Principles of Electromagnetics 6th Edition(Asian Edition) Solutions
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
INSTRUCTOR’S SOLUTIONS MANUAL FOR
PRINCIPLES OF ELEMENTS OF ELECTROMAGNETICS ASIAN EDITION INTERNATIONAL SIXTH EDITION
Matthew N. O. Sadiku Prairie View A&M University Sudarshan R. Nelatury Pennsylvania State University
S.V. Kulkarni IITYork Bombay New Oxford Oxford University Press
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
Copyright © 2015 by Oxford University Press POESM_Ch01.indd 2
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
1 CHAPTER 1 P. E. 1.1 A + B = (1,0,3) + (5,2,−6 ) = (6,2,−3) (a) A + B = 36 + 4 + 9 = 7 (b)
5 A − B = (5,0,15) − (5,2,−6) = (0,−2,21)
(c)
The component of A along ay is
(d)
Ay = 0
3 A + B = (3,0,9 ) + (5,2,−6 ) = (8,2,3) A unit vector parallel to this vector is (8,2,3) a11 = 64 + 4 + 9 = ±(0.9117a x + 0.2279a y + 0.3419a z )
P. E. 1.2 (a) rp = a x − 3a y + 5a z rR = 3a y + 8a z
(b)
The distance vector is rQR = rR − rQ = (0,3,8) − (2, 4, 6) = −2a x − a y + 2a z
(c)
The distance between Q and R is | rQR |= 4 + 1 + 4 = 3
P. E. 1.3
Consider the figure shown on the next page: 40 uZ = uP + uW = −350a x + ( −a x + a y ) 2 = −378.28a x + 28.28a y km/hr or
uz = 379.3∠175.72 km/hr Where up = velocity of the airplane in the absence of wind uw = wind velocity uz = observed velocity
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
22
N
y up
x
uW
uz
W
E
S P. E. 1.4 Using the dot product, A B −13 13 = =− AB 50 10 65 θ AB = 120.66
cos θ AB =
P. E. 1.5
(a) E F = ( E ⋅ a F )a F =
(E ⋅ F )F F
2
=
− 10(4,−10,5) 141
= − 0.2837a x + 0.7092a y − 0.3546a z ax (b) E × F = 0 4
a y az 3 4 = (55,16,−12 ) − 10 5
a E ×F = ± (0.9398,0.2734,−0.205)
P. E. 1.6
a + b + c = 0 showing that a, b, and c form the sides of a triangle.
a ⋅ b = 0, hence it is a right angle triangle.
1 1 1 a×b = b×c = c×a 2 2 2 1 1 4 0 −1 1 a×b = = (3,−17,12) 2 21 3 4 2 Area =
Area =
1 9 + 289 + 144 = 10.51 2
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
3 P. E. 1.7
( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 ) 2
(a) P1 P2 =
2
2
= 25 + 4 + 64 = 9.644
(
(b) rP = rP1 + λ rP2 − rP1
)
= (1,2,−3) + λ (− 5,−2,8) = (1 − 5λ ,2 − 2λ ,−3 + 8λ ).
(c) The shortest distance is d = P1 P3 sin θ = P1 P3 × a P1P2 1
=
−3 5
6
93 − 5 − 2 8 1 (− 14,−73,−27 ) = 8.2 = 93 P.E. 1.8 1.3 a) A – 3B Prob.1.1 r = 4a= 4a − 5xa– 2a + ay + 6az – 3(12ax + 18ay – 8az) OP
x
y
z
56a = y + 30az r –32ax –(4, −5,1) arOP = OP = = 0.6172a x − 0.7715a y + 0.1543a z b) | (2A r | + 5B)/|B| (16 + 25 + 1) OP
[2(4ax – 2ay + 6az) + 5(12ax + 18ay – 8az)] = 2 Prob. 1.2 (12 + 182 + 82)1/2 r = (−3, 2, 2) − (2, 4, 4) = (−5, −2, −2) 68ax + 86ay – 28az r = (−5, −2, −2) ar = = = −0.8704a x − 0.3482a y − 0.3482a z 23.06 r 25 + 4 + 4 = 2.94ax + 3.72ay – 1.214az c) 1.3 ax × A Prob. rMN = rN −axrM× =(4a (3,5, = 2a1)y −+(1, 6a−z)4, −2) = 2a x + 9a y + a z x–−
= 4(ax × ax) – 2(ax × ay) + 6(ax × az) Prob. 1.4 = 0 – 2az – 6ay = –6ay – 2az A − 2 B× =a (4, −6,3) − 2(−1,8,5) = (4, −6,3) − ( −2,16,10) d) (B x) ⋅ ay (a) −22, −7) ((12a =+(6, 18a – 8a ) × a ) ⋅ a x
y
z
x
y
(12(ax × ax) + 18(ay × ax) – 8(az × ax)) ⋅ ay (b) A B = (4, −6,3)(−1,8,5) = −4 − 48 + 15 = −37 (0 – 18az – 8ay) ⋅ ay = 0 – 8 = –8
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−2 4 4 3 −462−6 1 C = =A × B= = (−30, −18,12) Let Q P × (c) as expected. 2 −1 2 −20= ( −54,12, −10 ) Sadiku & Kulkarni Principles of Electromagnetics, 6e −30, − PC⋅ R= =±( −(4,12, −18,12) = −4 a y + 0.3244a z ) 10 ) ⋅ ( −1,1, 2 )(−=0.8111 4 + 12a−x 20 a⊥ =Q±×1.14 =± − 0.4867 Prob. 2 2 2 4 |C | + + 30 18 12 (a) Using the fact that −1 1 2 ( A × B ) × C = ( A ⋅ C )B − (B ⋅ C )A, or Q × P ⋅ R = R ⋅ Q × P = 4 3 2 = − ( −6 + 2 ) − ( −8 − 4 ) + 2 ( −4 − 6 ) = −4 get P.E.we 1.9 Prob. 1.12 2 −1 −2 7 (Q )×=−R6,5) )5A=−7−)(13 A× ×PQ A) × B(2, −)(=A(×(0,3,1) × A==0)(− A ⋅ A+)B120 + 70 = 206 (d) ( P 4,B−)12,10 4,⋅ A ⋅=( Q ⋅B −+10, = 16 (18
(b) A × ( A × ( A ×2B )−) 6= A45× −( 12 A ⋅ B10 ) A- ( A+ ⋅6Aa ) B y a +z 12a + 8a (e) ( P ×PQ×)Q × (=Q0× R3) = 1 = −21a x - 2a = 16 x y z = ( 4A ⋅ B−)10 - ( A7× A) − ( A ⋅ A) - ( A × B ) P ⋅Q 13 2− 2A−×1 −B=4) )−0.51 −7⎯⎯ cos θ P=⋅ R = =− A( − → θ PQ = 120.66o ( (f) cos θ PR =PQ = = = −0.9526 PQ 10 65 since AxA = 0P R 4 +1+ 4 1+1+ 4 3 6
Prob. 1.13 θ PR = 162.3 P.E. Prob.1.10 1.15 If A and B are parallel, then B = kA and A x B = 0. It is evident that k = -2 and that 1 1 4 1 −5 1 Area = a x| D a×yE |=a z = | (3 + 10)a x + (5 − 12)a y + (8 + 1)a z | P × Q 16 +21443+ 100 − 1 2 2 2 = 260 3 = 0.998 (g) sin=θ PQ A× B 1 = −2 3 = = 0 P Q 31 16 + 9 + 4 3 29 1 4 −7,9) − 6 |= 169 + 49 + 81 = 8.646 = −2| (13, 2 = 86.45 2 P. E. 1.7θ PQ as expected. 2 2 2 (a) P1 P2 = ( x2 − x1 ) + ( y2 − y1 ) + ( z2 − z1 ) Prob.1.11 1.11 P.E. Prob. 1.14 = the 25fact + 4 +that 64 = 9.644 (a) Using A B = (4, −6,1)(2, 0,5) = 8 − 0 + 5 = 13 ( A × B ) × C = ( A ⋅ C )B − (B ⋅ C )A, rP1 (b) (a) r|P B=|2r=P1 2+2 λ+ 5rP22 =−29 we get (1,2,−3B))|+2==λ−13 (−( A+5,×2−2B×,829 AA=B ) ×) A= 71 × (+A2×| B = (B ⋅ A)A − ( A ⋅ A)B = (1 − 5λ ,2 − 2λ ,−3 + 8λ ). (b) A × ( A × ( A × B ) ) = A × ( A ⋅ B ) A- ( A ⋅ A) B (b) (c) The shortest distance is A× B A ⋅PB )×-a( A × A) − ( A ⋅ A) - ( A × B ) a⊥ = ± d = P1 P3 sin θ ==( P 1 3 P1 P2 | A× B | = − A2 ( A × B ) 1 6 4 −−63 15 ==0 since Ax=A A ×93 B =− 5 − 2 8 = (−30, −18,12) Let C 2 0 5 1 Prob. 1.15 C= 30,,−−73 18,12) ((−14 ,−27 ) = 8.2 a⊥ = ± 1 = 93 ± 4 1 −5= ±(1−0.8111a x − 0.4867a y + 0.3244a z ) 1 2 2 Area = | C || D × E30 |= + 18 + 122 = | (3 + 10)a x + (5 − 12)a y + (8 + 1)a z | 2 2 −1 2 3 2
(
)
1 1 169 + 49 + 81 = 8.646 = | (13, −7,9) |= Prob. 1.12 2 Prob.1.1 2 P Q = (2, −6,5)(0,3,1) = 0 − 18 + 5 = −13 rOP = 4a x − 5a y + a z arOP =
rOP (4, −5,1) = = 0.6172a x − 0.7715a y + 0.1543a z | rOP | (16 + 25 + 1)
Prob. 1.2 r = (−3, 2, 2) − (2, 4, 4) = (−5, −2, −2) r (−5, −2, −2) ar = = = −0.8704a x − 0.3482a y − 0.3482a z r 25 + 4 + 4 Prob. 1.3 rMN = rN − rM = (3,5, −1) − (1, −4, −2) = 2a x + 9a y + a z
POESM_Ch01.indd 4
Copyright © 2015 by Oxford University Press Prob. 1.4 A − 2 B = (4, −6,3) − 2(−1,8,5) = (4, −6,3) − ( −2,16,10) (a)
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ar = Sadiku & Kulkarni
r (−5, −2, −2) = = −0.8704a x − 0.3482a y − 0.3482a z r 25 + 4 + 4
Principles of Electromagnetics, 6e
Prob. 1.3 5 rMN = rN − rM = (3,5, −1) − (1, −4, −2) = 2a x + 9a y + a z
Prob. 1.3 1.4 A − 2 B = (4, −6,3) − 2(−1,8,5) = (4, −6,3) − ( −2,16,10) (a) = (6, −22, −7)
4 (b) A B = (4, −6,3)(−1,8,5) = −4 − 48 + 15 = −37 (c)
A× B =
4
−6 3
−1
8
5
= (−30 − 24)a x + (−3 − 20)a y + (32 − 6)a z
= −54a x −23a y + 26a z
1.5 Prob. 1.4 3 5 1 = (−35 − 1)a x + (0 + 21)a y + (3 − 0)a z B×C = 0 1 −7 = −36a x + 21a y + 3a z A( B × C ) = (4, 2,1)(−36, 21,3) = −144 + 42 + 3 = −99
Prob. 1.6 1.5
(a)
B×C =
1 1 1 0 1 2
= a x − 2a y + a z
A( B × C ) = (1, 0, −1)(1, −2,1) = 1 + 0 − 1 = 0
(b)
A× B =
1 0 −1 1 1
1
= a x − 2a y + a z
( A × B )C = (1, −2,1)(0,1, 2) = 0 − 2 + 2 = 0
(c) A × ( B × C ) =
1 0 −1 = −2a x − 2a y − 2a z 1 −2 1
(d) ( A × B ) × C =
1 −2 1 = −5a x − 2a y + a z 0 1 2
Prob.1.7 Prob. 1.6 (a) T = (3, -2, 1) and S = (4, 6, 2)
(b) rTS = rs – rt = (4, 6, 2) – (3, -2, 1) = ax + 8ay + az (c) distance = |rTS| = 1 + 64 + 1 = 8.124 m Prob. 1.8 (a) If A and B are parallel, A=kB, where k is a constant.
(α ,3, −2) = k (4, β ,8) Equating coefficients gives
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Prob.1.7 (a) T = (3, -2, 1) and S = (4, 6, 2) Sadiku & Kulkarni
Principles of Electromagnetics, 6e
(b) rTS = rs – rt = (4, 6, 2) – (3, -2, 1) = ax + 8ay + az 6 (c) distance = |rTS| = 1 + 64 + 1 = 8.124 m
Prob. 1.8 1.7 (a) If A and B are parallel, A=kB, where k is a constant.
(α ,3, −2) = k (4, β ,8) Equating coefficients gives −2 = 8k
⎯⎯ →
5 k =−
1 4
α = 4k = −1 3 = βk ⎯⎯ → β = 3 / k = −12 This can also be solved using A X B = 0. (b) If A and B are perpendicular to each other, A• B = 0 ⎯⎯ → 4α + 3β − 16 = 0 Prob. 1.9 1.8
(a) A ⋅ B = AB cos θ AB A × B = ABsin θ AB an
( A⋅ B)
2
+ A × B = ( AB ) ( cos 2 θ AB + sin 2 θ AB ) = ( AB ) 2
2
2
(b) a x ⋅ (a y × a z ) = a x ⋅ a x = 1. Hence, a y × az a x ⋅ a y × az
=
ax = ax 1
ay az × a x = = ay 1 a x ⋅ a y × az ax × ay a x ⋅ a y × az
=
az = az 1
1.9 Prob. 1.10
(a) P + Q = ( 6, 2, 0 ) , P + Q − R = ( 7,1, −2 ) P + Q − R = 49 + 1 + 4 = 54 = 7.3485
2 −1 −2 (b) P .Q × R = 4 3 2 = 2 ( 6 − 2 ) + ( 8 + 2 ) − 2 ( 4 + 3) = 8 + 10 − 14 = 4 −1 1 2 Q×R =
4 3 2 = ( 4, −10, 7 ) −1 1 2
P .Q × R = ( 2, −1, −2 ) ⋅ ( 4, −10, 7 ) = 8 + 10 − 14 = 4
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
Prob. 1.10
(a) P + Q = ( 6, 2, 0 ) , P + Q − R = ( 7,1, −2 )
7
P + Q − R = 49 + 1 + 4 = 54 = 7.3485
2 −1 −2 (b) P .Q × R = 4 3 2 = 2 ( 6 − 2 ) + ( 8 + 2 ) − 2 ( 4 + 3) = 8 + 10 − 14 = 4 −1 1 2 Q×R =
4 3 2 = ( 4, −10, 7 ) −1 1 2
P .Q × R = ( 2, −1, −2 ) ⋅ ( 4, −10, 7 ) = 8 + 10 −614 = 4 (c) Q × P =
4 3 2 = ( −4,12, −10 ) 2 −1 −2
Q × P ⋅ R = ( −4,12, −10 ) ⋅ ( −1,1, 2 ) = 4 + 12 − 20 = −4 or
−1 Q × P ⋅ R = R ⋅Q × P = 4 2
1 2 3 2 = − ( −6 + 2 ) − ( − 8 − 4 ) + 2 ( − 4 − 6 ) = − 4 −1 −2
(d) ( P × Q ) ⋅ ( Q × R ) = ( 4, −12,10 ) ⋅ ( 4, −10, 7 ) = 16 + 120 + 70 = 206 (e) ( P × Q ) × ( Q × R ) =
4 −12 10 = 16ax + 12a y + 8az 4 −10 7
( −2 − 1 − 4 ) = −7 = −0.9526 P⋅R = P R 4 +1+ 4 1+1+ 4 3 6 7 θ PR = 162.3
(f) cos θ PR =
5 + 144 + 100 P ×2Q −6 16 a z = 0.998 = −21a x - 2a=y + 6260 (g) sin θPPQ× Q = = = 0 3 1 P Q 3 16 + 9 + 4 3 29 P ⋅ Q −13 θcos =PQ86.45 = = = −0.51 ⎯⎯ → θ PQ = 120.66o PQ θ PQ 10 65 Prob. 1.11 1.13 Prob. 1.10 If A and B are parallel, then B = kA and A x B = 0. It is evident that k = -2 and that A B = (4, −6,1)(2, 0,5) = 8 − 0 + 5 = 13 ax a y az (a) | B |2 = 22 + 52 = 29 A × B = 1 −2 3 = 0 A B + 2 | B |2 = 13 + 2 × 29 = 71 −2 4 −6
as expected. (b) A× B Prob. a⊥ = ±1.14 | Athe × Bfact | that (a) Using
( A × B ) × C = (4A ⋅−C6)B 1− =(B(−⋅ C30,)A−,18,12) Let C = A × B = 2 0 5 we get 30, ( A−×18,12) A ×C( A × B ) (=−− B ) × A = (B ⋅ A)A − ( A ⋅ A)B a⊥ = ± =± = ±(−0.8111a x − 0.4867a y + 0.3244a z ) 2 2 2 |C | + + 30 18 12 (b) A × ( A × ( A × B ) ) = A × ( A ⋅ B ) A- ( A ⋅ A) B = ( A ⋅ B ) - ( A × A) − ( A ⋅ A ) - ( A × B )
POESM_Ch01.indd 7
Copyright © 2015 by Oxford University Press Prob. 1.12 = − A2 ( A × B ) Q==0(2, −6,5)(0,3,1) = 0 − 18 + 5 = −13 since AP xA
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If A and B are parallel, then B = kA and A x B = 0. It is evident that k = -2 and that ax a y az Sadiku & Kulkarni Principles of Electromagnetics, 6e A × B = 1 −2 3 = 0 −2 4 −6 8 as expected.
1.11 Prob. 1.14 (a) Using the fact that
( A × B ) × C = ( A ⋅ C )B − (B ⋅ C )A,
we get
A × ( A × B ) = −( A × B ) × A = (B ⋅ A)A − ( A ⋅ A)B
(b) A × ( A × ( A × B ) ) = A × ( A ⋅ B ) A- ( A ⋅ A) B = ( A ⋅ B ) - ( A × A) − ( A ⋅ A ) - ( A × B )
since AxA = 0
= − A2 ( A × B )
8
Prob. 1.15 Prob. 1.12 1.16 P2 1 1 4 1 −5 1 Area = | D × E |= = | (3 + 10)a x + (5 − 12)a y + (8 + 1)a z | 2 2 −1 2 a 3 2 =
1 1 | (13, −7,9) |= 169 + 49 + 81 = 8.646 2 2 b P1
c
P3
a = rp 2 − rp1 = (1, −2, 4) − (5, −3,1) = ( −4,1,3) (a) b = rp 3 − rp 2 = (3,3,5) − (1, −2, 4) = (2,5,1) c = rp1 − rp 3 = (5, −3,1) − (3,3,5) = (2, −6, −4) Note that a + b + c = 0 ⎯⎯ → perpendicular a ⋅ b = −8 + 5 + 3 = 0 b ⋅ c = 4 − 30 − 4 ≠ 0 c ⋅ a = −8 − 6 − 12 ≠ 0 Hence P2 is a right angle. 1 1 −4 1 3 1 | a × b |= = | (1 − 15)a x + (6 + 4)a y + (−20 − 2)a z | 2 2 2 5 1 2
Area =
(b) =
1 1 | (−14,10, −22) |= 196 + 100 + 484 = 13.96 2 2
Prob. 1.17 Given rP = (−1, 4,8),
rQ = (2, −1,3),
rR = (−1, 2,3)
(a) | PQ |= 9 + 25 + 25 = 7.6811 (b) PR = −2a y − 5a z QP QR o ∠PQR = cos −1 = 42.57 | QP || QR | (d) Area of triangle PQR = 11.023 Copyright © 2015 by Oxford University Press (e) Perimeter = 17.31
(c)
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Hence P2 is a right angle. Sadiku & Kulkarni
Area =
(b)
Principles of Electromagnetics, 6e 1 1 −4 1 3 1 | a × b |= = | (1 − 15)a x + (6 + 4)a y + (−20 − 2)a z | 2 2 2 5 1 2 9 1 1 196 + 100 + 484 = 13.96 = | (−14,10, −22) |= 2 2
Prob. 1.13 1.17 Given rP = (−1, 4,8),
rQ = (2, −1,3),
rR = (−1, 2,3)
(a) | PQ |= 9 + 25 + 25 = 7.6811 (b) PR = −2a y − 5a z QP QR o ∠PQR = cos −1 = 42.57 | QP || QR | (d) Area of triangle PQR = 11.023 (e) Perimeter = 17.31
(c)
Prob.1.18 Prob. 1.14 Let R be the midpoint of PQ. 1 rR = {(2, 4, −1) + (12,16,9)} = (7,10,94) 2
OR = 49 + 100 + 16 = 165 = 12.845 OR 12.845 t= = = 42.82 ms 300 v Prob. 1.15 1.19
Ax A ⋅ ( A× B ) = Bx
Ay By
Az Ax Bz , ( A× B ) ⋅ C = Bx
Ay By
Az Bz
Cx
Cy
Cz
Cy
Cz
Cx
Hence, A ⋅ ( A× B ) = ( A× B ) ⋅ C Also, each equals the volume of the parallelopiped formed by the three vectors as sides. Prob. 1.16 1.20 (a) Let P and Q be as shown below:
y
Q
θ2
P θ1
x
P = cos 2 θ 1 + sin 2 θ 1 = 1, Q = cos 2 θ 2 + sin 2 θ 2 = 1,
Hence P and Q are unit vectors. (b) P ⋅ Q = (1)(1)cos(θ 2 -θ1 ) But P ⋅ Q = cos θ1 cos θ 2 + sin θ1 sin θ 2 . Thus, cos(θ 2 − θ1 ) = cos θ1 cos θ 2 + sin θ1 sin θ 2 © 2015 by Oxford University Press Let P1 = P = cos θCopyright 1 a x + sin θ 1 a y and POESM_Ch01.indd 9
Q1 = cos θ 2 a x − sin θ 2 a y .
