my =

gy_r_

ln

b1a

On the other hand, my - q B r in

Thus,

V v Brinbla

the magnetic field. and

m

—

V Br

B2r21n(b/a)

415

3.387 The equations of motion are, d dv dv, vz m — . - q Bvz , m —c . qE and m — = q vx B dt -t dt These equations can be solved easily. qE qE v = — t, y = First, — te2 y 2m vx2 + vz2 = constant = vo as before.

Then,

In fact, vx = vo cos cot and vz = vo sin wt as one can check. Integrating again and using x = z = 0, at t = 0 vo . vo ‘. x = — sin wt, z = — ki - cos cot) 03

CO

2x

Thus,

x = z = 0 for t = tn= n —(0 _ qE x

2n xn

2

27t

_ 2 n2mEn2

x 2

vx tan an = — , (vz = 0 at this moment) vy

Also,

-

mvo

B vo mvo B 1 x x 2nEn• 2nn qE m

-

qEtn

3.388 The equation of the trajectory is, vo vo x = —cosin cot, z = T() (1 - cos cot) , y =

qE — t 2 as before see (3.384). 2in

Now on the screen x = 1, so col

Sill (Or mil

— vo

or,

1_,. (Or go Sin

qE

(

--2- SM

VT„:17

ov vc,

sin

qE V

and

Z -

= / tan For small

or,

_ 1 col )

—co o

2

vo

2 m co

so,

—

vo

At that moment, Y=

1 col

,, wt

11;;T, - sin

Em (Ot

2 sin ' — = / tan — 2 2

1 [sin-1 1 2 vo

= / tan .V7g) .32 2 mE

416

3.389 In crossed field, E B

eE = evB, so v = Then,F = force exerted on the plate =

I

xm e

E

mIE = B

eB

3390 When the electric field is switched off, the path followed by the particle will be helical. is the velocity of the particle, parallel to B, and T, the and pitch, Al = v11 T, (where v ll time period of revolution.) = v c,os (90 - cp) T= v sin cp T = v sin cp 2 nni as T= 2::,) (1) qB DPP Now, when both the fields were present, qE = qvB sin (90 - q)), as no net force was effective on the system. E (2) or, vB cos cp From (1) and (2),

E 2 3tM Al a, 6 cm. tan Lu = B ---qB(P

3.391 When there is no deviation, - qE = q (i7X B) E B

or, in scalar from, E = vB (as 171B ) or, v =

(1)

Now, when the magnetic field is switched on, let the deviation in the field be x. Then, 1 (I_ 21

2

X -

t

2

,

m

where t is the time required to pass through this region. a t- v

also,

2 1 (qvB )(1 .

Thus X=

For the region where

2

m

1 _q_

v

2m

a2 B2 E

the field is absent, velocity in upward direction = r-)t= m

5- a B m

Ax _ x . Tip_ t,

Now,

m „ aB2 b b = -2when t' .: m E v

bB E

From (2) and (4), Ax_ 1 1 2m

or,

g_. m

a2 T,2 Li E

_q_a B2 b = In

2E Ax a B2 (a + 2b)

E

(4)

--▪ 3.392

417

-▪

(a) The equation of motion is, d2

—

Pri

q(E+v x131

dt2 71 k . v x B= x y z 0 0 B

Now,

By-jBx

So, the equation becomes, dv, dt

=

q B v„ ' m

Here, vx = x , vy = 57,

gE qB =

dt

m

dv vx , and

m

dt

=0

i. The last equation is easy to integrate; v = constant = 0,

since vx is zero initially. Thus integrating again,

z= constant = 0, and motion is confined to the x - y plane. We now multiply the second equation by i and add to the first equation. sot V + x

y

we get the equation, = to)

dt

E

. - twg,

=

qB m.

This equation after being multiplied by eiwcan be rewritten as, d — (ge"")= iwel cot — dt and integrated at once to give,

where C and a are two real constants. Taking real and imaginary parts. v =

+ C cos (0) t a) and v

= - C sin (cot +

x

Since vy = 0, when t = 0, we can take a = 0, then vx = 0 at t = and we get, E , E = — (1 - cos wt) and v = — sin cot. vx B Y B Integrating again and using x = y .. 0, at t = 0, we get E( x (t) = — t - s' B This is the equation of a cycloid. (b) The velocity is zero, when cot =

„ E „ , y (t) = — k.i. - cos cot). B 2 n n. We see that 2

v2=

v2 + v2 . x

Y

E(

(2 - 2 cos wt)

0 gives, C =

418

or,

ds

v=

2E

=

dt

B

wt sin 2

The quantity inside the modulus is positive for 0 < cot < 2 n. Thus we can drop the modulus and write for the distance traversed between two successive zeroes of velocity.