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θ2 Sadiku & Kulkarni
P θ1
Principles of Electromagnetics, 6e
x
P = cos 2 θ 1 + sin 2 θ 1 = 1, Q = cos 2 10 θ 2 + sin 2 θ 2 = 1,
Hence P and Q are unit vectors. (b) P ⋅ Q = (1)(1)cos(θ 2 -θ1 ) But P ⋅ Q = cos θ1 cos θ 2 + sin θ1 sin θ 2 . Thus, cos(θ 2 − θ1 ) = cos θ1 cos θ 2 + sin θ1 sin θ 2 Let P1 = P = cos θ 1a x + sin θ 1a y and Q1 = cos θ 2 a x − sin θ 2 a y . P1 and Q1 are unit vectors as shown below: y
P1 θ1
θ1+θ2 θ2
10
x
Q1
P1 ⋅ Q1 = (1)(1) cos(θ 1 + θ 2 ) But P1 ⋅ Q1 = cosθ 1 cosθ 2 − sin θ 1 sin θ 2 , cos(θ 2 + θ 1 ) = cosθ 1 cosθ 2 − sin θ 1 sin θ 2 Alternatively, we can obtain this formula from the previous one by replacing θ2 by -θ2 in Q.
(c) 1 1 | P − Q |= | (cos θ1 − cos θ 2 ) ax + (sin θ1 − sin θ 2 ) a y 2 2
=
1 cos 2 θ1 + sin 2 θ1 + cos 2 θ 2 + sin 2 θ 2 − 2 cos θ1 cos θ 2 − 2sin θ1 sin θ 2 2
1 1 2 − 2(cos θ1 cos θ 2 + sin θ1 sin θ 2 ) = 2 − 2 cos(θ 2 − θ1 ) 2 2 Let θ 2 − θ1 = θ , the angle between P and Q. =
1 1 | P − Q |= 2 − 2 cos θ 2 2 But cos 2A = 1 – 2 sin 2A. 1 1 | P − Q |= 2 − 2 + 4sin 2 θ / 2 = sin θ / 2 2 2 Thus,
θ −θ 1 | P − Q |=| sin 2 1 | 2 2 Copyright © 2015 by Oxford University Press POESM_Ch01.indd 10
Prob. 1.21
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1 1 2 − 2(cos θ1 cos θ 2 + sin θ1 sin θ 2 ) = 2 − 2 cos(θ 2 − θ1 ) 2 2 Sadiku & Kulkarni Let θ 2 − θ1 = θ , the angle between P and Q. =
Principles of Electromagnetics, 6e
11
1 1 | P − Q |= 2 − 2 cos θ 2 2
But cos 2A = 1 – 2 sin 2A. 1 1 | P − Q |= 2 − 2 + 4sin 2 θ / 2 = sin θ / 2 2 2 Thus,
θ −θ 1 | P − Q |=| sin 2 1 | 2 2 Prob. 1.17 1.21
w(1,−2,2) = (1,−2,2), r = rp − ro = (1,3,4) − ( 2,−3,1) = ( −1,6,3) 3 1 −2 2 u = w×r = = ( −18,−5,4) −1 6 3 w=
u = −18a x − 5a y + 4az 1.22 Prob. 1.18 r1 = (1,1,1),
cos θ =
r2 = (1, 0,1) − (0,1, 0) = (1, −1,1)
r1 ⋅ r2 (1 − 1 + 1) 1 = = r1r2 3 3 3
⎯⎯ → θ = 11 70.53o
Prob. 1.19 1.23
T ⋅ S ( 2, −6,3) ⋅ (1, 2,1) −7 = = = −2.8577 S 6 6 (S ⋅ T )T = − 7(2,−6,3) (b) S T = ( S ⋅ a T )a T = 72 T2 = − 0.2857a x + 0.8571a y − 0.4286a z (a) Ts = T ⋅ as =
(c) sin θ TS =
T×S
=
T S
2 − 6 3 (− 12,1,10) 245 = = = 0.9129 1 2 1 7 6 7 6
θ TS = 65.91 1.24 Prob. 1.20 Let A = AB + AB ⊥
AB = ( A ⋅ a B )a B = Hence,
A⋅ B B B⋅B
AB⊥ = A − AB = A −
A⋅ B B B⋅B
Prob.1.25 (a) H (1,3, −2) = 6a x + a y + 4a z
aH = POESM_Ch01.indd 11
(6,1, 4) = 0.8242a x + 0.1374a y + 0.5494a z 36 + 1 + 16Copyright © 2015 by Oxford University Press
(b) | H |= 10 = 4 x 2 y 2 + ( x + z ) 2 + z 4
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Let A Sadiku & Kulkarni B
A = AB + AB ⊥ = ( A ⋅ a B )a B =
Hence,
A⋅ B B B⋅B
AB⊥ = A − AB = A −
Principles of Electromagnetics, 6e
12
A⋅ B B B⋅B
Prob. 1.21 Prob.1.25 (a) H (1,3, −2) = 6a x + a y + 4a z
(6,1, 4) = 0.8242a x + 0.1374a y + 0.5494a z 36 + 1 + 16
aH =
(b) | H |= 10 = 4 x 2 y 2 + ( x + z ) 2 + z 4 or 100 = 4 x 2 y 2 + x 2 + 2 xz + z 2 + z 4 1.26 Prob. 1.22 C = 5a x + a z
(a) B × C =
1 1 0 = a x − a y − 5a z 5 0 1
A( B × C ) = (4, −1,1)(1, −1, −5) = 4 + 1 − 5 = 0 (b) AB = ( AaB )a B =
( A ⋅ B ) B (4 − 1)(1,1, 0)12 = = 1.5a x + 1.5a y | B |2 1+1
1.23 Prob. 1.27
(a) At (1, -2, 3), x = 1, y = -2, z = 3. G = a x + 2a y + 6a z , H = −6a x + 3a y − 3a z G = 1 + 4 + 36 = 6.403 H = 36 + 9 + 9 = 7.348
(b) G H = −6 + 6 − 18 = −18
(c)
GH −18 = = −0.3826 6.403 × 7.348 GH = 112.5o
cos θGH =
θGH
Prob. 1.28 1.24 rPQ = rQ − rP = (−2,1, 4) − (1, 0,3) = ( −3,1,1)
At P, H = 0a x − 1a z = −a z The scalar component of H along rPQ is D = H arPQ =
H • rPQ | rPQ |
=
−1 = −0.3015 9 +1+1
Prob. 1.29 (a) At P, x = -1, y = 2, z = 4 D = 8a x − 4a y - 2a z , E = −10a x + 24a y + 128a z Copyright © 2015 C = D + E = −2a x + 20 a y + 126 a z by Oxford University Press POESM_Ch01.indd 12
C a x
−2
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rPQ = rQ − rP = (−2,1, 4) − (1, 0,3) = ( −3,1,1) At P, H = 0a x − 1a z = −a z
Sadiku & Kulkarni
Principles of Electromagnetics, 6e
The scalar component of H along rPQ is D = H arPQ =
H • rPQ | rPQ |
−1 = −0.301513 9 +1+1
=
1.25 Prob. 1.29 (a) At P, x = -1, y = 2, z = 4 D = 8a x − 4a y - 2a z , E = −10a x + 24a y + 128a z C = D + E = −2a x + 20a y + 126a z
(b)
C a x = C cos θ x
⎯⎯ → cos θ x =
C a x −2 = = −0.01575 C 22 + 202 + 1262
θ x = 90.9o
13
Prob. 1.26 1.30 (a) At (1,2,3), E = (2,1,6)
E = 4 + 1 + 36 = 41 = 6.403 (b) At (1,2,3), F = (2,-4,6) E F = ( E ⋅ aF ) aF =
( E ⋅ F )F F
2
=
36 ( 2,−4,6) 56
= 1.286a x − 2.571a y + 3.857az (c) At (0,1,–3), E = (0,1,–3), F = (0,–1,0) E×F = a E ×F = ±
0 1 −3 = (−3,0,0) 0 −1 0 E×F = ± ax E×F
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Principles of Electromagnetics, 6e
14 14 CHAPTER 2 P. E. 2.1 (a) At P(1,3,5),
ρ=
x = 1,
2
x + y
2
y = 3,
10 ,
=
z =5,
z = 5,
φ = tan −1 y / x = tan −1 3 = 71.6o
P( ρ , φ , z ) = P( 10, tan −1 3,5) = P (3.162, 71.6o ,5) Spherical system: r=
x2 + y2 + z2 =
35 = 5.916
θ = tan −1 x 2 + y 2 z = tan −1 10 5 = tan −1 0.6325 = 32.31° P (r ,θ , ϕ ) = P (5.916,32.31°, 71.57°) At T(0,-4,3),
x=0
y =-4,
z =3;
ρ = x + y = 4, z = 3, ϕ = tan y / x = tan − 4 / 0 = 270° T ( ρ, ϕ , z) = T (4,270° ,3). 2
−1
2
−1
Spherical system: r=
x 2 + y 2 + z 2 = 5,θ = tan −1 ρ / z = tan −1 4 / 3 = 5313 . °.
T (r ,θ , ϕ ) = T (5,5313 . ° ,270° ).
At S(-3-4-10),
x =-3, y =-4, z =-10; −4 ρ = x 2 + y 2 = 5, φ = tan −1 = 233.1° −3 S ( ρ , φ , z ) = S (5, 233.1,− 10).
Spherical system: r = x 2 + y 2 + z 2 = 5 5 = 11.18. 5 = 153.43°; −10 S (r , θ , φ ) = S (11.18,153.43°, 233.1°).
θ = tan −1 ρ z = tan −1
(b)
In Cylindrical system, Qx =
ρ ρ + z2 2
;
ρ = x2 + y 2 ; yz = z ρ sin φ , z ρ sin φ
Qy = 0;
Qz = −
ρ 2 + z2
;
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
15 Qρ cos φ Qφ = − sin φ Qz 0 Qρ = Qx cos φ =
ρ cos φ ρ +z 2
0 Qx 0 0 ; 1 Qz
sin φ cos φ 0
2
,
Qφ = −Qx sin φ =
− ρ sin φ
ρ 2 + z2
Hence, Q=
ρ ρ + z2 2
(cos φ a ρ − sin φ a φ − z sin φ a z ).
In Spherical coordinates: r sin θ Qx = = sin θ ; r 1 Qz =− r sin φ sin θ r cosθ = − r sin θ cosθ sin φ . r Qr sin θ cos φ sin θ sin φ cosθ Qx Qθ = cosθ cos φ cosθ sin φ − sin θ 0 ; Qφ − sin φ cos φ 0 Qz Qr = Qx sin θ cos φ + Qz cosθ = sin 2 θ cosφ − r sin θ cos2 θ sin φ . Qθ = Qx cosθ cos φ − Qz sin θ = sin θ cosθ cos φ + r sin 2 θ cosθ sin φ . Qφ =− Qx sin φ = − sin θ sin φ. ∴ Q = sin θ ( sin θ cosφ − r cos2 θ sin φ ) a
At T :
r
+ sin θ cosθ (cosφ + r sinθ sin φ )aθ − sinθ sin φ aφ .
4 12 a x + a z = 0.8a x + 2.4a z ; 5 5 4 Q ( ρ , φ , z ) = (cos 270° a ρ − sin 270° aφ − 3sin 270°a z 5 = 0.8aφ + 2.4 a z ; Q ( x, y , z ) =
4 45 4 3 20 4 Q (r ,θ , φ ) = (0 − (−1))ar + ( )(0 + (−1))aθ − (−1)aφ 5 25 5 5 5 5 36 48 4 = a r − aθ + aφ = 1.44ar − 1.92aθ + 0.8 aφ ; 25 25 5
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Principles of Electromagnetics, 6e
16 16
Note, that the magnitude of vector Q = 2.53 in all 3 cases above. P.E. 2.2 (a)
Ax cosφ −sinφ 0 ρz sinφ A = sinφ cosφ 0 3ρ cosφ y Az 0 0 1 ρ cosφ sinφ A = (ρz cosφ sinφ −3ρ cosφ sinφ) ax + (ρz sin2 φ +3ρ cos2 φ) ay + ρ cosφ sinφ az.
y x y But ρ = x2 + y2 , tanφ = , cosφ = , sinφ = ; 2 2 2 2 x x +y x +y Substituting all this yields: 1 A= [(xyz −3xy)ax + (zy2 + 3x2 ) ay + xy az ]. 2 2 x +y Bx sin θ cos φ B = sin θ sin φ y Bz cos θ
cos θ cos φ cos θ sin φ − sin θ
Since r = x 2 + y 2 + z 2 , tan θ = and sin θ = and sin φ =
x2 + y 2 2
2
x +y +z y x2 + y2
,
2
− sin φ cos φ 0
r2 0 sin θ
x2 + y2 y , tan φ = ; z z
, cos θ = cos φ =
z 2
x + y2 + z2 x x2 + y2
;
;
y 1 = ( r 2 x − y ). r r x 1 B y = r 2 sin θ sin φ + sin θ cos φ = ry + = (r 2 y + x ). r r 1 Bz = r 2 cos θ = r z = ( r 2 z ). r Bx = r 2 sin θ cos φ − sin θ sin φ = rx −
Hence, B=
1 2
2
x +y +z
2
[{x ( x 2 + y 2 + z 2 ) − y} a x + { y ( x 2 + y 2 + z 2 ) + x} a y + z ( x 2 + y 2 + z 2 )a z ].
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Principles of Electromagnetics, 6e
17 P.E.2.3 (a) At: (1, π / 3, 0),
H = (0, 0.06767,1) 1 a x = cos φ a ρ − sin φ aφ = (a ρ − 3 aφ ) 2 H • a x = −0.0586.
(b)
At: (1, π / 3, 0),
aθ = cos θ a ρ − sin θ a z = − a z . aρ
aφ az 0.06767 1 = − 0.06767 a ρ . 0 1
H × az = 0 0
(c)
(d)
( H • a ρ ) a ρ = 0 aρ .
H × az =
aρ
aφ
az
0
0.06767
1
0
0
1
= 0.06767 a ρ .
H × a z = 0.06767
P.E. 2.4 (a) A • B = (3, 2, − 6) • ( 4, 0,3) = − 6.
(b)
A× B =
3 2 −6 = 6 ar − 33aθ − 8aφ . 4 0 3
Thus the magnitude of A × B = 34.48.
(c) At (1, π / 3, 5π / 4), θ = π / 3, a z = cos θ a r − sin θ aθ =
1 3 ar − aθ . 2 2
3 3 1 ( Aa z )a z = − 3 ar − aθ = −0.116ar + 0.201aθ 2 2 2
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Principles of Electromagnetics, 6e
18 18
P.E. 2.5
In spherical coordinates, the distance between two points is given by eq. 2.33: By solving the above equation, we obtain
P.E. 2.6 In Cartesian system the dot product of two vectors and is given by (1) Now using the eq. 2.21 to represent spherical coordinates in Cartesian system and the dot product is determined by using Eq. (1) Dot product of the given vectors is 3 Using the above equation
Hence by using the above derived equation we can directly calculate dot product of vectors in spherical system 33 33
P.E. 2.7 2.7 Prob. 2.24 P.E. Prob. 2.24 At P (0, 2,−−5), 5), 90°°;; At P(0, 2, φφ ==90 cosφφ −−sin sinφφ 00 BBxx cos BB == sin sinφφ cos cosφφ 00 y y BBzz 00 00 11
00 == 11 00
−−11 00 00
00 00 11
BB== −−aaxx −−55aayy −−33aazz
BBρρ BB φφ BBzz
−−55 11 −−33
(2,4,10) 4,10)++((−−1,1,−−5, 5,−−3) 3) ((aa)) AA++ BB == (2, == aaxx−−aayy ++77aazz.. −52 A• B cosθθAB = A • B == −52 ((bb)) cos AB = AB 4200 AB 4200 −52 = cos cos−−11(( −52 )) == 143.36 143.36°°.. θθAB AB = 4200 © 2015 by Oxford University Press Copyright 4200 Chapter02.indd 18 POESM_Ch02.indd 18
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Bz Sadiku & Kulkarni
0 0 1 Bz 0 −1 0 −5 = 1 0 0 1 0 0 1 −3
Principles of Electromagnetics, 6e
19
B = −a x − 5a y − 3a z
(a) A + B = (2, 4,10) + ( −1, −5, −3) = ax − a y + 7 az . −52 A• B = AB 4200 −52 = cos −1 ( ) = 143.36°. 4200
(b) cos θ AB =
θ AB
(c) AB = A • a B =
A• B 52 =− = −8.789. B 35 18
Prob. 2.24 Prob.At2.1(1, 60o , −1), ρ = 1, φ = 60o , z = −1, 2 + sin y 2 60 = o )4a+ρ 36 (a) ρA== (−x2 − + (4=+6.324 2 cos 60o )aφ − 3(1)( −1)a z 1 y −1 6 3a z o ρ + 5aφ=+71.56 (a) φ = = tan−−2.866 = atan x 2 o o P is 60 (6.324, 71.56 B = 1cos a ρ + sin 60o ,a−φ 4) + a z = 0.5aφ + 0.866aφ + a z
A B = −1.433 + 4.33 + 3 = 5.897 r = x 2 +2 y 2 + z 2 = 4 + 36 + 16 = 7.485 AB = 2.866 + 26 + 9 0.25 + 1 + 0.8662 = 9.1885 2 + y2 −1B x 5.897 −1 6.324 o −1 6.324 o A ==tan 90→ + tan (b) cos θθAB==tan = 0.6419 =⎯⎯ θ AB = 50.07=o 147.69 z −4 4 AB 9.1885 o o P is (7.483,147.69 , 71.56 ) Let D = A × B. At (1,90o , 0), ρ = 1, φ = 90o , z = 0
(b) A = − sin 90o a ρ + 4aφ = −a ρ + 4aφ Prob.2.2 (a) Given P(1,-4,-3), convert to cylindrical and spherical values; B = 1cos 90o2 a ρ +2 sin 90o a2 φ + a z =2 aφ + a z ρ = x + y = 1 + (−4) = 17 = 4.123. y −4 = tan −1 = 284.04°. x 1 ∴ P ( ρ , φ , z ) = (4.123, 284.04°, − 3)
φ = tan −1
Spherical : r = x 2 + y 2 + z 2 = 1 + 16 + 9 = 5.099.
ρ
4.123 = 126.04°. −3 z P (r , θ , φ ) = P (5.099, 126.04°, 284.04°).
θ = tan −1
(b)
= tan −1
y 0 = tan −1 = 0o x 3 o Q( ρ , φ , z ) = Q(3, 0 ,5)
ρ = 3,
φ = tan −1
r = 9 + 0 + 25 = 5.831,
θ = tan −1
Q(r , θ , φ ) = Q(5.831,30.96 , 0 ) o
(c) POESM_Ch02.indd 19
o
ρ z
= tan −1
3 = 30.96o 5
6 Universityo Press Copyright © 2015 by−1Oxford
ρ = 4 + 36 = 6.325,
φ = tan o
−2
= 108.4
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Spherical : Sadiku & Kulkarni
ρ
4.123 = 126.0420 °. −3 z P (r , θ , φ ) = P (5.099, 126.04°, 284.04°).
θ = tan −1
Principles of Electromagnetics, 6e
r = x 2 + y 2 + z 2 = 1 + 16 + 9 = 5.099.
(b)
= tan −1
y 0 = tan −1 = 0o 3 x o Q( ρ , φ , z ) = Q(3, 0 ,5)
ρ = 3,
φ = tan −1
ρ
θ = tan −1
r = 9 + 0 + 25 = 5.831,
z
Q(r , θ , φ ) = Q(5.831,30.96 , 0 ) o
(c)
ρ = 4 + 36 = 6.325,
o
φ = tan −1
R ( ρ , φ , z ) = R(6.325,108.4o , 0)
r = ρ = 6.325,
θ = tan −1
ρ
= tan −1
3 = 30.96o 5
6 = 108.4o −2
= tan −1
z R (r , θ , φ ) = R(6.325,90 ,108.4o ) o
6.325 = 90o 0
19
Prob. 2.3 (a) x = ρ cos φ = 2 cos 30° = 1.732; y = ρ sin φ = 2sin 30° = 1; z = 5; P1 ( x, y, z ) = P1 (1.732,1, 5).
(b)
x = 1cos 90° = 0;
y = 1sin 90° =1;
z = − 3.
P2 ( x, y, z ) = P2 (0, 1, − 3).
(c)
(d)
x = r sin θ cos φ = 10sin(π / 4) cos(π / 3) = 3.535; y = r sin θ sin φ = 10sin(π / 4) sin(π / 3) = 6.124; z = r cos θ = 10 cos(π / 4) = 7.0711 P3 ( x, y, z ) = P3 (3.535, 6.124, 7.0711).
x = 4sin 30° cos 60° =1 y = 4sin 30° sin 60° = 1.7321 z = r cos θ = 4 cos 30° = 3.464 P4 ( x, y, z ) = P4 (1,1.7321,3.464).
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Principles of Electromagnetics, 6e
21 20 Prob. 2.4 x = ρ cos φ = 5cos120o = −2.5
y = ρ sin φ = 5sin120o = 4.33 (a) z =1 Hence Q = (−2.5, 4.33,1) r = x 2 + y 2 + z 2 = ρ 2 + z 2 = 25 + 1 = 5.099 (b)
θ = tan −1
x2 + y2 ρ 5 = tan −1 = tan −1 = 78.69o z z 1
φ = 120o Hence Q = (5.099, 78.69o ,120o ) Prob. 2.5 T (r ,θ , φ )
⎯⎯ →
r = 10, θ = 60o , φ = 30o
x = r sin θ cos φ = 10sin 60o cos 30o = 7.5 y = r sin θ sin φ = 10sin 60o sin 30o = 4.33 z = r cos θ = 10 cos 60o = 5 T ( x, y, z ) = (7.5, 4.33,5)
ρ = r sin θ = 10sin 60o = 8.66 T ( ρ , φ , z ) = (8.66,30o ,5)
Prob. 2.6 (a) x = ρ cos φ ,
y = ρ sin φ ,
V = ρ z cos φ − ρ sin φ cos φ + ρ z sin φ 2
(b) U = x2 + y 2 + z 2 + y 2 + 2z 2 = r 2 + r 2 sin 2 θ sin 2 φ + 2r 2 cos 2 θ = r 2 [1 + sin 2 θ sin 2 φ + 2 cos 2 θ ]
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Principles of Electromagnetics, 6e
22 21 Prob. 2.7 (a)
Fρ cos φ F = − sin φ φ Fz 0
Fρ = Fφ = Fz = _
F =
1
ρ +z 2
[ ρ cos 2 φ + ρ sin 2 φ ] =
2
1
ρ 2 + z2 4
ρ + z2 2
1
ρ + z2 2
sin φ 0 cos φ 0 0 1
x 2 ρ + z2 y ρ 2 + z2 4 2 2 ρ +z
ρ ρ + z2 2
;
[− ρ cos φ sin φ + ρ cos φ sin φ ] = 0; ;
( ρ aρ + 4 az )
In Spherical:
Fr sinθ cosφ sinθ sinφ cosθ Fθ = cosθ cosφ cosθ sin φ −sinθ cosφ 0 −sinφ Fφ
x r y r 4 r
r r 4 4 Fr = sin2 θ cos2 φ + sin2 θ sin2 φ + cosθ = sin2 θ + cosθ ; r r r r 4 4 Fθ =sinθ cosθ cos2 φ + sinθ cosθ sin2 φ − sinθ = sinθ cosθ − sinθ ; r r Fφ =− sinθ cosφ sinφ + sinθ sinφ cosφ = 0; _ 4 4 ∴ F = (sin2 θ + cosθ ) ar + sinθ (cosθ − )aθ r r
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Principles of Electromagnetics, 6e
23 22
(b)
Gρ cos φ G = − sin φ φ Gz 0
ρ2
Gρ =
sin φ
0 cos φ 0 0 1
xρ 2 2 ρ + z2 2 yρ ρ 2 + z2 zρ 2 2 2 ρ + z
[ ρ cos φ + ρ sin φ ] = 2
ρ 2 + z2
ρ3
2
ρ 2 + z2
;
Gφ = 0; zρ 2
Gz =
ρ 2 + z2 ρ2
G=
ρ 2 + z2
;
( ρ aρ + z az )
Spherical : G= Prob. 2.8 B = ρ ax +
ρ2
y
ρ
( xa x + ya y + za z ) =
r
r 2 sin 2 θ rar = r 2 sin 2 θ ar r
a y + za z
Bρ cos φ B = − sin φ φ Bz 0 Bρ = ρ cos φ +
sin φ cos φ 0 y
ρ
Bφ = − ρ sin φ +
0 ρ 0 y / ρ 1 z
sin φ
y
ρ
cos φ
Bz = z But y = ρ sin φ Bρ = ρ cos φ + sin 2 φ , Bφ = − ρ sin φ + sin φ cos φ Hence, B = ( ρ cos φ + sin 2 φ )a ρ + sin φ (cos φ − ρ )aφ + za z
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
24 23 Prob. 2.9 Ax cos φ A = sin φ y Az 0
− sin φ cos φ 0
At P, ρ = 2,
φ = π / 2,
0 2 0 3 1 4 z = −1
Ax = 2 cos φ − 3sin φ = 2 cos 90o − 3sin 90o = −3 Ay = 2sin φ + 3cos φ = 2sin 90o + 3cos 90o = 2 Az = 4 Hence, A = −3a x + 2a y + 4a z Prob. 2.10 (a) Ax cos φ Ay = sin φ Ay 0
− sin φ cos φ 0
0 ρ sin φ 0 ρ cos φ 1 −2 z
Ax = ρ sin φ cos φ − ρ cos φ sin φ = 0 Ay = ρ sin 2 φ + ρ cos 2 φ = ρ = x 2 + y 2
Hence,
Az = −2 z
A = x 2 + y 2 a y − 2 za z (b)
Bx sin θ cos φ cos θ cos φ − sin φ B = sin θ sin φ cos θ sin φ cos φ y Bz cos θ − sin θ 0 Bx = 4r sin θ cos 2 φ + r cos θ cos φ By = 4r sin θ sin φ cos φ + r cos θ sin φ
4r cos φ r 0
Bz = 4r cos θ cos φ − r sin θ But
2
2
sin θ =
2
r= x +y +z ,
sin φ =
y 2
x + y2
,
cos φ =
x2 + y2 , r x
cos θ =
z r
x2 + y 2
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Principles of Electromagnetics, 6e
25 24
x2 zx + 2 2 2 x +y x + y2 xy zy + By = 4 x 2 + y 2 2 2 2 x +y x + y2 Bx = 4 x 2 + y 2
x
Bz = 4 z B=
2
x +y 1 2
x +y
2
2
− x2 + y2
x(4 x + z )a x + y (4 x + z )a y + (4 xz − x 2 − y 2 )a z
Prob. 2.11 Gx cos φ G = sin φ y Gz 0
− sin φ 0 ρ sin φ cos φ 0 − ρ cos φ 0 1 ρ Gx = ρ cos φ sin φ + ρ sin φ cos φ = 2 ρ sin φ cos φ G y = ρ sin 2 φ − ρ cos 2 φ = ρ (1 − cos 2 φ ) − ρ cos 2 φ = ρ − 2 ρ cos 2 φ Gz = ρ
ρ = x 2 + y 2 , cos φ = Gx = 2 x 2 + y 2
But
x
ρ
=
xy = x + y2 2
Gy = x 2 + y 2 − 2 x 2 + y 2
x 2
x +y
2
,sin φ =
y
ρ
=
y 2
x + y2
2 xy x2 + y2 x2 2 x2 2 2 = + − x y x2 + y 2 x2 + y 2
Gz = x 2 + y 2 Thus, G=
2x2 ax + x2 + y 2 − x2 + y 2 x2 + y2
2 xy
a y + x2 + y 2 az
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
26 25 Prob. 2.12
H x sin θ cos φ cos θ cos φ − sin φ cos θ H = sin θ sin φ cos θ sin φ cos φ sin θ y H z cos θ − sin θ 0 0 H x = sin θ cos θ cos φ + sin θ cos θ cos φ = 2sin θ cos θ cos φ
H y = cos θ sin θ sin φ + sin θ cos θ sin φ = 2sin θ cos θ sin φ H z = cos 2 θ − sin 2 θ
But,
x2 + y 2 sin θ = , r y sin φ = , 2 x + y2 Hx =
2z x2 + y2 x2 + y2 + z 2
Hy =
2z x2 + y2 x2 + y 2 + z 2
z cos θ = , r cos φ = x x2 + y2 y x2 + y2
x 2
x + y2
=
2 xz x + y2 + z2
=
2 yz x + y2 + z2
2
2
z 2 − x2 − y2 x2 + y 2 + z 2 1 H= 2 2 xza x + 2 yza y + [ z 2 − x 2 − y 2 ]a z ) 2 2 ( x +y +z
Hz =
Prob. 2.13 x = ρ cos φ (a)
B = ρ cos φ a z
x = r sin θ cos φ (b) B = r sin θ cos φ a z ,
Bx = 0 = By , Bz = r sin θ cos φ
Br sin θ cos φ sin θ sin φ cos θ 0 0 Bθ = cos θ cos φ cos θ sin φ − sin θ cos φ 0 r sin θ cos φ Bφ − sin φ Br = r sin θ cos θ cos φ = 0.5r sin(2θ ) cos φ Bθ = −r sin 2 θ cos φ ,
Bφ = 0
B = 0.5r sin(2θ ) cos φ ar − r sin 2 θ cos φ aθ
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
27 26 Prob. 2.14 (a)
a x • a ρ = (cos φ a ρ − sin φ aφ ) • a ρ = cos φ a x • aφ = (cos φ a ρ − sin φ aφ ) • aφ = − sin φ a y • aρ = (sin φ a ρ + cos φ aφ ) • aρ = sin φ _
_
a y • aφ = (sin φ a ρ + sin φ aφ ) • aφ = cos φ (b) and (c) In spherical system : a x = sin θ cos φ ar + cos θ cos φ aθ − sin φ aφ . a y = sin θ sin φ a r + cos θ sin φ aθ − cos φ aφ . a z = cos θ a x − sin θ aθ .
Hence, a x • a r = sin θ cos φ ; a x • aθ = cos θ cos φ ; a y • a r = sin θ sin φ ; a y • aθ = cos θ sin φ ; _
_
a z • a r = cos θ ; _
_
a z • aθ = − sin θ ;
Prob. 2.15 (a) r=
x2 + y2 + z2 =
ρ θ = tan −1 ; z
or
ρ=
ρ 2 + z2 .
φ = φ.
x 2 + y 2 = r 2 sin 2 θ cos2 φ + r 2 sin 2 θ sin 2 φ .
= r sin θ ; z = r cosθ ;
φ = φ.
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Principles of Electromagnetics, 6e
28 27
(b) From the figures below, cosθ aρ
z
z
aρ
aρ
az
ar -az
θ
θ
sin θ aρ
cos θ a z
sin θ ( − az )
ρ a r = sin θ a ρ + cos θ a z ;
ρ
aθ = cos θ a ρ − sin θ a z ;
aφ = aφ .
Hence, a r sin θ 0 cos θ a ρ aθ = cos θ 0 − sin θ aφ 0 1 0 aφ az From the figures below, a ρ = cos θ aθ + sin θ a r ; a z = cos θ ar − sin θ aθ ; aφ = aφ . z
sin θ ar
z
az
cosθ aθ
sin θ ( − aθ )
−aθ
aρ
cosθ ar
ar
θ
θ
aθ ρ
ar
ρ
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Principles of Electromagnetics, 6e
29 28
a ρ sin θ aφ = 0 cos θ az
cos θ 0 − sin θ
0 1 0
ar aθ az
Prob. 2.16 If A and B are perpendicular to each other, AB = 0
A B = ρ 2 sin 2 φ + ρ 2 cos 2 φ -ρ 2 =ρ 2 (sin 2 φ + cos 2 φ )-ρ 2 =ρ 2 − ρ 2 =0 As expected. Prob. 2.17 (a ) A + B = 8a ρ + 2aφ − 7a z
(b) A B = 15 + 0 - 8 = 7 (c ) A × B =
3 2
1
5 0 −8
=-16a ρ + (5 + 24)aφ − 10a z =-16a ρ + 29aφ − 10a z A⋅ B 7 7 = = AB 9 + 4 + 1 25 + 64 14 89 =0.19831
(d ) cosθ AB =
θ AB =78.56o Prob. 2.18 (a) Ax cos φ A = sin φ y Az 0
− sin φ 0 cos φ 0 0 1
ρ cos φ 0 ρ z 2 sin φ
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Principles of Electromagnetics, 6e
30 29
Ax = ρ cos 2 φ =
x2 + y 2
Ay = ρ sin φ cos φ = Az = ρ z 2 sin φ = ρ z 2 1
A =
2
x +y
2
x2 = x2 + y2
x2 + y 2 y
ρ
x2 x2 + y2
xy = x + y2
xy
2
x2 + y 2
= yz 2
[ x 2 a x + xya y + yz 2 x 2 + y 2 a z ]
At (3,-4,0) x=3, y=-4, z=0; 1 A = [9a x − 12 a y ] 5 A=3
(b)
Ar sin θ cos φ Aθ = − cos θ cos φ Aφ − sin φ
sin θ sin φ cos θ sin φ cos φ
cos θ − sin θ 0
x = r sin θ cos φ , y = r sin θ sin φ , Ar =
x2 ρ xy ρ yz 2
z = r cos θ , ρ = r sin θ .
r sin θ cos φ r sin θ cos φ sin φ sin θ cos φ + sin θ sin φ + r sin θ r sin θ r 3 sin θ cos 2 θ sin φ cos θ 2
2
2
2
2
= r sin 2 θ cos3 φ + r sin 2 θ sin 2 φ cos φ + r 3 sin θ sin φ cos3 θ Aθ = r sin θ cos 2 φ cos θ cos φ + r sin θ cos φ sin φ cos θ sin φ − r 2 cos 2 θ sin φ sin θ = r sin θ cos θ cos φ − r 2 sin θ cos 2 sin φ = r sin θ cos θ [cos φ − r cos θ sin φ ] Aφ = − r sin θ cos 2 φ sin φ + r sin θ cos φ sin φ cos φ = 0. ∴ A=
r sin θ [cos φ sin θ + sin θ cos φ sin 2 φ +r 2 cos 3 θ sin φ ] ar + r sin θ cos θ [cos φ − r 2 cos θ sin θ sin φ ] aθ
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Principles of Electromagnetics, 6e
31 30
At (3 − 4, 0), r = 5, θ = π / 2, φ = 306.83 cos φ = 3 / 5, sin φ = −4 / 5. 3 A = 5[12 * + 5(0)( −4 / 5)] ar + 5(1)(0)aθ 5 = 3ar A = 3.
Prob. 2.19 Ax cos φ − sin φ 0 Aρ A = sin φ cos φ 0 A y φ Az 0 0 1 Az x y 0 − 2 2 2 2 x +y x +y y x 0 = 2 2 x2 + y 2 x +y 0 0 1 Ax sin θ cos φ A = sin θ sin φ y Az cos θ
cos θ cos φ cos θ sin φ
x x2 + y 2 + z 2 y 2 2 2 = x +y +z z x2 + y 2 + z 2
− sin θ
Aρ A φ Az
− sin φ cos φ 0
Ar Aθ Aφ
xz x2 + y 2 x2 + y 2 + z 2 yz x2 + y 2 x2 + y 2 + z 2 −
x2 + y 2 x2 + y 2 + z 2
x2 + y 2 x 2 2 x +y 0 −y
Ar Aφ Aφ
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Principles of Electromagnetics, 6e
32 31
Prob. 2.20 (a) Using the results in Prob.2.14, Aρ = ρz sin φ = r 2 sin θ cosθ sin φ Aφ = 3ρ cos φ = 3r sin θ cos φ Az = ρ cos φ sin φ = r sin θ cos φ sin φ
Hence, Ar sin θ Aθ = cos θ Aφ 0
0 cos θ r 2 sin θ cos θ sin φ 0 − sin θ 3r sin θ cos φ 1 0 r sin θ cos φ sin φ A( r , θ , φ ) = r sin θ sin φ cos θ ( r sin θ + cos φ ) ar + sin φ ( r cos 2 θ − sin θ cos φ ) aθ + 3cos φ aφ
At (10, π / 2,3π / 4),
r = 10, θ = π / 2, φ = 3π / 4
A = 10(0ar + 0.5aθ −
(b)
Br = r 2 = ( ρ 2 + z 2 ),
3 aφ ) = 5aθ − 21.21aφ 2
Bθ = 0,
Bρ sin θ B = 0 φ Bz cos θ
Bφ = sin θ =
cos θ 0 − sin θ
ρ ρ 2 + z2
0 Br 1 Bθ 0 Bφ
ρ + B( ρ , φ , z ) = ρ 2 + z 2 ρ aρ + 2 a z a φ z ρ + z2 At (2, π / 6,1),
ρ = 2, φ = π / 6, z = 1
B = 5(2aρ + 0.4aφ + a z ) = 4.472a ρ + 0.8944aφ + 2.236a z Prob. 2.21
(a) d =
(b)
(6 − 2) 2 + ( − 1 − 1) 2 + (2 − 5) 2 =
29 = 5.385
d 2 = 32 + 52 − 2(3)(5) cos π + ( − 1 − 5) 2 = 100 d = 100 = 10
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Principles of Electromagnetics, 6e
33 32
(c)
d 2 = 102 + 52 − 2(10)(5) cos = 125 − 100(cos
π
4 d = 99.12 = 9.956.