S

4E — coB

=

,

cot) - cos — 2

Putting cot = 2 n, we get

(c)

8 mE

8E

S

=

coB

4

/32

The drift velocity is in the x-direction and has the magnitude,

< vx> = C E— (1 - cos cot)> = B.

3.393

When a current

I flows along the axis, a magnetic field B., -

Ro I 2n p

2 2 p mi X +

ii s

set up where

y2. In terms of components, lx Rb'

Bx=

,

2 n p" Suppose a p.d.

n

D„=

—4- and Bz = 0 2 n p`

V is set up between the inner cathode and the outer anode. This means a

potential function of the form In p/b

cp = V

In a/b

' a>p>

b,

as one can check by solving Laplace equation. The electric field corresponding to this is,

tx

Vx

Ex =

p2 ln

, E, = -

2 't/I

aIb

p ln

'11

, Ez = 0.

a 1 b

6

The equations of motion are,

I e IVz

d

— mv

dt

+

I e I go

p Inalb + e Ivy

dt nivY and

2np

2 xz

I eliA0I

yz

p`' ln a 1 b+ 2 7t p2 Imvz = -lel

RO I

[to 1 2 (X

+ y;)

z It p (- I

-

I e I

e l) is the charge on the electron.

Integrating the last equation,

mvz-

-(el

Rol — in p / a 2n

mi.

2n Ida P

419 since vz .

0 where p - a. We now substitute this i in the other two equations to get 15_1 ( 1 mv2 + 1 nwl

dt . [

leIV

2

1

lnalb

x

2

Y

el2 osoi)2 inp/b1

m

P

leIV_Iel2r/2 m

leIV_Iel2 m

0 ln

2x

In 16b

a In — b

1-i+2361

23r

(1 2

1 d 2 - - - 1 1 P 2p

b

t

ink clink

b dt

b

2 it

Integrating and using v2 = 1 2 — Mv = 2

=

0, at p = b, eIV I a In — b

we get, 2

lnf)--sln lel 2

Hii°11)

The RHS must be positive, for all a > p > b. The condition for this- is, 2 1 I I (tto 1

V

2——

2m

—

2n

a in

b

3.394 This differs from the previous problem in (a 4-,b) and the magnetic field is along the z—direction. Thus Bx = By = 0, Bz = B Assuming as usual the charge of the electron to be — I e I , we write the equation of motion d mv . dt x

and

lelVx p

2

b In — a

f

-IeIBy,

m

vz .0 = > z = 0

d

leIV mv = --L +I elB; Y b dt p2 ln ;

The motion is confined to the plane z = 0. Eliminating B d (1 2). di riv

—1 y2m Or,

2

lejvxi+h ln b I a

p2

leivi jtpzi =

I

ln

b/a

so, as expected, since magnetic forces do not work, v=

, at p = b.

from the first

two equations,

-

420

On the other hand, eliminating

V,

we also get,

d — m(xv —yv )= leIB(x.i+yj?) dt Y x ielB

(xvy — Yvx)

i.e.

2 p

+ constant

2m a. Thus,

The constant is easily evaluated, since v is zero at p

(xv

e IB 2m

Y — Yv1)

At

p

Thus,

b,

( p2 — a2)> 0

(xvy— yvx) s vb

(b2 — a2)

vb a

2 mb or,

B s b2 — a 2b

or,

{2771 B

B s b2—a2

3.395

lel

The equations are as in 3.392.

g

dvx

dv

qE

qp_ cos wt — m

v dt

m

v x and

dvz dt

dt

211 m ,

co

with

dr.i

vx + ivy, we get,

— co cos oat — i B

au

or multiplying by eiwt, d • — (getw ) dt

icor or integrating,

e

2i (e

+ 1)

2B

E 471

(eiw

or,

Ens i

2 iag e

ne + — iwt 2B

+2icotei'`)+C

eitg

4B

since

Thus,

a ( s inn cot — cot cos wt), y 2 k 2 co a

x

t sin cot.