π
cos
6
π 4
cos
− sin
π 6
π
− 2(10)(5) sin
π
π 4
sin
π 6
cos(7
π 4
−
3π ) 4
sin ) = 125 − 100 cos 75o = 99.12 4 6
Prob. 2.22 We can convert Q to cylindrical system and then use equation 2.32
At Q, r = 4 θ =
π
φ=
π
2 2 o ρ = r sin θ = 4sin 90 = 4
φ=
π
2 z=r cos θ = 4 cos 90o = 0
Q is (4, π / 2, 0). d 2 = ρ 22 + ρ12 − 2 ρ1 ρ 2 cos(φ2 − φ1 ) + ( z2 − z1 ) 2 = 102 + 42 − 2(10)(4) cos(π / 4 − π / 2) + 0 = 59.431 d = 7.709
Prob. 2.23 (a)
An infinite line parallel to the z-axis.
(b)
Point (2,-1,10).
(c)
A circle of radius r sin θ = 5 , i.e. the intersection of a cone and a sphere.
(d)
An infinite line
(e)
A semi-infinite line parallel to the x-y plane.
(f)
A semi-circle of radius 5 in the y-z plane.
parallel to the z-axis.
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AB
θ AB = cos −1 (
Sadiku & Kulkarni
4200 −52 ) = 143.36°. 4200
(c) AB = A • a B =
Prob. 2.24 At (1, 60o , −1),
Principles of Electromagnetics, 6e
A• B 52 34 =− = −8.789. B 35
ρ = 1, φ = 60o , z = −1,
(a) A = (−2 − sin 60o )a ρ + (4 + 2 cos 60o )aφ − 3(1)( −1)a z = −2.866 a ρ + 5aφ + 3a z B = 1cos 60o a ρ + sin 60o aφ + a z = 0.5aφ + 0.866aφ + a z A B = −1.433 + 4.33 + 3 = 5.897 AB = 2.8662 + 26 + 9 0.25 + 1 + 0.8662 = 9.1885 A B 5.897 cos θ AB = = = 0.6419 ⎯⎯ → θ AB = 50.07o AB 9.1885 Let D = A × B. At (1,90o , 0),
ρ = 1, φ = 90o , z = 0
(b) A = − sin 90o a ρ + 4aφ = −a ρ + 4aφ
34
B = 1cos 90o a ρ + sin 90o aφ + a z = aφ + a z
aρ aφ D = A × B = −1 4 0 1 aD =
az 0 = 4a ρ + aφ − a z 1
D (4,1, −1) = = 0.9428a ρ + 0.2357aφ − 0.2357a z D 16 + 1 + 1
Prob. 2.25 At T (2,3, −4) x2 + y 2 13 = tan −1 = 137.97 z −4 −4 13 cos θ = = −0.7428,sin θ = = 0.6695 29 29 y 3 φ = tan −1 = tan −1 = 56.31 x 2 2 3 cos φ = sin φ = 13, 13
θ = tan −1
a z = cos θ ar − sin θ aθ = −0.7428ar − 0.6695aθ . ar = sin θ cos φ a x + sin θ sin φ a y + cos θ a z . = 0.3714a x + 0.5571a y − 0.7428a z .
Prob. 2.26 G a y = G y = Gr sin θ sin φ + Gθ cos θ sin φ + 0 = 6r 2 sin θ sin 2 φ + r 2 cos θ sin φ At (2, −3,1), x = 2, y = −3, z = 1 r 2 = x 2 + y 2 + z 2 = 4 + 9 +Copyright 1 = 14 © 2015 by Oxford University Press POESM_Ch02.indd 34
ρ
x2 + y2
13
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13, Sadiku & Kulkarnia z
13
= cos θ ar − sin θ aθ = −0.7428ar − 0.6695aθ .
ar = sin θ cos φ a x + sin θ sin φ a y + cos θ a z . = 0.3714a x + 0.5571a y − 0.7428a z .
Principles of Electromagnetics, 6e
35
Prob. 2.26 G a y = G y = Gr sin θ sin φ + Gθ cos θ sin φ + 0 = 6r 2 sin θ sin 2 φ + r 2 cos θ sin φ At (2, −3,1), x = 2, y = −3, z = 1 r 2 = x 2 + y 2 + z 2 = 4 + 9 + 1 = 14
ρ
sin θ =
r
x2 + y2
=
x2 + y2 + z 2
z 1 = , r 14
cos θ =
13 14
=
sin φ =
y
ρ
=
−3 13
13 9 1 −3 35− 3.1132 = 52.925 + 14 = 56.04 14 13 14 13
G y = 6(14)
Prob. 2.27 _
G = cos 2 φ a x +
_ 2r cos θ sin φ _ a y + (1 − cos 2 φ ) a z r sin θ
_
_
_
= cos 2 φ a x + 2 cot θ sin φ a y + sin 2 φ a z
Gr sin θ cos φ Gθ = cos θ cos φ − sin φ Gφ
sin θ sin φ cos θ sin φ cos φ
cos θ − sin θ 0
cos 2 φ 2 cot θ sin φ 2 sin φ
Gr = sin θ cos3 φ + 2 cos θ sin 2 φ + cos θ sin 2 φ = sin θ cos3 φ + 3cos θ sin 2 φ Gθ = cos θ cos3 φ + 2 cot θ cos θ sin 2 φ − sin θ sin 2 φ Gφ = − sin φ cos 2 φ + 2 cot θ sin φ cos φ _
G = [sin θ cos3 φ + 3cos θ sin 2 φ ] ar + [cos θ cos3 φ + 2 cot θ cos θ sin 2 φ − sin θ sin 2 φ ]aθ + sin φ cos φ (2 cot θ − cos φ ) aφ
Prob. 2.28 (a) J z = ( J • a z )a z . At (2, π / 2, 3π / 2), a z = cos θ ar − sin θ aθ = − aθ .
J z = − cos 2θ sin φ aθ = − cos π sin(3π / 2) aθ = − aθ . (b) Jφ = tan
θ 2
ln r aφ = tan
π 4
ln 2 aφ = ln 2aφ = 0.6931a φ .
(c) J t = J − J n = J − J r = − aθ + ln 2 aφ = − aθ + 0.6931aφ . (d )
_
_
POESM_Ch02.indd 35
_
_
_
J P = (J • ax ) ax _
_ by Oxford University _ _ Copyright © 2015 Press
a x = sin θ cos φ a r + cos θ cos φ aθ − sin φ aφ = aφ . At (2, π / 2, 3π / 2),
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Gφ = − sin φ cos 2 φ + 2 cot θ sin φ cos φ _
Sadiku & Kulkarni G
= [sin θ cos3 φ + 3cos θ sin 2 φ ] ar
Principles of Electromagnetics, 6e
+ [cos θ cos3 φ + 2 cot θ cos θ sin 2 φ − sin θ sin 2 φ ]aθ 36 + sin φ cos φ (2 cot θ − cos φ ) aφ
Prob. 2.28 (a) J z = ( J • a z )a z . At (2, π / 2, 3π / 2), a z = cos θ ar − sin θ aθ = − aθ .
J z = − cos 2θ sin φ aθ = − cos π sin(3π / 2) aθ = − aθ . (b) Jφ = tan
θ 2
ln r aφ = tan
π 4
ln 2 aφ = ln 2aφ = 0.6931a φ .
(c) J t = J − J n = J − J r = − aθ + ln 2 aφ = − aθ + 0.6931aφ . (d )
_
_
_
_
J P = (J • ax ) ax _
_
_
_
_
a x = sin θ cos φ a r + cos θ cos φ aθ − sin φ aφ = aφ . At (2, π / 2, 3π / 2), _
_
J P = ln 2 aφ .
Prob. 2.29
H a x = H x H x cos φ H = sin φ y H z 0
− sin φ cos φ 0
0 ρ 2 cos φ 0 − ρ sin φ 36 1 0
H x = ρ 2 cos 2 φ + ρ sin 2 φ At P, ρ = 2, φ = 60o , z = −1 H x = 4(1/ 4) + 2(3 / 4) = 1 + 1.5 = 2.5
Prob. 2.30 (a) 5 = r ⋅ a x + r ⋅ a y = x + y (b)
10 = rxa z =
a plane
x y z =| ya x − xa y |= x 2 + y 2 = ρ 0 0 1
a cylinder of infinite length
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
37 CHAPTER 3 P. E. 3.1 (a) DH =
φ = 60°
φ
π
r sin θ dφ
= 45°
θ = 90 °
(b) FG =
rdθ
r=5
θ = 60 °
(c)
π
π
= 3(1)[ − ] = = 0.7854. r = 3,90 3 4 4 o
5π π π = 5( − ) = = 2.618. 2 3 6
θ = 90 ° φ = 60 °
AEHD =
r
2
sin θ dθ dφ
θ = 60 ° φ = 45 °
r=3
φ = 60 ° 90 ° = 9 ( − cos θ )|θθ == 60 ° φ |φ = 45 °
1 π 3π = 9 ( )( ) = = 1178 . . 2 12 8
(d)
ABCD =
r = 5 θ = 90
rdθ dr =
r = 3 θ = 60
r2 r = 5 π π 4π ( − )= = 4.189. 2 r =3 2 3 3
(e) Volume =
r =5
r =3
=
φ = 60°
φ
= 45°
θ =90
θ
r 2 sin θ dr dθ dφ =
= 60
r3 3
r =5 r =3
( − cos θ )
θ = 90° θ = 60°
φ
φ = 60° φ = 45°
1 1 π (98)( ) 3 2 12
=
49π = 4.276 . 36
P.E. 3.2 y
3
2 60o 1
x
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Principles of Electromagnetics, 6e
38 38
A• dl = ( + L
1
2
+ ) A• dl = C1 + C2 + C3 3
2
Along (1), C1 = A• dl = ρ cos φ d ρ |φ =0 = 0
ρ2 20
= 2.
Along (2), dl = ρ dφ aφ , A• dl = 0, C2 = 0 0
Along (3), C3 = ρ cos φ d ρφ =60° = −
ρ2
2
2
0
2
A• dl = C
1
1 ( ) = −1 2
+ C2 + C3 = 2 + 0 − 1 = 1
l
P.E. 3.3 (a)
∇U =
∂U ∂U ∂U ax + ay+ az ∂x ∂ y ∂z
= y (2 x + z ) a x + x( x + z ) a y + xy a z (b)
∇V =
∂V ∂V 1 ∂V aρ + aφ + az ∂ρ ρ∂φ ∂z
= ( z sin φ + 2 ρ ) a ρ + ( z cos φ −
z2
ρ
sin 2φ ) aφ + ( ρ s inφ + 2 z cos 2 φ ) a z
(c) ∇f =
∂f 1∂ f 1 ∂f ar + aθ + aφ ∂r r ∂θ r sin θ ∂φ cos θ sin φ sin θ sin φ ln r (cos θ cos φ ln r + r 2 ) + 2rφ )ar − aθ + aφ r r r sin θ sin θ sin φ ln r cos θ sin φ cot θ cos φ ln r = + 2rφ a r − aθ + + r cos ecθ aφ r r r
=(
P.E. 3.4
∇Φ = ( y + z ) a x + ( x + z ) a y + ( x + y ) a z At (1, 2,3), ∇Φ = (5, 4,3) (2, 2,1) 21 = = 7, 3 3 where (2, 2,1) = (3, 4, 4) − (1, 2,3)
∇Φ • a1 = (5, 4,3) •
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Principles of Electromagnetics, 6e
39 P.E. 3.5
Let f = x 2 y + z − 3,
g = x log z − y 2 + 4,
∇ f = 2 xy a x + x 2 a y + a z , ∇ g = log z a x − 2 y a y +
x az z
At P ( − 1 , 2,1), ∇f (−4 a x + a y + a z ) ∇g (−4 a y − a z ) =± , =± ng = ± | ∇f | | ∇g | 18 17 ( − 5) cos θ = n f . n g = ± 18 × 17 T ake positive value to get acute angle. 5 = 73.39 ° θ = cos − 1 17.49 3 nf = ±
P.E. 3.6 (a) ∇ • A = At
∂ Ax ∂ Ay ∂ Az + + = 0 + 4 x + 0 = 4 x. ∂x ∂y ∂z
(1, −2,3), ∇ • A = 4.
(b)
∇•B = =
1 ∂
ρ ∂ρ 1
ρ
( ρ Bρ ) +
2 ρ z sin φ −
1
ρ
1 ∂ Bφ
ρ ∂φ
+
∂ Bz ∂ρ
3ρ z 2 sin φ = 2 z sin φ − 3 z 2 sin φ
= (2 − 3 z ) z sin φ . At (5, (c)
π 2
,1) ,
∇ • B = (2 − 3)(1) = −1.
1 ∂ 2 1 ∂ 1 ∂ Cφ r C C θ ( ) ( sin ) + + r θ r2 ∂ r r sin θ ∂θ r sin θ ∂φ 1 = 2 6r 2 cos θ cos φ r = 6 cos θ cos φ
∇•C =
At (1,
π π
, ), 6 3
∇ • C = 6 cos
π 6
cos
π 3
= 2.598.
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
40 40 P.E. 3.7 This is similar to Example 3.7. Ψ = D • dS = Ψ t + Ψ b + Ψ c S
Ψ t = 0 = Ψ b since D has no z-component Ψ c = ρ cos φρ dφ dz = ρ 2
2
3
φ = 2π
cos φ dφ 2
φ =0
z =1
dz ρ = 4
z =0
= (4)3 π (1) = 64π Ψ = 0 + 0 + 64π = 64π By the divergence theorem,
D • dS = ∇ • Ddv S
V
∇• D =
1 ∂
ρ ∂ρ
= 3ρ cos 2 φ +
z
ρ
Ψ = ∇ • Ddv = V
(3ρ cos 2 φ +
V
2π
1
0
0
0
= 3(
ρ ∂φ
z sin φ +
∂ Az dz
cos φ .
4
= 3 ρ 2 d ρ
1 ∂
( ρ 3 cos 2 φ ) +
2 cos φ dφ dz +
z
ρ
cos φ ) ρ dφ dzd ρ
4
2π
0
0
d ρ cos φ dφ
1
zdz 0
3
4 ) π (1) = 64π . 3
P.E. 3.8 (a)
∇ × A = a x (1 − 0) + a y ( y − 0) + a z (4 y − z ) = a x + y a y + (4 y − z ) a z _
At (1, −2,3) , ∇ × A = a x − 2 a y −11 a z
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
41 (b) ∇ × B = a ρ (0 − 6 ρ z cos φ ) + aφ ( ρ sin φ − 0) + a z
1
ρ
(6 ρ z 2 cos φ − ρ z cos φ )
= −6 ρ z cos φ a ρ + ρ sin φ aφ + (6 z − 1) z cos φ a z At (5,
π 2
, − 1) , ∇ × B = 5 aφ
(c)
_
1 2r cos θ sin φ 3 1/ 2 a φ aθ (r −1/ 2 cos θ − 0) + (− − r ) + (0 + 2r sin θ cos φ ) r sin θ r sin θ 2 r 3 = r −1/ 2 cot θ a r − (2 cot θ sin φ + r −1/ 2 ) aθ + 2sin θ cos φ aφ 2
∇ × C = ar
π π
, ), ∇ × C = 1.732 a r − 4.5 aθ + 0.5 aφ 6 3
At (1,
P.E. 3.9
A • dl = (∇ × A) • dS L
S
But (∇ × A) = sin φ a z +
z cos φ
ρ
a ρ and
d S = ρ dφ d ρ a z
(∇ × A) • dS = ρ sin φ dφ d ρ S
=
ρ2 2
|
2
0
60°
(− cos φ ) |
0
1 = 2(− + 1) = 1. 2
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Principles of Electromagnetics, 6e
42 42 P.E. 3.10
ax ∂ ∇ × ∇V = ∂x ∂V ∂x =( P.E. 3.11 (a) ∇ 2U =
ay ∂ ∂y ∂V ∂y
az ∂ = ∂z ∂V ∂z
∂ 2V ∂ 2V ∂ 2V ∂ 2V ∂ 2V ∂ 2V ) ax + ( )ay +( )az =0 − − − ∂ y∂ z ∂ y∂ z ∂ x∂ z ∂ z∂ x ∂ x∂ y ∂ y∂ x
∂ ∂ 2 ∂ (2 xy + yz ) + ( x + xz ) + ( xy ) ∂x ∂y ∂z
= 2 y.
(b) ∇ 2V = =
1 ∂
ρ ∂ρ 1
ρ
ρ ( z sin φ + 2 ρ ) +
( z sin φ + 4 ρ ) −
= 4 + 2 cos 2 φ − (c)
∇2 f =
2z2
ρ2
ρ2
ρ
2
(− ρ z sin φ − 2 z 2
∂ ∂ sin φ cos φ ) + ( ρ sin φ + 2 z cos 2 φ ) ∂ρ ∂z
( z ρ sin φ + 2 z 2 cos 2φ ) + 2 cos 2 φ .
cos 2φ .
1 ∂ 21 1 ∂ [r cos θ sin φ + 2r 3φ ] + 2 [− sin 2 θ sin φ ln r ] 2 r ∂r r r sin θ ∂ θ +
=
1
1
1 [− cos θ sin φ ln r ] r sin 2 θ 2
1 cos θ sin φ (1 − 2 ln r − csc2 θ ln r ) + 6φ 2 r
Copyright © 2015 by Oxford University Press POESM_Ch03.indd 42
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
43 P.E. 3.12 If B is conservative , ∇ × B = 0 must be satisfied.
∇× B =
ax
ay
az
∂ ∂x y + z cos xz
∂ ∂y x
∂ ∂z x cos xz
47
Prob. 3.11 = 0 a x + (cos xz − xz sin xz1− cos xz + xz sin2 xz ) a y + (1 − 1) a z = 0 2 z 2 zdxdz dx zdz (1) ψ = AB dSis=a conservative = = =2 Hence field. 0 2 S 0 0 P.E. 3.13 ψ = B dS, dS = dxdya z Prob.S 3.1 (a) 2 1 2 x3 1 y 2 2 ψ = d S3x= ρydxdy dφ dz= 3 3 0 2 0 = (1)(2) = 2 y =0 x =0 5
Prob. 3.12 S = d S = ρ dφ dz = 2 dz 0 (a) = dxdydz P.E.dv 3.14
π 2
π
π dφ
π
= 2(5)[ − ] = 2 3
10π = 5.236 6
3
(b) 2 1 1 1 1 z ���� dSxydxdydz d=φ� xdx cylindrical, = ρ d�ρ �� Inxydv Cartesian coordinates ��� � = ydy dz� � � � 3 � 1�. v
z =0 y =0 z =0 3 2 2
π
0
0
0
3 π ρin2 cylindrical For representing this vector coordinates, we use eq. (2.13): 2 = field ( ) = 3.142 S = d S x= 1 ρ yd ρ1 dφ 4
z �= (1/ 0.5� 0 3� � � �� 1� =s�� 4 2)(1/ 2cos2)(2) 2 10 2 0 0 0 � �� θs�� � � (c) In spherical, d �S��=� r�2 sin dφ�dθcos � 0� � 0 0 1 �� 0 2π 2π (b) � � π 2 3 s�� �� � �cos ������ � � �cos � �3� � ��� � s�� � ������ � ��3� � ����� ρ dφ dz sin θ dθ dφ = 100 (2π )(− cos θ ) 3| = 200 π (0.5 + 0.7071) = 758.4 S = dvd=Sρ=d 100 =
π π π π Using the following relationships, from eq.zdz (2.8),dφ�� � � cos ��, ρ zdv = ρ z ρ d ρ d φ dz = ρ d ρ ρ obtain φ 2
(d)
3
4
v
= 0 z = 0 =1
0
3
2
1
2
4
0
0
� � s�� ��, we
d S = r3dr3 dθ 2 2 ρ z 1 = π (π ) = (9 − )(2π ) = 54.45 � � �cos ��3�� ��� � �s�� ���� cos ��� �� 4 2 �� 0 2 � ��� s�� 3 1cos r 2 4 3π π 8π S = dS = rdr dθ = |0�( − ) = = 4.189 �0 ���3�� cos2�� � s�� ��� � �cos ���� cos ��� �� 2 ��� 3 s�� ����� 6 π Prob. 3.13 � � 3 ������� � �3� cos � � �� s�� � cos � � � s�� � cos ���� Let I = Adv = r�sin φ�ar dv � � v
� ��3� cos � s�� � � ��s�� � � �cos ���� v
ar = sin θ cos φ a x + sin θ sin φ a y + cos θ a z
� in cylindrical coordinates is �5, 53.13°, 5�
Ar = r sin θ sin φ cos φ a x + r sin θ sin 2 φ a y + r cos θ sin φ a z dv = r 2 sin θ1dθ∂dφ dr ��� �
POESM_Ch03.indd 43
1 ∂�� ∂�� � � ∂� � ∂� ∂z 1 ������������ ���� cos� � � �� s�� � cos �� � 1 � ���� cos �s��� � � �3�� cos� �� Copyright ©�2015 by Oxford University Press � �� s�� � cos � � �� s�� � cos �� ���� � �
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Sadiku & Kulkarni
� �cos ��3�� cos ��� � ��� s�� ��� � �s�� ���� cos ��� ��
Principles of Electromagnetics, 6e
� ���3�� cos ��� � ��� s�� ����� s�� ��� � �cos ���� cos ��� ��
������� � �3�� cos� � � �� s�� � cos � � � s�� � cos ���� 44 � ��3�� cos� � s�� � � ��s��� � � �cos� ���� � in cylindrical coordinates is �5, 53.13°, 5�
1 ∂�� ∂�� 1 ∂ ���� � � � � ∂� � ∂� ∂z 1 ������������ ���� cos� � � �� s�� � cos �� � 1 � ���� cos �s��� � � �3�� cos� �� � � �� s�� � cos � � �� s�� � cos �� � 1�.00 (after substituting the values of � and �) ��� �
For representing the given vector field in spherical coordinates we use eq. (2.27): �� s�� � cos � s�� � s�� � ��� � � �cos � cos � cos � s�� � �� � s�� � cos �
cos � 3� � � �� � � s�� �� � � 0 0
� � ��s�� � cos ���3� � � ��� � �s�� � s�� ������ � ��cos � cos ���3� � � ��� � �cos � s�� ������ � ���� s�� ���3� � � ���� � ��cos ����� ��
Using the following relationships, from eq. (2.22), � � � s�� � cos ��, � � � s�� � s�� ��, we obtain � � ��3� � sin� �cos� �� � �3�sin� � cos � sin ����� � ��3� � cos �sin� �cos� �� � ��� sin � cos � cos � sin �� � � � � �3�sin � � ��3� sin � �cos ����cos sin ����� �� sin � cos��cos sin � ���� � � � � ��3� � cos �cos �� � ��� sin � cos � cos � sin �� � �sin � � � ���3� sin �cos � sin �� � ���� sin �sin� �� � �� sin � cos � cos � sin ����� � �� sin �� cos�� ������ � ���3� sin �cos � sin �� � ���� sin �sin� �� � �� sin � cos� �����
� in spherical system is �7.071, 45° , 53.13° �
1 ��� 1 ∂ system is �7.071, � 1 � in spherical 45° , 53.13° � �� sin � � � � � � � �� � �� � � � sin � �� � �� � sin � �� � � � � � � � � ��3� �1�� 1sin∂ �cos � � �� 1sin � cos � sin ��+ 1 �� sin� �cos� � � � �� sin � � � ��� � �� �� � � � �� � �� � � �sin�� ��� � cos� � �� 3�sin sin � �� � � �� cos��� sin � cos � sin � ��sin �� �cos� � cos � sin �� � � � � � � � � ��3� � sin� �cos� � � ���1�� sin �cos � � �� sin � cos � sin ��+ � � ��� ��� � sin� �sin� � cos � � 3� � sin� �cos� � � 4��sin � cos � sin � � � ��� � �� � cos� � sin� � cos�� � � 3�sin� � cos � sin � � �� sin �cos� � cos � sin �� � ���sin � cos � sin � � 1� ������ s��s�i���in� ��� �����s o� �, � �n� �� ��� � sin� �sin� � cos � � 3� � sin� �cos� � � 4� sin � cos � sin � � � ��� �
Note: this it is clear that the divergence of a vector field is the �� sin From � cos � sinexample, �� � 1� ������ s��s�i���in� ��� �����s o� �, � �n� �� same irrespective of the coordinate system used. Note: From this example, it is clear that the divergence of a vector field is the same irrespective of the coordinate system used. Copyright © 2015 by Oxford University Press POESM_Ch03.indd 44
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1 ∂ � 1 ��� 1 � �� �� � � �� sin � � � � �� � sin � �� � sin � �� Principles of Electromagnetics, 6e � � � � � � 43��+ � ��3� � sin� �cos� � � � �1�� sin �cos � � �� sin � cos � sin
��� �
Sadiku & Kulkarni
�
� ��� �
�� � cos� � sin� � cos� � � 3�sin� � cos � sin � � �� sin �cos� � cos � sin �� �
� 45 P.E. 3.12 ��� � sin� �sin� � cos � � 3� � sin� �cos� � � 4� sin � cos � sin � � � ��� � If B is conservative � , ∇ × B = 0 must be satisfied.
�� sin � cos � sin � � 1� ������ s��s�i���in� ��� �����s o� �, � �n� ��
ax
ay
az
y + z cos xz
x
x cos xz
Note: From this example, it∂is clear∂that the divergence of a vector field is the ∂ ∇ ×irrespective B= same of the coordinate ∂x ∂y ∂system z 50used.
P.E. 3.15 ∇ F = a x − 2a y + a z = 0 a x + (cos xza −−xz2asin+xza − cos xz + xz sin xz ) a y + (1 − 1) a z = 0 ∇F x y z 0.4082a x − 0.8165a y + 0.4082a z aHence n = B is=a conservative=field. | ∇F | 1+ 4 +1 Prob. 3.18
Method 1: Prob. ∂3.1 T 1 ∂T 1 ∂T ∇T = ar + aθ + aφ (a) ∂dr S = ρrd∂φθdz r sin θ ∂φ = sin θ cos φ ar + cos θ cos φ aθ − sinπφ aφ 5
2 o π π 10π At P, Sr == 2,dθ S= =60ρo , φ d=φ30 dz = dz 2 o o 0o π dφo = 2(5)[ o2 − 3 ] = 6 = 5.236 ∇T = sin 60 cos 30 ar + cos 60 cos 330 aθ − sin 30 aφ
(b) = 0.75ar - 0.433aθ − 0.5aφ In cylindrical, dS = ρ d ρ dφ 2 | ∇T |= 0.752 + 0.433 + 0.52 = 1 π 3 2 3 π itsρ direction The magnitude of T4is 1 and is along ∇T. S = d S = ρ d ρ dφ = ( ) = 3.142 4 2 1 1 0 Method 2: (c) In spherical, d S = r 2 sin θ dφ dθ T = r sin θ cos φ = x ∇T = a x S = d S = 100 | ∇T |= 1
2π 3
2π
π
0
2π 3
sin θ dθ dφ = 100 (2π )(− cos θ ) π| 4
= 200π (0.5 + 0.7071) = 758.4
4
(d) 3.19 Prob. ∂fd S = ∂rfdr dθ ∂f ∇f = a x + a y + πa z = (2 xy − 2 y 2 )a x + ( x 2 − 4 xy )a y + 3 z 2 a z ∂x ∂y 4 ∂z 2 r2 4 π π 8π = dS = x =rdr = z =|0−(3 − ) = = 4.189 At pointS(2,4,-3), 2, yd=θ 4, 2 2 3 6 π 0 ∇f = (16 − 32)a x + (4 − 32) 3 a y + 27a z = −16a x − 28a y + 27 a z a x + 2a y − a z
1 (1, 2, −1) 1+ 4 +1 6 The directional derivative is 1 99 ∇f a = (−16, −28, 27) (1, 2, −1) = − = −40.42 6 6 a=
=
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
46 44 Prob. 3.2 (a) dl = ρ dφ ;
ρ=3 π 2
3π π π L = dl = 3 dφ = 3( − ) = = 2.356 2 4 4 π 4
(b)
dl = r sin θ dφ;
r = 1, θ = 30° ; π 3
L = dl = r sin θ
(c)
π
dφ = (1) sin 30° [( ) − 0] = 3
0
0.5236.
dl = rdθ π
L=
π π 4π d l = r dθ = 4( − ) = = 4.189 2 6 3 π 2
6
Prob. 3.3 2
S = dS =
π /2
φ π
L = dl = L
π /6
ρ = 10
z =0 = / 4
2
π /2
0
π /4
= 10 dz Prob. 3.4
ρ dφ dz
ρ dφ
φ =0
dφ = 10(2)(π / 2 − π / 4) = 5π = 15.71
= 4(π / 6) = 2.094
ρ =4
Prob. 3.5 (a ) dV = dxdydz 1
2
3
0
1
−3
V = dxdydz = dx dy dz = (1) (2 − 1)(3 − − 3) = 6
(b) dV = ρ dφ d ρ dz 5
4
π
2
−1
π
V = ρ d ρ dz dφ =
ρ2
| 2
5 2
π 1 2π (4 − −1)(π − ) = (25 − 4)(5)( ) = 35π = 110 3 2 3
3
Copyright © 2015 by Oxford University Press POESM_Ch03.indd 46
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
47 45
(c) dV = r 2 sin θ drdθ dφ π 3
π 2
π 2
π 6
2
3
π /3 π r3 3 π ( − cos θ ) ( − ) | | π /2 2 3 1 6
V = r 2 dr sin θ dφ = 1
=
1 1 π 26 π (27 − 1)( )( ) = = 4.538 3 2 3 18
Prob. 3.6
H •dl = ( x dx + y 2
2
dy )
L
But on L, y = x 2 dy = 2 xdx 1
2 4 H •dl = ( x + x .2 x)dx = L
Prob. 3.7 (a)
0
1
F • dl = ( x y =0
2
− z )dy|
x3 x6 1 1 1 + 2 | = + = 0.6667 3 6 0 3 3
2
x = 0, z = 0
+
x=2
2 xydx |
x =0
y =1, z = 0
+
z =3
(−3xz
z =0
2
)dz|
x = 2, y =1
x2 2 z3 3 − 3(2) | | 2 0 3 0 = 0 + 4 − 54 = − 50 = 0 + 2(1)
(b)
Let x = 2t. y = t , z = 3t dx = 2dt , dy = dt , dz = 3dt ; 1
F • dl = (8t
2
− 5t 2 − 162 t 3 ) dt
0
1 = (t 3 − 40.5t 4 ) = −39.5 0
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
48 46 Prob. 3.8 W = F • dl = L
π /4
z ρ dφ
+ z = 0, ρ = 2
φ =0
3
ρ cos φ dz ρ
= 2,φ =π / 4
z =0
= 0 + 2 cos(π / 4)(3) = 6 cos 45o = 4.243 J
Prob. 3.9
H • dl =
0
( x − y )dx
x =1
y = 0, z = 0
+ ( x 2 + zy )dy + 5yzdz 0
2
1
0
= xdx + ( y −
+
1
5 yzdz x = 0, y = 0
z =0
x = 0, z = 1 − y / 2 0
y2 )dy + (10 z − 10 z 2 )dz 2 1
= −1.5
Prob. 3.10
Method 1:
B ⋅ dl = − L
1
y =0
But z = y
B ⋅ dl
yzdy
1
xzdz
z =0
x =1
+ (− yzdy + xzdz )
x =1
⎯⎯ → dz = dy on the last segment (or integral).
= 0+
L
=
z=0
+
0 z2 1 1 y3 y 2 0 + (− y 2 + y )dy = + (− + ) 2 0 y =1 2 3 2 1
1 1 1 1 + − = = 0.333 2 3 2 3
Method 2:
B ⋅ dl = ∇ × B dS L
S
∂ ∇ × B = ∂x xy
∇ × B dS = S
∂ ∂y -yz 1
∂ ∂z = ya x − za y − xa z , xz y
y =0 z =0
1
ydzdy = y 2 dy = 0
dS = dydza x
y3 1 1 = = 0.333 3 0 3
Copyright © 2015 by Oxford University Press POESM_Ch03.indd 48
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Sadiku & KulkarniProb.
3.11
Principles of Electromagnetics, 6e 1
2
0
0
ψ = A dS = zdxdz = dx zdz = (1) S
z 2 =2 2 0 47 49 2
P.E. 3.13 Prob. 3.11 ψ = B dS, dS = dxdya1z 2 z2 2 ψ = S A dS = zdxdz = dx zdz = (1) =2 2 1 3 2 0 2 1 2 S 0 0 x y ψ = 3x 2 ydxdy = 3 = (1)(2) = 2 0 0 3 2 y =0 x =0 P.E. 3.13 ψ = B dS, dS = dxdya z Prob. 3.12 S (a) dv = dxdydz 2 1
ψ=
xydv = y =0 x =0
v
x3 1 y 2 2 = (1)(2) =2 z 3 0 12 0 1 xydxdydz = xdx ydy dz
2 =3 23x1 ydxdy 1 z =0 y =0 z =0
0
0
0
1
1
z
Prob. 3.122 1 y2 1 2 (a) dv == xdxdydz z = (1/ 2)(1/ 2)(2) = 0.5 2 0 2 0 0 2
1
1
v xydv = z =0 y =0 z =0 xydxdydz = 0 xdx 0 ydy 0 dz (b) dv = ρ d ρ dφ dz x 2 1π y22 13 2 3 2 π = z = (1/ 2)(1/ 2)(2) 2= 0.5 ρ zdv = ρ z ρ d ρ d φ dz = ρ d ρ zdz 2 0 2 0 0 dφ φ = 0 z = 0 ρ =1
v
(b)
1
0
0
ρ 3 3 z2 2 1 = (π ) = (9 − )(2π ) = 54.45 dv = ρ d ρ d3φ dz 1 2 0 3 π
2
3
3
2
π
zdv = ρ z ρ d ρ dφdz = ρ d ρ zdz dφ v ρ3.13 Prob. φ = 0 z = 0 ρ =1 1 0 0 Let I = A3dv = 2r sin φ ar dv ρ 3 zv 2 1 =v (π ) = (9 − )(2π ) = 54.45 2 θ0 sin φ a y + cos 3 θ az ar = sin θ cos φ3a x1+ sin 2
Ar = r sin θ sin φ cos φ a x + r sin θ sin 2 φ a y + r cos θ sin φ a z Prob. 3.13 dv = r 2 sin dφ = dr r sin φ a dv θdv Let I =θ dA r v
v
ar = sin θ cos φ a x + sin θ sin φ a y + cos θ a z Ar = r sin θ sin φ cos φ a x + r sin θ sin 2 φ a y + r cos θ sin φ a z dv = r 2 sin θ dθ dφ dr
Copyright © 2015 by Oxford University Press POESM_Ch03.indd 49
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
50 48 I = Adv = r 3 sin 2 θ sin φ cos φ dθ dφ dra x v
+ r 3 sin 2 θ sin 2 φ dθ dφ dra y + r 3 sin θ cos θ sin φ dθ dφ dra z = ax
1
π
r dr sin 3
r =1
0
+a y
1
r =1
+a z
1
2
θ dθ sin φ cos φ dφ 2π
0
0
π
4
2π
r dr sin θ cos θ dθ sin φ dφ 3
0
= 0a x + 0a z + a y ay
0
π
r 3 dr sin 2 θ dθ sin 2 φ dφ
r =1
=
2π
0
2π 1 r 1 1 (1 cos 2 ) (1 − cos 2φ )dφ d − θ θ 4 00 2 2 0
(π / 2)(π ) =
π
4
π2 8
a y = 1.234a y
Prob. 3.14 ∂V1 ∂V ∂V a x + 1 a y + 1 az ∂x ∂y ∂z = (6 y − 2 z )a x + 6 xa y + (1 − 2 x )a z
(a) ∇V1 =
∂V2 ∂V 1 ∂V2 aρ + aφ + 2 a z ∂ρ ∂z ρ ∂φ = (10 cos φ − z )a ρ − 10sin φ aφ − ρ a z
(b) ∇V2 =
∇V3 =
(c)
∂V3 1 ∂V3 1 ∂V3 ar + aθ + aφ ∂r r ∂θ r sin θ ∂φ 2 1 2 cos φ ar + 0 + − sin φ aφ 2 r r sin θ r 2 2sin φ aφ = − 2 cos φ ar − 2 r r sin θ =−
Copyright © 2015 by Oxford University Press POESM_Ch03.indd 50
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Sadiku & Kulkarni
Principles of Electromagnetics, 6e
51 49 Prob. 3.15
(a)
∇U =
∂U ∂U ∂U ax + ay + az ∂x ∂y ∂z
= e x + 2 y cosh za x + 2e x + 2 y cosh za y + e x + 2 y sinh za z ∂T 1 ∂T ∂T aρ + aφ + az ρ ∂φ ∂ρ ∂z 3z 3z 3 = − 2 cos φ a ρ − 2 sin φ aφ + cos φ a z
∇T = (b)
ρ
∇W = (c)
ρ
ρ
∂W 1 ∂W 1 ∂W ar + aθ + aφ ∂r r ∂θ r sin θ ∂φ
5sin φ 1 5cos θ = − + 4r sin φ ar + 2r 2 cos φ − aφ 2 r r r sin θ
Prob. 3.16
r = x2 + y 2 + z 2 ,
r n = ( x 2 + y 2 + z 2 ) n /2
Method 1: ∇r n =
∂r n ∂r n ∂r n n ax + ay + a z = ( x 2 + y 2 + z 2 ) n /2−1 (2 x)a x + ∂x ∂y ∂z 2
= n( x 2 + y 2 + z 2 )
n−2
2
( xa x + ya y + za z ) = nr n − 2 r
Method 2: ∂r n r n ∇r = ar = nr n −1 = nr n − 2 r ∂r r Prob. 3.17 ∇ T = 2x a x + 2 y a y − a z At
(1,1, 2 ) , ∇ T = (2, 2, −1).
The mosquito should move in the direction of
2ax + 2a y − az
Copyright © 2015 by Oxford University Press POESM_Ch03.indd 51
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Sadiku & KulkarniP.E.