421

qE . where a =

, and we have used x. y

0, at t

0.

The trajectory is an unwinding spiral. 3396 We know that for a charged particle (proton) in a magnetic field, mv2

Bev or my Ber

r

eB —, m

But,

E. 1 2 1 — mv = — m 2 2

Thus So,

2 r2

AE= mco2 r Ar = 4 n2 v2 mr At-

On the other hand AE = 2 eV, where V is the effective acceleration voltage, across the Dees, there being two crossings per revolution. So, V a 2 n2 v2 mr bole 2

3.397

my — Bev, or, my = Ber

(a) From — r and

(Ber)2

T

2n (b) From -w

2m

1 my 2— 12 MeV 2

r v 1 ..V-r 2 2m

we get,

3.398

—

f z'min nr mr r (a) The total time of acceleration is,

—

15 MHz

1

n, 2v where n is the number of passages of the Dees. 2 B e2 r 2 But, T = neV — 2m t=

So, (b)

But,

B2 e 2mV

n

or,

t

= eB 1 m

x

B2 er2

n B r2

n2 my r2

2mV

2V

eV

The distance covered is, s =

Va =

2

, 21;

12 7 -

' v ,,

-

30 µs

422

But,

n=

Thus,

s 2 n r„

3.399 In the nth orbit,

vn

2 B e2 r2 2 eV m =

4

2 it2 M V2 r2 eV

mr2

n3 V2

3eV

= 1.24 km

n — n To = — • We ignore the rest mass of the electron and write

vn= c. Also IV= cp = cBer„. 2 7tW.. n

Thus,

Bec2 2 n Wv

or,

n

9

1 .1

Bec2

3.400 The basic condition is the relativistic equation, 2 MV

r

— Bqr.

— v2/c2

Or calling,

0)=

m ' co = Bq r o

030

—

we get,

mv

or, my —

Bqv,

03(2). r2 V 1 +

m°

2

is the radius of the instantaneous orbit. The time of acceleration is,

1 = 2vn

t

N

N

N

n-1

= NI

Wn q Bc2

n n-1

N is the number of crossing of either Dee. But, Tvn = mo

c2

n AW +— 2 , there being two crossings of the Dees per revolution.

So,

Also,

——

r=r WN

naN

N 2q Bc

423

Hence fmally,

wo

w =

V

i 4. (72 B2 x m20 c2

N2

WW2 4 q2 B2 c2

wo wo

_

N i i ÷ AAIL)2.

ViTiTi t '

4 qBc2

4 ni20 cal X

a Aw t q B AW

a= 3.401

n mo 2 c2

When the magnetic field is being set up in the solenoid, and electric field will be induced in it, this will accelerate the charged particle. If

B

is the rate, at which the magnetic field

is increasing, then. Tt r2 i si. 2

1

nrE

or

E. - rB

•

2

dv Thus,

m dt

1 • - - Bq or = 2r ' V

qBr 22 mm '

After the field is set up, the particle will execute a circular motion of radius p, where 1

my = B q p, 3.402

or p = i

The increment in energy per revolution is

e 4),

r

so the number of revolutions is,

N=W e 4:b The distance traversed is, s = 2 nrN 3.403

On the one hand, r

ar_ . eE . dt

e

AI .

e d — — f 27.tr' B (r) dr' 2nr dt 0

2nr dt

On the other ,

p = B (r) er, r = constant. • dp d = er — B (r) ■ er B (r) dt dt • er B (r) - e n r2 d < B > 2 ;tr. dt

SO,

Hence,

i3 (r) = -Ili< B >

So,

This equations is most easily satisfied by taking

B (r0) -

1 - < B >. 2

r 0

3.404 T he condition,

1

B (r0) =

1

> = f 13• 2nr dr/nro 2 = — (ro) .. < dt At 2.13 11 < B > = dt At

A). nr2d. 2icr2B dt dt At

So, energy increment per revolution is, d(1) 2 7t r2eB e dt 12 At 3.407 (a) Even in the relativistic case, we know that : p = Ber 14, = Nic2p2 + m20 c4 c2 = mo 2 - / Thus, m0 C ( V1 -1- (Ber I m0c)2 -1) (b) The distance traversed is, W 2nr -e cf).= 2.7tr

W

= 27tr2eB / At on using the result of the previous problem.

W At Ber '