3.15 ∇ F = a x − 2a y + a z an =
Principles of Electromagnetics, 6e
a − 2a y + a z ∇F 52 a + 0.4082a = x = 0.4082a x − 0.8165 y z | ∇F | 1+ 4 +1
Prob. 3.18
Method 1: ∂T 1 ∂T 1 ∂T ∇T = ar + aθ + aφ ∂r r ∂θ r sin θ ∂φ = sin θ cos φ ar + cos θ cos φ aθ − sin φ aφ At P, r = 2,θ = 60o , φ = 30o ∇T = sin 60o cos 30o ar + cos 60o cos 30o aθ − sin 30o aφ = 0.75ar - 0.433aθ − 0.5aφ | ∇T |= 0.752 + 0.4332 + 0.52 = 1 The magnitude of T is 1 and its direction is along ∇T. Method 2: T = r sin θ cos φ = x
∇T = a x | ∇T |= 1 Prob. 3.19 ∂f ∂f ∂f ∇f = a x + a y + a z = (2 xy − 2 y 2 )a x + ( x 2 − 4 xy )a y + 3 z 2 a z ∂x ∂y ∂z At point (2,4,-3), x = 2, y = 4, z = −3 ∇f = (16 − 32)a x + (4 − 32)a y + 27a z = −16a x − 28a y + 27a z
a x + 2a y − a z
1 (1, 2, −1) 1+ 4 +1 6 The directional derivative is 1 99 (1, 2, −1) = − = −40.42 ∇f a = (−16, −28, 27) 6 6 a=
=
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Principles of Electromagnetics, 6e
53 51 Prob. 3.20
(a) Let f = ax + by + cz – d = 0 ∇f = aa x + ba y + ca z aa + ba y + ca z ∇f = x | ∇f | a 2 + b2 + c2 Let g = α x + β y + γ z − δ ∇g α a x + β a y + γ a z an 2 = = | ∇g | α2 + β2 +γ 2 an1 =
cos θ = an1 an 2 =
θ = cos −1
aα + bβ + cγ a 2 + b2 + c 2 α 2 + β 2 + γ 2 aα + bβ + cγ
(a + b 2 + c 2 )(α 2 + β 2 + γ 2 ) 2
a = 1, b = 2, c = 3 (b) α = 1, β = 1, γ = 0
θ = cos −1
1+ 2 + 0 (12 + 22 + 32 )(12 + 12 + 02 )
= cos −1
3 = cos −1 0.5669 = 55.46o 28
Prob. 3.21
∂Ax ∂Ay ∂A z + + = 3y − x ∂x ∂y ∂z 1 1 ∇ ⋅ B = 2 ρ z 2 + ρ 2sin φ cos φ + 2 ρ sin 2 φ
(a) ∇ ⋅ A =
(b)
ρ
ρ
= 2 z + sin 2φ + 2 ρ sin 2 φ 2
(c) ∇ ⋅ C =
1 2 3r + 0 = 3 r2
Prob. 3.22
∂Ax ∂Ay ∂Az + + = 2 xy + 0 + 2 y = 2 y (1 + x) ∂x ∂y ∂z (a) At (−3, 4, 2), x = −3, y = 4 ∇ A = 2(4)(1 − 3) = −16 ∇ A =
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Principles of Electromagnetics, 6e
54 52
∇ B =
1 ∂ 1 ∂Bφ ∂Bz 1 ∂ ρ Bρ + 3ρ 2 sin φ + 0 + 8 z cos 2 φ + = ρ ∂ρ ρ ∂φ ρ ∂ρ ∂z
(
)
(
)
= 6sin φ + 8 z cos 2 φ
(b) At
(5,30o ,1), z = 1, φ = 30o
∇ B = 6sin 30o + 8(1) cos 2 30o = 3 + 6 = 9 1 ∂ 2 1 ∂Cφ 1 ∂ 4 ( r Cr ) + 0 + (r cos θ ) + 0 = 4r cos θ = 2 2 r ∂r r sin θ ∂φ r ∂r (c) At (2, π / 3, π / 2), r = 2, θ = π / 3 ∇ C = 4(2)cos(π / 3) = 4 ∇ C =
Prob. 3.23
2
∇• H = k ∇• ∇ T = k ∇ T 2
∇ T=
∂ 2T ∂ 2T πx π y π2 π2 h 50sin cos (− + = + )= 0 2 2 4 4 ∂ x2 ∂ y 2
Hence, ∇• H = 0 Prob. 3.24 We convert A to cylindrical coordinates; only the ρ-component is needed. Aρ = Ax cos φ + Ay sin φ = 2 x cos φ − z 2 sin φ
But x = ρ cos φ , Aρ = 2 ρ cos 2 φ − z 2 sin φ
Ψ = A ⋅ dS = Aρ ρ dφ dz = 2 ρ 2 cos 2 φ − ρ z 2 sin φ dφ dz S
= 2(2) 2
π /2
0
1
1
1 (1 + cos 2φ )dφ dz − 2 z 2 dz 2 0 0
π /2
sin φ dφ 0
π /2 π /2 1 z3 1 (− cos φ ) = 4(φ + sin 2φ ) −2 = 2π − 2 / 3 = 5.6165 0 0 2 3 0
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Principles of Electromagnetics, 6e
55 53 Prob. 3.25
z
Z=1
y
Z=-1
x
(a)
D • dS = [
z =−1
+ + z =1
ρ
]D • dS
=5
= − ρ 2 cos 2 φ dφ d ρ + ρ 2 cos 2 φ dφ d ρ + 2 ρ 2 z 2 dφ dz| = 2(5) 2
2π
1
0
−1
z
dφ z dz = + 50(2π ) ( 3 2
3
ρ =5
|1−1 )
200 π = 209.44 3 1 ∂ (2 ρ 2 z 2 ) = 4 z 2 (b) ∇ • D = =
ρ ∂ρ
1
5
2π
−1
0
0
2 2 ∇ • Ddv = 4 z ρ d ρ dφ dz = 4 z dz ρ d ρ dφ
= 4x
3
z 3
1
−1
ρ
5
2
2
(2π ) = 0
200π = 209.44 3
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Principles of Electromagnetics, 6e
56 54 Prob. 3.26
H dS = S
π / 2 2π
10 cosθ r θ φ
2
sin θ dθ dφ
=0 =0
2π
π /2
0
0
= 10(1)2 dφ
sin θ = 20π 2 2
r =1 π /2
sin θ cos θ dθ = 10(2π )
sin θ d (sin θ ) 0
π /2 = 10π = 31.416 0
Prob. 3.27 H ⋅ d S = ∇ ⋅ Hdv S
v
H ⋅ dS = − 2 xydydz x = 0 + 2 xydydz x = 1 − ( x
2
+ z 2 )dxdz
S
+ ( x 2 + z 2 )dxdz
y=2
− 2 yzdxdy
2
3
1
2
1
2
1
−1
0
1
0
1
z = −1
+ 2 yzdxdy
y =1 z =3
= 0 + 2 ydy dz + 2 dx ydy + 6 dx ydy = 12 + 3 + 9 = 24
∇⋅H =
∂H x ∂H y ∂H z + + = 2y + 0 + 2y = 4y ∂x ∂y ∂z 1
2
3
0
1
−1
∇ ⋅ Hdv = 4 ydxdydz = 4 dx ydy dz
V
= 4(1)
y 2 (3 + 1) = 24 2 1 2
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Principles of Electromagnetics, 6e
57 55 Prob. 3.28 ψ2
Side 1:
ψ3
ψ1
ψ = D dS =ψ 1 +ψ 2 +ψ 3 S
= 0+
2π
3
10 z × ρ dφ d ρ
φ =0 ρ =0
z=4
+
4
2π
5ρ × ρ dφ dz ρ = 3
z =0 φ =0
ρ2 3 = 10(4)(2π ) + 5(9)(2π )(4) = 360π + 360π = 2261.95 2 0
Side 2:
ψ = ∇ Bdv,
1 ∂ (5 ρ 2 ) + 0 + 10 = 10 + 10 = 20 ρ ∂ρ
∇ B =
v
2π
ψ = 20dv = 20
4
3
φ =0 z =0 ρ =0
ρ2 3 2 0
ρ dφ d ρ dz = 20(2π )(4)
= 720π = 2261.95
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Principles of Electromagnetics, 6e
58 56 Prob. 3.29 Let ψ = A dS = ∇ Adv v
S
∂Ay
∂A ∂Ax + + z = 2( x + y + z ) ∂x ∂y ∂z ∇ A = 2( ρ cos φ + ρ sin φ + z ), dv = ρ d ρ dφ dz ∇ A =
ψ = 2( ρ cos φ + ρ sin φ + z ) ρ d ρ dφ dz 2π
2ρ 2 1 z 2 4 1 (2π ) = (16 − 4)(2π ) = 0 + 0 + 2 ρ d ρ zdz dφ = 2 0 2 2 2 0 2 0 1
4
= 12π = 37.7 Prob. 3.30
∇• A =
1 ∂ 1 ∂ (r 4 ) + (r sin 2 θ cos φ ) 2 r ∂r r sin θ ∂θ
= 4 r + 2 cos θ cos φ
∇ • Adv = 4r
3
sin θ dθ dφ dr + 2r 2 sin θ cos θ cos φ dθ dφ dr
π π /2 π r4 3 2r 3 3 cos 2 θ π / 2 ( cos ) ( ) ( ) sin θ φ − + − | |0 2 3 |0 2 |0 |02 4 0 1 π = 81(1)( ) + 18(0 + )(1 − 0) 2 2 81π = + 9 = 136.23 2
=4
z
y
x
A • dS = [φ +φ π + +θ π =0
= /2
r =3
] A • dS
= /2
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Principles of Electromagnetics, 6e
59 57
Since A has no φ − component, the first two integrals on the right hand side vanish.
A • dS =
π /2
π /2
φ θ =0
=0
r 4 sin θ dθ dφ|
π
3
+
r =3
r =0
π /2
π /2
φ
=0
r 2 sin 2 θ cos φ drdφ|
θ =π / 2
π /2
= 81 ( ) (− cos θ )| + 9(1) sin φ| 0 0 2 81π = + 9 = 136.23 2 Prob. 3.31 Let ψ =
F • dS = ψ
t
+ ψ b + ψ o +ψ i
where ψ t , ψ b , ψ o , ψ i are the fluxes through the top surface, bottom surface, outer surface ( ρ = 3), and inner surface respectively. For the top surface, dS = ρ dφ d ρ a z ,
z = 5;
F • dS = ρ 2 z dφ dz. Hence:
ψt =
3
ρ
=2
2π
ρ 2 z dφ dz|z =5 =
φ
=0
For the bottom surface,
190 π = 198.97 3
z = 0, dS = ρ dφ d ρ (− a z )
F • dS = − ρ 2 z dφ d ρ = 0. Hence, ψ b = 0. For the outer curved surface, ρ = 3, dS = ρ dφ dz a ρ F • dS = ρ 2 sin φ dφ dz. Hence,
ψa =
5
2π
dz ρ φ sin φ dφ|ρ 3
z =0
=0
=3
=0
For the inner curved surface, ρ = 2, d S = ρ dφ dz ( −a ρ ) F • dS = − ρ 3 sin φ dφ dz. Hence,
ψa = −
5
z =0
ψ =
dz ρ 3
2π
sin φ dφ|ρ
φ =0
=2
=0
190 π 190 π + 0+0+ 0= = 198.97 3 3
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Principles of Electromagnetics, 6e
60 58
ψ=
F • dS = ∇ • FdV
∇•F =
1 ∂
ρ∂ρ
= 3 ρ sin φ −
V
z
ρ
( ρ 3 sin φ ) +
1 ∂
ρ ∂φ
( z cos φ ) + ρ
sin φ + ρ z
(3ρ sin φ − ρ sin φ + ρ ) ρ dφ d ρ dz
∇ • Fdv = 5
2π
3
0
0
2
= 0 + 0 + dz dφ ρ 2 d ρ =
190 π = 198.97 3
Prob. 3.32
ax ∂ (a) ∇xA = ∂x xy
ay ∂ ∂y y2
az ∂ = za y − xa z ∂z − xz
1 1 ∇xB = 2 ρ z 2sin φ cos φ − 0 a ρ + (2 ρ z − 2 z sin 2 φ )aφ + ( 2 ρ sin 2 φ − 0 ) a ρ ρ (b) = 4 z sin φ cos φ a ρ + 2( ρ z − z sin 2 φ )aφ + 2sin 2 φ a z
= 2 z sin 2φ a ρ + 2 z ( ρ − sin 2 φ )aφ + 2sin 2 φ a z ∇xC = (c)
1 r sin θ
1 ∂ 2 ∂ 2 2 ∂θ (r cos θ sin θ ar − r ∂r (r cos θ aθ
cos 2 θ r (2 cos θ )(− sin θ ) sin θ + cos θ (cos 2 θ ) ar − (2r )aθ r sin θ r (cos3 θ − 2sin 2 θ cos θ ) = ar − 2 cos 2 θ aθ sin θ
=
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Principles of Electromagnetics, 6e
61 59
Prob. 3.33 (a) ax ∂ ∇× A = ∂x x2 y
ay az ∂ ∂ = ∂y ∂z y 2 z −2 xz
− y2 a x + 2z a y − x2 a z
∇ •∇× A = 0 (b) ∇× A = (
1 ∂ Az
ρ ∂φ
−
∂ Aφ ∂ A ∂ Az 1 ∂ ( ρ Aρ ) ∂ Aρ − ) aρ + ( ρ − ) aφ + ( )az ∂z ∂z ∂ρ ρ ∂ρ ∂φ
= (0 − 0) a ρ + ( ρ 2 − 3z 2 ) aφ +
1
ρ
(4 ρ 3 − 0) a z
= ( ρ 2 − 3z 2 ) aφ + 4 ρ 2 a z ∇•∇× A = 0
(c)
1 sin φ 1 −1 cos φ 1 ∂ cos φ ) − 0 aθ + ( − 0 aφ 0 − 2 ar + 2 r sin θ r r sin θ r r ∂r r sin φ cos φ cos φ ar + 3 aθ + 3 aφ =− 3 r sin θ r sin θ r
∇•∇× A =
− sin φ sin φ +0+ 4 =0 4 r sin θ r sin θ
∇× A =
∇ •∇× A= 0
Prob. 3.34 ∇ × H = 0a ρ + 1aφ +
1
ρ
(2 ρ cos φ − ρ cos φ )a z = aφ + cos φ a z
1 1 1 1 ∇ × ∇ × H = − sin φ − 0 a ρ + 0aφ + (1 − 0)a z = - sin φ a ρ + a z ρ ρ ρ ρ
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Principles of Electromagnetics, 6e
62 60
Prob. 3.35 Method 1: We can express A in spherical coordinates. a r A = 3 ar = 2r , r r −2 a 1 ∇ × A = ∇ × 2r = ∇ 2 × ar = 3 ar × ar = 0 r r r
Method 2: x y z a + 3 a y + 3 az 3 x r r r ∂ ∂ ∂ 3 ∂x ∂y ∂z 3 ∇× A = = − z ( x 2 + y 2 + z 2 ) −5/ 2 (2 y ) − − y ( x 2 + y 2 + z 2 ) −5 / 2 (2 z ) a x + ... 2 x y z 2 3 3 3 r r r =0 A=
Prob. 3.36
y 1 1
2 3
0
1
2
x
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Principles of Electromagnetics, 6e
63 61
(a)
F • d l = ( L
+ + ) F • d l
1
2
3
_
_
_
For 1, y = x dy = dx, dl = dx a x + dy a y , 1
F •d l = 1
x 3 dx − xdx = −
0
1 4 _
_
_
For 2, y = − x + 2, dy = − dx, dl = dx a x + dy a y , 2
3 2 F • d l = (− x + 2 x − x + 2)dx = 2
1
17 12
For 3, 0
F •d l = x 3
2
2
1
F •d l = − 4
ydx| +
L
y =0
=0
17 7 + 0 = 12 6
(b) ∇ × F = − x2 a z ;
dS = dxdy (− a z ) 1 x
2 (∇ × F ) • d S = − (− x )dxdy =
=
1
2 x y| dx +
x
0
0
2 x dydx + 0 0
2
2 x y| 1
− x+2
0
dx =
2
− x+2
1
y =0
x1 | + 4 0
x 2 dydx 2
x (− x + 2)dx = 2
1
7 6
(c) Yes
Prob. 3.37
A • dl = =
1
ρ
=2
ρ sin φ d ρ
φ =0
+
π /2
φ
=0
ρ 2 ρ dφ
ρ =1
+
2
ρ
ρ sin φ d ρ
=1
φ = 90
+ o
0
φ π
= /2
ρ 3 dφ
ρ =2
π
1 π + (4 − 1) + 8(− ) = −9.4956 2 2 2
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Principles of Electromagnetics, 6e
64 62 62 62 63
Prob. Prob. 3.38 Prob.3.38 3.38 (c)
63
1 y∂yy 1 ∂ cos θ cos φ 1 sin θ cos φ ∇ ⋅∇V = ∇ V = 2 (− sin θ cos φ ) + 2 (sin θ ) + 2 2 (− ) r ∂r r sin θ ∂θ r r sin θ r cos φ(c) cos φ = 0+ 3 (1 − 2sin 2 θ ) − 3 1 ∂ 1 ∂ cos θ cos φ 1 sin θ cos φ r sin θ ∇ ⋅∇V = ∇ 2V = r2 sin (θ− sin θ cos φ ) + 2 (sin θ ) + 2 2 (− ) o o o r ∂r r sin θ ∂θ r r sin θ r 2sin θ cos φ 45 45 45 =− cos φ cos φ r3 = 0+ 3 (1 − 2sin 2 θ ) − 3 r sin θ 222 000 xxxr sin θ 2sin θ cos φ Prob. 3.41 =− r3 r 2
2 22 0 00 r sin θ [(cos φ − sin φ ) a xπ+π/4π/4/4(cos φ + sin φ ) a y ] r sin θFF +++ 333zzzsin sin +++ 222ρρρzd zd zdρρρ sinφφφρρρdddφφφ zdρρρ F•••dldldl=== 0Prob. 0222ρρρzdzd zzz===111 0 00 zzz===111 ρρρ===2,2, 3.41 2,zzz===111 2 22 0 = r (cos φ − sin φ ) a x + r (cos + sin ) φ φ a y r r sin θ [(cos φ − sin φ ) a x + (cos φ + sin φ ) a y ] Q= 22 r sin θ πππ/ //444 2 22000 2 22 2 Qr == sin sin θ===(4 ρρ θ +cos φφφ)θ)) sin φ+++ρρρcos cos cos 1.757 ++((−(φ −−666cos −−0) =ρ (4 +++6(6( cos (4 −Q 0) 6(−−−cos cosπππ/ //444+++1)1) 1)+++(0(0 (0−−−4)4) 4)===1.757 1.757 x0) 22 φ + sin φ ) a y 0 r (cos φ − sin0φ00) a x + r2(cos 00= Qθ = 1 11cos θ cos φ cos θ sin φ − sin θ Qy ∇ F=== [3[3− ∇ +++...... sin φ 0 Q ∇xxφF xF sinφφφ−−−000]]aaaz zcos ... Q ρρρ 3zzzsin Qr z sin θ cos φ sin θ zsin φ cos θ Qx 2 22 π π/π4//44 //44y4 Qθ33 3zzz= cos θ cos φ cos θ sin φ − sin θ πππ /Q x x d d d d d d φ φ ρ ρ φ φ ρ ρ φ φ ( ( ) ) sin sin 3(2)( 3(2)( cos cos ) ) ∇ ∇ F F ⋅ ⋅ S S = = = = − − θ θ sin cos Q = a + a + a r r r x d d d φ ρ φ ρ φ ( ) sin 3(2)( cos ) ∇ F ⋅ S = = − r 0=0φρ ρρθ −φsin φ zzz=== cos φ 0 0 0 0 Q ρ ρ=ρ0==00 φφ=φQ z =0
Q=
(a)
1 ,−−cos 30 1 θ a + ra a6( ρππ =πQ °θ=a2(+ )r=cos dl = ρ d=φ==6( r+1) φ− cos ++ 1) 1.757 1.757 6( cos 1)=r==sin 1.757 =sin θ φ r 2 (a)30° = 3 z = r cos Prob. Prob. 3.39 Prob.3.39 3.39 1 2 −d−y−lyy=2 ρ dφ−a ρ = r sin 30° = 2( ) = 1 − −y−= yy −y−yφy , = + ρ Q r z ∇∇ xe xe 16 xe ∇⋅ ⋅A ⋅ AA== 8xe xe +++888xe xe ===16 16xe xe φ=88 2 − −y−yy − −y−yy 2 π ∇∇ ((∇ 16 16 xe ∇(∇ ∇⋅ ⋅A ⋅A A)))===16 16eee aa2ax xx−−−16 16xe xe aaayy 30° = 3 2 z = r ycos ρ ρ Q l • = + d z dφ = 2(1)(2π ) = 4π 0 Qφ = r = ρ 2 + z 2 aaax xx aaay yy aaaz zz (b) 2π ∂∂∂ ∂l∂∂= ∂∂∂ 2 2 ρ d−φ−y−yy= 2(1)(2 − −y y • dcos ∇∇ θ aφ ρ ===+((−(z−−16 ∇ ⋅A ((=∇ 16 16 A))θ)===a r − 2 aQθ + ∇x×x∇ ∇(∇ ∇⋅cot ⋅A xQ 16eee +++16 16eee −)yπa))aa)z zz====400π0 ∂∂∂xxx ∂∂∂yyy 0 ∂∂∂zzz 2 − −y−yy (b) r eeesin θ−d−−16 dxe θ16 φ−a−y−yyr 000 For S1 , dS =16 16 xe 16 16xe 2∇ ∇∇ ×x∇ Q cot xsin == 000θ. θ Should Should be expected expected since ..daθr d−φ|2 aθ + cos θ aφ x∇ ∇VθVV==cot Shouldbe be expected since (∇ × Q ) • dSsince = r∇
S1
r =2
S301 ,° dS = r sin θ dθ dφ a r For Prob. Prob. 3.40 Prob.3.40 3.40 2π sin sin cos cos cos 2φ sinθθθcos cosφφφ = 4 cos cos cos sin φsin θ cot θ dθ dφ cos =sin dφθθθcos ) φ•φφdaθaSθaθdθ−=θ−−sin r4φ2π (a) (a) aaar rr+++(∇ ×2Q aaaφ φφ |r = 2 (a)∇∇ ∇VVV===−−− 2 22 2 2 2 2 0 0 rrr rrr rrr S1
(b(b (b))) ∇∇ ∇xx∇ x∇ ∇VVV===000
2
2π
30°
0
0
= 4 dφ
cos θ dθ = 4π
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Principles of Electromagnetics, 6e
65 63
(c)
1 ∂ 1 ∂ cos θ cos φ 1 sin θ cos φ (− sin θ cos φ ) + 2 (sin θ ) + 2 2 (− ) 2 r ∂r r sin θ ∂θ r r sin θ r cos φ cos φ = 0+ 3 (1 − 2sin 2 θ ) − 3 r sin θ r sin θ 2sin θ cos φ =− r3
∇ ⋅∇V = ∇ 2V =
Prob. 3.41 r r sin θ [(cos φ − sin φ ) a x + (cos φ + sin φ ) a y ] Q= r sin θ = r (cos φ − sin φ ) a x + r (cos φ + sin φ ) a y
Qr sin θ cos φ Qθ = cos θ cos φ Qφ − sin φ
sin θ sin φ cos θ sin φ cos φ
cos θ − sin θ 0
Qx Q y Qz
Q = r sin θ a r + r cos θ aθ + r a φ
(a) dl = ρ dφ aφ ,
1 2
ρ = r sin 30° = 2( ) = 1
z = r cos 30° = 3 Qφ = r = 2π
Q • dl =
ρ 2 + z2 ρ 2 + z 2 ρ dφ = 2(1)(2π ) = 4π
0
(b)
∇ × Q = cot θ a r − 2 aθ + cos θ aφ For S1 , dS = r 2 sin θ dθ dφ a r
(∇ × Q ) • dS = r
2
S1
sin θ cot θ dθ dφ|
2π
30°
0
0
= 4 dφ
r =2
cos θ dθ = 4π
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Principles of Electromagnetics, 6e
66 64
(c) dS = r sin θ dθ dr aθ
For S2 ,
(∇ × Q) • dS = − 2 r sin θ dφ dr|θ
S2
2
2π
0
0
= −2sin 30 rdr
= 30°
dφ
= − 4π (d) For S1 , dS = r 2 sin θ d φ dθ a r
S1
Q • dS = r 3 sin 2 θ dθ d φ| 2π
30
0
0
= 8 dφ = 4π [
(e)
r=2
°
π 3
sin
−
2
θ dθ
3 ] = 2.2767 2 _
For S 2 , dS = r sin θ dφ dr aθ
S2
Q • dS = r 2 sin θ cos θ dφ dr|
θ =30°
=
(f)
4π 3 = 7.2552 3
1 ∂ 3 r ∂ (r sin θ ) + (sin θ cos θ ) + 0 2 r ∂r r sin θ ∂ θ = 2sin θ + cos θ cot θ
∇•Q =
∇ • Qdv = (2sin θ + cos θ cot θ )r = =
r3 2 (2π ) 3 0
30
(1 + sin
2
2
sin θ dθ dφ dr
θ )dθ
0
4π 3 (π − ) = 9.532 3 2
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Principles of Electromagnetics, 6e
67 65 Check :
∇ • Qdv = (
S1
+ Q • dS S2
π 3 3 = 4π [ − + ] 3 2 3 =
4π 3 [π − ] 3 2
(It checks!)
Prob. 3.42 Since u = ω × r ,
∇ × u = ∇ × ( ω × r ). From Appendix A.10,
∇ × ( A× B ) = A(∇ • B ) − B (∇ • A) + ( B • ∇) A− ( A• ∇) B
∇ × u = ∇ × (ω × r ) _
∇ × (ω × r ) = ω (∇ • r ) − r (∇ • ω ) + (r • ∇) ω − (ω • ∇) r = ω (3) − ω = 2 ω or ω =
1 ∇ × u. 2
Alternatively, let u=
x = r cos ω t ,
y = r sin ω t
∂x ∂ y ax + ay ∂t ∂t
= − ω r sin ω t a x + ω r cos ωt a y = −ω y ax + ω x a y
∂ ∂ ∇×u = ∂ x ∂ y −ω y ω x i.e., ω =
∂ ∂ z = 2ω a z = 2ω 0
1 ∇×u 2
Note that we have used the fact that ∇ • ω = 0,
(r • ∇)ω = 0,
(ω • ∇)r = ω
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Principles of Electromagnetics, 6e
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Prob. 3.43
∇ B =
1 ∂ 1 ∂Bφ ∂Bz 1 ( ρ Bρ ) + + = 2 ρ cos φ + ρ cos φ − 4 ρ ∂ρ ρ ∂φ ∂z ρ = (2 + ρ ) cos φ − 4
∂B 1 ∂Bz ∂Bφ ∂B ∂B 1 ∂ − aρ + ρ − z aφ + ( ρ Bφ ) − ρ a z (a) ∇ × B = ρ ∂ρ ∂z ∂ρ ∂φ ρ ∂φ ∂z 1 = 0a ρ + 0aφ + 3ρ 2 sin φ + ρ sin φ a z ρ = (3ρ + 1) sin φ a z (b) ∇ F =
1 ∂ 2 1 ∂Fφ 1 ∂ 4 1 ( ) 0 ( −2r sin φ ) r F + + = 2 (r sin θ ) + r 2 r ∂r r sin θ ∂φ r ∂r r sin θ 2sin φ = 4r sin θ − sin θ ∂ ∂Fθ ∂θ ( Fφ sin θ ) − ∂φ ∂F 1 ∂ + (rFθ ) − r aφ r ∂r ∂θ
∇×F =
1 r sin θ
1 1 ∂Fr ∂ ar + r sin θ ∂φ − ∂r (rFφ ) aθ
1 1 2 ∂ ∂θ (2r sin θ cos φ ) − 0 ar + r sin θ r cos φ − 4r cos φ aθ + 0aφ r cos φ = 2 cot θ cos φ ar + − 4 cos φ aθ sin θ
=
1 r sin θ
Prob. 3.44 (a)
∂V ∂V ∂V ax + V ay+ V az ∇ (V∇V)=∇ V ∂y ∂z ∂x =
∂ ∂V V ∂x ∂x
∂ ∂V + V ∂y ∂y
∂ ∂V + V ∂z ∂z 2
2
∂ 2V ∂ 2V ∂ 2V ∂V ∂V ∂V = V 2 +V 2 +V 2 + + + ∂x ∂y ∂z ∂x ∂y ∂z = V ∇ 2V + | ∇V |2
2
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Principles of Electromagnetics, 6e
69 67 ∂ ∂ ∂ ∇ × VA = ∂x ∂y ∂z VAx VAy VAz
(b)
∂ ∂ ∂ ∂ ∂ ∂ = (VAz ) − (VAy ) a x + (VAx ) − (VAz ) a y + (VAy ) − (VAx ) a z ∂z ∂x ∂y ∂z ∂y ∂x ∂A ∂V ∂A ∂V = Az + V z − Ay − V y ax ∂y ∂z ∂z ∂y ∂A ∂V ∂A ∂V + Ax + V x − Az −V z ay ∂z ∂x ∂x ∂z ∂A ∂V ∂A ∂V + Ay + V y − Ax − V x az ∂x ∂y ∂y ∂x ∂A ∂A ∂A ∂A ∂A ∂A ∇ × VA = V z − y a x + x − z a y + y − x a z ∂z ∂x ∂y ∂z ∂x ∂y ∂V ∂V ∂V ∂V ∂V ∂V + Az − Ay − Az − Ax a x + Ax az a y + Ay ∂z ∂z ∂x ∂x ∂y ∂y = V ∇ × A + ∇V × A
Prob. 3.45 (a) ∂B ∂B ∂B ∇ B = x + y + z = 2 xy + 1 + 1 = 2 + 2 xy ∂x ∂y ∂z (b) ∂ ∂ ∂ ∂y ∂z = (−1 + 0)a x + (0 − 0)a y + (4 x + x 2 )a z ∇ × B = ∂x x 2 y (2 x 2 + y ) ( z − y ) = −a x + x(4 − x)a z (c) ∇(∇ B) = ∇(2 + 2 xy ) = 2 ya x + 2 xa y
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(d) ∂ ∂ ∇ × ∇ × B = ∂x ∂y −1 0
∂ ∂z = 0a x - (4 - 2x)a y + 0a z 2 (4 x − x )
= 2( x − 2)a y
Prob. 3.46 (a) V1 = x 3 + y 3 + z 3
∂ 2V1 ∂ 2V1 ∂ 2V1 ∇ V1 = + + ∂x 2 ∂y 2 ∂z 2 ∂ ∂ ∂ ( 3x 2 ) + 3 y 2 ) + (3z 2 ) = ( ∂x ∂y ∂x 2
= 6 x + 6 y + 6z = 6( x + y + z) (b) V2 = ρ z 2 sin 2φ ∇ 2V2 = =
1 ∂
ρ ∂ρ z
2
ρ
=(
( ρ z 2 sin 2φ ) −
sin 2φ −
−3z 2
ρ
4z
2
ρ
4z2
ρ
sin 2φ +
∂ (2 ρ z sin 2φ ) ∂z
sin 2φ + 2 ρ sin 2φ
+ 2 ρ ) sin 2φ
(c) V3 = r 2 (1 + cos θ sin φ ) ∇ 2V3 =
1 ∂ [2r 3 (1 + cos θ sin φ )] 2 r ∂r
1 ∂ 1 ( − sin 2 θ sin φ )r 2 + 2 2 r 2 ( − cosθ sin φ ) r sinθ ∂θ r sin θ 2 sinθ cosθ sin φ = 6(1 + cosθ sin φ ) − cosθ sin φ − sinθ sin 2 θ cosθ sin φ = 6 + 4 cosθ sin φ − sin 2 θ +
2
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Prob. 3.47 (a) U = x 3 y 2 e xz
∂ ∂ ∂ 4 2 xz (3 x 2 y 2 e xz + x 3 y 2 ze xz ) + (2 x 3 y e xz ) + (x y e ) ∂x ∂y ∂z
∇ 2U =
= 6 xy 2 e xz + 3 x 2 yze xz + 3 x 2 y 2 ze xz + x3 y 2 z 2 e xz + 2 x3e xz + x5 y 2 e xz = e xz (6 xy 2 + 3 x 2 y 2 z + 3 x 2 y 2 z + x3 y 2 z 2 + 2 x3 + x5 y 2 ) At (1, −1,1), ∇ 2U = e1 (6 + 3 + 3 + 1 + 2 + 1) = 16e= 43.493
(b)
V = ρ 2 z (cos φ + sin φ ) 1 ∂ ∇ 2V = [2 ρ 2 z (cos φ + sin φ )] − z (cos φ + sin φ ) + 0
ρ ∂ρ
= 4 z (cos φ + sin φ ) − z (cos φ + sin φ ) = 3 z (cos φ + sin φ )
π
At (5, (c)
6
, − 2), ∇ 2V = −6(0.866 + 0.5)= −8.196
W = e − r sin θ cos φ ∇ 2W =
1 ∂ e− r ∂ 2 −r φ ( − r e sin θ cos ) + cos φ (sin θ cos θ ) 2 2 ∂θ r ∂r r sin θ
−
e − r sin θ cos φ r 2 sin 2 θ
1 (−2re − r sin θ cos φ ) + e − r sin θ cos φ 2 r e − r cos φ e − r cos φ + 2 (1 − 2sin 2 θ ) − 2 r sin θ r sin θ 2 2 ∇ 2W = e − r sin θ cos φ (1 − − 2 ) r r =
At
(1, 60°,30°),
∇ 2W = e −1 sin 60 cos 30(1 − 2 − 2) = −2.25e −1 = − 0.8277
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Prob. 3.48 (a ) Let V = 1nr = 1n x 2 + y 2 + z 2
∂V 1 1 x = ( 2 x ) ( x 2 + y 2 + z 2 )−1/2 = 2 ∂x r 2 r ∇V =
∂V ox
ax +
∂V oy
ay +
∂V oz
az =
x a x + ya y + z a z r
2
=
r r2
r 1 = ar in spherical coordinates. r2 r 1 ∂ 2 1 ∂ ∇ 2 (1nr ) = ∇ ∇ (1nr ) = ∇ A = 2 (r A r ) = 2 (r ) r ∂r r ∂r 1 = 2 r (b) Let ∇V = A =
Prob. 3.49 ∂V ∂V ∂V ∇V = ax + ay + az ∂x ∂y ∂z = y 2 z 3a x + 2 xyz 3a y + 3xy 2 z 2 a z At P(1,2,3,) x = 1, y = 2, z = 3 ∇V = 4(27)a x + 2(2)(27)a y + 3(4)(9)a z = 108( a x + a y + a z )
∂ 2V ∂ 2V ∂ 2V ∇V = 2 + 2 + 2 ∂x ∂y ∂z ∂ ∂ ∂ = ( y 2 z 3 ) + ( 2 xyz 3 ) + ( 3xy 2 z 2 ) ∂x ∂y ∂z 2
= 0 + 2 xz 3 + 6 xy 2 z = 2 xz ( z 2 + 3 y 2 )
At P(1,2,3,) x = 1, y = 2, z = 3. ∇ 2V = 2(1)(3)(9 + 3 × 4) = 6(9 + 12) = 126
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Prob. 3.50 ∂V 1 ∂V ∂V aρ + aφ + a z = 2 ρ z cos φ a ρ - ρ z sin φ aφ + ρ 2 cos φ a z ∇V = ρ ∂φ ∂ρ ∂z 1 ∂ 2V ∂ 2V 1 ∂ 1 + 2 = 2 ρ 2 z cos φ ) − 2 ρ 2 z cos φ + 0 ( + 2 2 ∂z ρ ∂ρ ρ ρ ∂φ = ( 4 − 1) z cos φ = 3z cos φ
∇ 2V =
1 ∂ ∂V ρ ρ ∂ρ ∂ρ
Prob. 3.51
∂V 1 ∂V 1 ∂V ar + aθ + aφ ∂r r ∂θ r sin θ ∂φ 10 5sin φ = − 3 cos φ ar − 3 aφ r r sin θ
∇V = (a)
1 ∂ 2 ∂V 1 ∂ ∂V 1 ∂ 2V r ( ) + (sin ) + θ r 2 ∂r r 2 sin θ ∂θ r 2 sin 2 θ ∂φ 2 ∂r ∂θ 1 ∂ 2 10 cos φ 1 5cos φ r (− = 2 ) + 0 + 2 2 (− ) 3 r ∂r r r sin θ r2 10 cos φ 5cos φ ∇ ∇V = − 4 2 r4 r sin θ
∇ ∇V = ∇ 2V =
(b)
(c)
∇ × ∇V = 0, see Example 3.10.
Prob. 3.52 ∂U ∂U ∂U ax + ay + a z = 4 yz 2 a x + (4 xz 2 + 10 z )a y + (8 xyz + 10 y )a z ∇U = ∂x ∂y ∂z ∂ ∂ ∂ ∇ ∇U = (∇U x ) + (∇U y ) + (∇U z ) = 0 + 0 + 8 xy = 8 xy ∂x ∂y ∂z
∇ 2U =
∂ 2U ∂ 2U ∂ 2U + + = 0 + 0 + 8 xy = 8 xy ∂x 2 ∂y 2 ∂z 2
Hence, ∇ 2U = ∇ ∇U
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Prob. 3.53
Method 1 ∇ 2G
ρ
= ∇ 2Gρ −
2 ∂Gφ Gρ − ρ 2 ∂φ ρ 2
1 ∂ 2 ρ sin φ 8ρ sin φ 2 ρ sin φ +0+ − ( 2 ρ sin φ ) − 2 ρ ∂ρ ρ ρ2 ρ2 2sin φ 2sin φ 8sin φ 2sin φ 6sin φ = − + − =
=
ρ
∇ 2G
φ
ρ
= ∇ 2Gφ +
ρ
ρ
ρ
2 ∂Gρ Gφ − ρ 2 ∂φ ρ 2
1 ∂ 1 4 ρ cos φ 4 ρ cos φ − (4 ρ cos φ ) − 4 ρ cos φ + 0 + ρ ρ2 ρ2 ρ ∂ρ 4 cos φ 4 cos φ 4 cos φ 4 cos φ = − + − =0 =
ρ
∇ 2G
z
ρ
= ∇ 2 Gz = =
1
ρ
ρ
ρ
1 ∂ ∂ ρ ( z 2 + 1) + 0 + (2 z ρ ) ρ ∂ρ ∂z
( z 2 + 1) + 2 ρ
Adding the components together gives ∇ 2G =
6sin φ
ρ
1 a ρ + 2 ρ + ( z 2 + 1) a z ρ
Method 2:
∇ 2G = ∇(∇ G ) − ∇ × (∇ × G ) 1 ∂ 1 (2 ρ 2 sin φ ) + (−4 ρ sin φ ) + 2 z ρ = 2 z ρ ρ ∂ρ ρ ∇(∇ G ) = ∇V = 2 za ρ + 2 ρ a z
Let V = ∇ G =
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Principles of Electromagnetics, 6e
75 73
Let
1 1 ∂ A = ∇ × G = 0 − 0 a ρ + 0 − ( z 2 + 1) aφ + (4 ρ 2 cos φ ) − 2 ρ cos φ a z ρ ∂ρ ρ 2 = − ( z + 1)aφ + 6 cos φ a z
6 1 ∂ ∇ × ∇ × G = ∇ × A = − sin φ + 2 z a ρ + (0 − 0)aφ + ( − ρ ( z 2 + 1)) − 0 a z ρ ∂ρ ρ 6 1 = 2 z − sin φ a ρ − ( z 2 + 1)a z ρ ρ 2 ∇ G = ∇V − ∇ × A 6 1 = 2 zaρ + 2 ρa z − 2 z − sin φ a ρ + ( z 2 + 1)a z ρ ρ 6 1 = sin φ a ρ + 2 ρ + ( z 2 + 1) a z ρ ρ
Prob. 3.54 ∂ ∂ ∂ ∇ A = ( xz ) + ( z 2 ) + ( yz ) = z + y ∂x ∂y ∂z ∇(∇ A) = a y + a z
∇ 2 A = ∇ 2 Ax a x + ∇2 Ay a y + ∇2 Az a z = 0 + 2a y + 0 = 2a y ∇(∇ A) - ∇2 A = − a y + a z ∂ ∇ × A = ∂x xz
∂ ∂y z2
∂ ∇ × ∇ × A = ∂x −z
(1)
∂ ∂z = − za x + xa y yz ∂ ∂y x
∂ ∂z = − a y + a z 0
(2)
From (1) and (2), ∇ × ∇ × A = ∇( ∇ A) - ∇ 2 A
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Prob. 3.55 ∂A ∂A ∂A ∇ A = x + y + z = 1 + 1 + 1 = 3 ≠ 0 ∂x ∂y ∂z
∂ ∇ × A = ∂x x ∇ B =
∂ ∂y y
∂ ∂z = 0 z
1 ∂ 1 ∂Bφ ∂Bz ( ρ Bρ ) + + = 4 cos φ − 4 cos φ = 0 ρ ∂ρ ρ ∂φ ∂z
∂B 1 ∂Bz ∂Bφ ∂B ∂B 1 ∂ ∇× B = − a ρ + ρ − z aφ + ( ρ Bρ ) − ρ a z ∂z ∂ρ ∂φ ρ ∂ρ ρ ∂φ ∂z 1 = 0aρ + 0aφ + [ −8 ρ sin φ + 2 ρ sin φ ] a z = −6sin φ a z ≠ 0
ρ
1 ∂ 2 2sin θ (r sin θ ) + 0 + 0 = ≠0 2 r ∂r r 1 ∂ 1 ∂ 2 2 r ∇×C = ( sin θ ) − 0 + 0 − (r sin θ ) aθ a r r sin θ ∂θ r ∂r 1 + [ 0 − cos θ ] aφ r cosθ = 2 cos θ ar - 2sinθ aθ aφ ≠ 0 r (a) B is solenoidal. (b) A is irrotational. ∇ C =
Prob. 3.56 (a) ax ay ∂ ∂ ∇×G = ∂x ∂y 16 xy − z 8 x 2
az ∂ ∂z −x
= 0 a x + (−1 + 1) a y + (16 x − 16 x) a z = 0 Thus, G is irrotational.
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Principles of Electromagnetics, 6e
77 75
(b) Assume that ψ represents the net flux.
ψ = G • dS = ∇ • Gdv
∇ • G = 16 y + 0 + 0 = 16 y 1
1
1
0
0
0
ψ = 16 ydxdydz = 16 dx dz ydy = 16(1)(1)( (c)
y2 1 )=8 2 0
y 1
0
G • dl L
=
1 y =1
x =1
(16 xy − z )dx|
x =0
y =0 z =0
= 0 + 8(1) y| + 16(1) 1
0
= 8−8 = 0
+
8x dy|
y =0
2
x =1 z =0
+
x y =0
x =0
(16 xy − z )dx|
x =1
y =1 z =0
+
8 x dy|
y =1
2
x =0 z =0
x2 0 | +0 2 1
This is expected since G is irrotational, i.e.
G • dl = (∇ × G ) • dS =
0
∇ • T = −6 + 0 = −6
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Prob. 3.57 ∇⋅F = 0
ax ∂ ∇xF = ∂x yz
ay ∂ ∂y xz
az ∂ = ( x − x )a x + ( y − y )a y + ( z − z )a z = 0 ∂z xy
Hence F is both solenoidal and conservative.
Prob. 3.58 ∇× H = 0
H dl = (∇ × H) dS = 0 L
S
Prob. 3.59 From Appendix A.10, ∇ ( A × B ) = B (∇ × A) − A (∇ × B ) If A and B are irrotational, ∇× A = 0 = ∇× B i.e. ∇ ( A × B ) = 0 which implies that A × B is solenoidal.
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79 77 CHAPTER 4 P. E. 4.1
5 ×10−9 [(1, −3, 7) − (2, 0, 4)] [(1, −3, 7) − (2, 0, 4)]3 −9 1×10 −9 (a) F = ( 2 10 )[(1, 3, 7) ( 3, 0,5)] − × − − − 10−9 + 3 4π [(1, −3, 7) − ( −3, 0,5)] 36π =[
45(−1, −3,3) 18(4, −3, 2) ] nN − 193/ 2 293/ 2
= −1.004a x − 1.284a y + 1.4 a z nN (b)
E=
F = −1.004a x − 1.284a y + 1.4a z V/m Q
P. E. 4.2 Let q be the charge on each sphere, i.e. q=Q/3. The free body diagram below helps us to establish the relationship between various forces. P
θ
T A
F1 d/2
F2 mg
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Principles of Electromagnetics, 6e
80 78
At point A, T sin θ cos 30° = F1 + F2 cos 60° =
q2
4πε 0 d
2
+
q2
1 ( ) 4πε 0 d 2 2
3 q2 = 8πε 0 d 2 T cos θ = mg Hence,
sin θ =
But
Thus,
or
tan θ cos 30° =
q2 =
h d tan θ = = l 3l
Q2 =
d 3 l2 −
d2 3
d 3 ( ) 3q 2 3 2 = 8πε 0 d 2 mg d2 2 l − 3 4π ε0 d 3 m g 3 l2 −
but q =
3q 2 8πε 0 d 2 mg
Q 3
d2 3
⎯⎯ →
q2 =
Q . Hence, 9
12 π ε 0 d 3 m g d2 l − 3 2
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81 79 P.E. 4.3
_
d2 l eE = m 2 dt _
_
d 2 x _ d 2 y _ d 2z _ ax + 2 a y + 2 az ) dt 2 dt dt E0 = 200 kV / m _
eE0 ( − 2a x + a y ) = m( where
d 2z =0 ⎯⎯ → z = ct + c2 dt 2 d2x − 2eE0 t 2 m 2 = − 2eE0 ⎯⎯ → x= + c3 t + c4 2m dt d2y eE0 t 2 m 2 = eE0 ⎯⎯ → y= + c5 t + c6 2m dt At t = 0, ( x, y, z ) = (0, 0, 0) c1 = 0 = c4 = c6 dx dy dz , , ) = (0, 0, 0) dt dt dt ⎯⎯ → c1 = 0 = c3 = c5 At t = 0
Also, (
Hence,
( x, y ) =
i.e. 2 | y | = | x |
eE0t 2 (−2,1) 2m
Thus the largest value of is 80 cm = 0.8 m P.E. 4.4 (a)
Consider an element of area dS of the disk. The contribution due to dS = ρ dφ d ρ is dE =
ρ s dS ρ s dS = 2 4πε 0 r 4πε 0 ( ρ 2 + h 2 )
The sum of the contribution along ρ gives zero.
ρs Ez = 4π ε 0
a
ρ
=0
2π
hρ h ρ d ρ dφ φ =0 ( ρ 2 + h2 )3/ 2 = 2 ε 0s
a
ρ
−0
ρ dρ ( ρ + h 2 )3/ 2 2
a
a h ρs hρs 2 2 −3/ 2 2 2 2 −1/ 2 ( ) ( ) ( 2( ) | = + = − + ρ h d ρ ρ h 0 4 ε 0 0 4ε 0
=
ρs h [1 − 2 ] 2ε 0 (h + a 2 )1/ 2
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Principles of Electromagnetics, 6e
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(b)
As a E=
⎯⎯ →
∞,
ρs az 2ε 0
(c) Let us recall that if a/h a 2πρ At ( 0, 1 cm, 0 ) , 3 × 1× 10−2 = Hφ = 2π × 4 × 10−4 H = 11.94 aφ A/m At
300 8π
( 0, 4 cm, 0 ) , 3 2π × 4 × 10−2 11.94 aφ A/m
Hφ = H =
=
300 8π
Prob. 7.22 For 0 < ρ < a
L
H dl = I enc = J dS
H φ 2πρ =
2π
φ =0
ρ
ρ
=0
Jo
ρ
ρ dφ d ρ
= J o 2πρ Hφ = J o For ρ > a 2π
a
Hdl = J dS = φ ρ =0
Jo
=0
ρ
ρ dφ d ρ
H ρ 2πρ = J o 2π a Hφ =
Joa
ρ
J o , 0a,
H ⋅ dl = I
2π
2 ko 2ko ρ2 a (2π ) ρ d ρ dφ = = J ⋅ dS = a 2 0 ρ =0 φ =0 a a
enc
H φ 2πρ = 2π ko a
⎯⎯ → Hφ =
a H = ko aφ , ρ
ρ >a
ko a
ρ
Prob. 7.24 ∂ J = ∇ × H = ∂x y2
∂ ∂y x2
∂ ∂z = (2 x − 2 y )a z 0
At (1,-4,7), x =1, y = -4, z=7, J = [ 2(1) − 2(−4) ] a z = 10a z A/m 2 Prob. 7.25 (a) J = ∇× H =
1 ∂ 1 ∂ ( ρ Hφ )a z = (103 ρ 3 )a z ρ ∂ρ ρ ∂ρ
= 3ρ ×103 a z A/m
2
(b) Method 1: 2
2π
0
0
I = J dS = 3ρ ρ dφ d ρ103 = 3 × 103 ρ 2 d ρ dφ S
= 3 × 103 (2π )
Method 2:
ρ 2 3
3 2
= 16π × 103 A = 50.265 kA
2π
I = H dl =103 ρ 2 ρ dφ = 103 (8)(2π ) = 50.265 kA L
0
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Prob. 7.26 Let H = H1 + H 2 where H1 and H 2 are due to the wires centered at x = 0 and x = 10cm respectively. For H1 , ρ = 50 cm, aφ = al × a ρ = a z × a x = a y
(a)
H1 =
5 50 ay = a −2 π y 2π ( 5 × 10 )
For H 2 , ρ = 5 cm, aφ = − a z × −a x = a y , H 2 = H1 H = 2H1 =
100
π
ay
= 31.83 a y A/m 2a y − a x 2a + a y For H1 , aφ = a z × x = 5 5 −a x + 2a y 5 H1 = = − 3.183a x + 6.366a y −2 2π 5 5 ×10 5 For H 2 , a ρ = − a z × a y = a x
(b)
H2 =
5
2π ( 5 )
a x = 15.915a x
H = H1 + H 2 = 12.3 a x + 6.366a y A/m Prob. 7.27 (a) B =
μo I aφ 2πρ
At (-3,4,5), ρ=5. B=
4π × 10−7 × 2 aφ = 80aφ nW/m 2 2π (5)
μI Ψ = B • dS = o 2π (b)
d ρ dz
ρ
6 4 4π ×10−7 × 2 = ln ρ z 2 0 2π
= 16 ×10−7 ln 3 = 1.756 μ Wb
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208 204
Prob. 7.28 (a) I = J dS =
2π
a
ρ2
J o (1 −
a2
φ =0 ρ =0
) ρ d ρ dφ
ρ2 ρ4 = 2π J o − 2 2 4a 1 = π a2 Jo 2 (b)
H dl = I
enc
a 0
2π
a
0
0
= J o dφ ( ρ −
ρ3 a2
)d ρ
2π 2 a 2 = Jo a − 2 2
= J dS
For ρ < a, H φ 2πρ = J dS ρ2 ρ4 = 2π J o − 2 2 4a H ρ 2πρ = 2π J o Hρ =
ρ2
ρ2 2 − 4 a2
Jo ρ ρ2 2 − 4 a2
For ρ > a,
Hdl =
J o dS = I
1 H φ 2πρ = π a 2 J o 2 2 a Jo Hφ = 4ρ Jo ρ ρ2 2 − , ρ < a a2 4 Hence Hφ = aJ o , ρ >a 4ρ
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Principles of Electromagnetics, 6e
209 205
Prob. 7.29
B =
μ0 I aφ 2πρ
ψ =
=
Prob. 7.30
B ⋅ dS =
d+a
μ0 I dρ dz z = 0 2πρ
ρ =d
b
μ0 Ib d + a In 2π d
For a whole circular loop of radius a, Example 7.3 gives H=
Ia 2 a z 2 a 2 + h 2
3/2
→0 Let h ⎯⎯ I az 2a For a semicircular loop, H is halfed I H= az 4a μI B = μo H = o a z 4a H=
Prob. 7.31 ∂Bx ∂By ∂Bz + + =0 ∂x ∂y ∂z showing that B satisfies Maxwell’s equation.
(a) ∇ • B =
(b)
dS = dydza x 4
Ψ = B • dS =
1
z =1 y = 0
y 2 dydz =
4 y3 1 ( z ) = 1 Wb 1 3 0
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Principles of Electromagnetics, 6e
210 206
(c) ∇ × H = J
J = ∇×
⎯⎯ →
B
μo
∂ ∂ ∂ ∇ × B = ∂x ∂y ∂z = −2 za x − 2 xa y − 2 ya z y 2 z 2 x2 2 J = − ( za x + xa y + ya z ) A/m 2
μo
Prob. 7.32
h ( ρ − a) 6 where H1 and H 2 are due to the wires centered at x = 0 and x = 10cm respectively. On the slant side of the ring, z =
ψ =
B.dS
μo I
2πρ
=
h ( ρ −a) b
dρ dz
=
μo I 2π
=
μo Ih a + b b − a ln as required. a 2π b
a+b
ρ
dz dρ
ρ
z=0
=a
=
μo Ih 2π b
a 1− dρ =a ρ
a+b
ρ
If a = 30 cm, b = 10 cm, h = 5 cm, I = 10 A,
ψ =
4π × 10 −7 × 10 × 0.05 4 0.1 − 0.3 ln −2 3 2π 10 × 10
(
)
= 1.37 × 10 −8 Wb
Prob. 7.33
ψ =
BdS
= μo
0.2
50o
z=0 φ =0
106
ρ
sin 2φ ρ dφ dz
cos 2φ ψ = 4π × 10 × 10 ( 0.2) − 2 −7
(
50o
6
=
0.04π 1 − cos 100
=
0.1475 Wb
o
)
0
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Principles of Electromagnetics, 6e
211 207
Prob. 7.34 π /4 2
ψ = B dS =
ρ
20
φ = 0 =1
S
= 20(1)
π /4
0
ρ
2
π /4
1
0
sin φ ρ d ρ dφ = 20 d ρ 2
sin
2
φ dφ
π /4 1 1 (1 − cos 2φ )dφ = 10(φ − sin 2φ ) 0 2 2
π
1 = 10( − ) = 2.854 Wb 4 2
Prob. 7.35
ψ = B dS , dS = r 2sinθ dθ dφ ar S
ψ =
2π
2 cos θ r 2sinθ dθ dφ = 2 dφ r =1 r3 0 π /3
= 2(2π ) sin θ d (sin θ ) = 4π 0
π /3
cos θ sin θ dθ 0
sin 2 θ π / 3 = 2π sin 2 (π / 3) 0 2
= 4.7123 Wb
Prob. 7.36 B = μo H =
μo J × R dv 4π v R 3
Since current is the flow of charge, we can express this in terms of a charge moving with velocity u. Jdv = dqu.
μo 4π
qu × R R 3 In our case, u and R are perpendicular. Hence, B=
μo qu 4π ×10−7 1.6 ×10−19 × 2.2 ×106 1.6 ×10−20 = × = B= 4π R 2 4π (5.3 ×10−11 ) 2 (5.3) 2 ×10 −22 = 12.53 Wb/m 2
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212 208
Prob. 7.37 (a ) ∇A = − ya sin ax ≠ 0 ∂ ∂x
∇×A =
∂ ∂y
y cos ax
∂ ∂z y + e-x
0
= a x + e − x a y − cos axa z ≠ 0 A is neither electrostatic nor magnetostatic field
1 ∂ 1 ∂ ρ Bρ = ( 20) = 0 ρ ∂ρ ρ ∂ρ ∇× B = 0 B can be E-field in a charge-free region.
(b)
∇⋅ B =
(c )
∇⋅ C =
(
)
1 ∂ 2 (r sinθ ) = 0 r sin θ ∂φ 1 ∂ 1∂ 3 ∇×C = r 2 sin 2 θ ar (r sinθ )aθ ≠ 0 r sin θ ∂θ r ∂r C is possibly H field.
(
)
Prob. 7.38 (a) ∇⋅ D = 0 ∂ ∂x
∇× D =
y2 z
∂ ∂y
∂ ∂z
2(x + 1)yz -(x + 1)z 2
= 2(x + 1)ya x + . . . ≠ 0 D is possibly a magnetostatic field. (b)
∇⋅ E = ∇× E =
∂ sin φ 1 ∂ =0 ( ( z + 1) cos φ ) + ∂z ρ ρ ∂ρ 1
ρ2
cos θ a ρ + . . . ≠ 0
E could be a magnetostatic field. (c )
∇⋅ F =
1 ∂ 1 ∂ sinθ ( 2cosθ ) + ≠ 0 2 r ∂r rsinθ ∂θ r 2
1 ∂ 2 sin θ −1 aθ ≠ 0 r sin θ + r ∂r r 2 F can be neither electrostatic nor magnetostatic field.
∇×F =
(
)
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Principles of Electromagnetics, 6e
213 209
Prob. 7.39 A=
μo Idl μo ILaz = 4π r 4π r
This requires no integration since L 1, 120π 4.6 [ x + 1.393 + 0.667 ln( x + 1.444) ]
50 =
We solve this iteratively and obtain: x = 1.8628, w = xh = 14.9024 mm For this w and h, ε eff = 3.4598 (b)
β=
ω ε eff c
β = 450 =
=
π 4
=
ω ε eff
πc 4 ε eff 2π f
c =
3 × 108 8 ×
3.4598 × 8 × 109
= 0.00252 m
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Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
381 379 Prob. 11.68
w = 1.5cm, h = 1cm,
(a)
εr - 1 0.6 6 +1 = 1.6 + + 2 1 + 12h/w 1 + 12/1.5 2
ε eff =
377 1.8 (1.5 + 1.393 + 0.667 In (2.944))
Z0 =
(b)
w = 1.5 h
α c = 8.686 Rs =
1
281 3.613
= 77.77Ω
Rs wZo =
σc σ
=
= 1.8
μπ f σc
=
19 × 2.5 × 109 × 4π ×10-3 1.1 ×107
= 2.995 × 10-2
αc =
u =
8.686 × 2.995 × 10−2 1.5 × 10−2 × 77.77 c
ε eff
u c = f f ε eff
=
3 × 108 2.5 × 109 1.8
0.8 ( 2.2 ) 2 ×10−2 1.2 1.8 8.944 × 10−2
=
96.096 14.3996
→ λ =
α d = 27.3 ×
= 0.223dB/m
α d = 6.6735 dB/m (c)
α = α c + α d = 6.8965 dB/m α = 20dB → =
20
α
=
20 = 2.9 m 6.8965
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= 8.944 × 10−2
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Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
382 380 Prob. 11.69 0.1 5 × 1.2 ln w ' = 0.5 + = 0.5 + 0.1279 = 0.6279 3.2 0.1 Zo =
377 4 × 1.2 8 × 1.2 ln 1 + + 3.354 2π 2 π × 0.6177 4 × 0.6279
= 42.43ln {1 + 2.49(3.822 + 3.354} = 42.43ln(20.255) = 127.64 Ω
Prob. 11.70 Suppose we guess that w/h < 2
A =
w h
75 3.3 60 2 =
+
1.3 0.11 0.23 + = 1.117 3.3 2.3
8e A 24.44 = = 3.331 → w = 3.331h = 4mm 2A 7.337 e - 2
If we guess that w/h > 2, B=
60π 2 Zo ε r
=
60π 2 = 5.206 75 2.3
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Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
383 381
w 2 1.3 0.61 = 4.266 − ln 9.412 + ln 4.206 + 0.39 − h 4.6 2.3 π = 1.665 2 h
ε eff =
3.3 + 2
3 × 108 u = 1.953
1.3 12 2 1 + 3.331
= 1.953
= 2.1467 × 108 m/s
Prob. 11.71
Z L − Z o 100 − 150 = = −0.2 250 Z L + Zo RL = −20log | Γ |= 13.98 dB
Γ=
Prob. 11.72 Γ=
Z L − Z o 120 − 50 = = 0.4118 Z L + Zo 170
RL = −20log10 | Γ |= −20log10 (0.4118) = 7.706 dB
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Principles of Electromagnetics, 6e
384 CHAPTER 12 P. E. 12.1 (a) For TE10, fc = 3 GHz, 1 − ( f c / f ) 2 = 1 − (3 / 15) 2 = 0.96 , β o = ω / uo = 4π f / c
β=
4π f c
0.96 =
4π × 15 × 109 0.96 = 615.6 rad/m 3 × 108
ω 2π × 15 × 109 = = 1.531 × 108 m/s β 615.6 60π μ = 192.4Ω η'= = 60π , ηTE = ε 0.96 u=
(b) For TM11, fc = 3 7.25 GHz, 1 − ( f c / f ) 2 = 0.8426 4π f 4π × 15 × 109 (0.8426) β= (0.8426) = = 529.4 rad/m c 3 × 108 u=
ω 2π × 15 × 109 = = 1.78 × 108 m/s 529.4 β
ηTM = 60π (0.8426) = 158.8Ω P. E. 12.2 (a) Since Ez ≠ 0 , this is a TM mode
Ezs = Eo sin(mπ x / a)sin(nπ y / b)e − j β z mπ = 40π a i.e. TM21 mode.
Eo = 20,
m=2,
nπ = 50π b
n=1
u' 3 × 108 2 2 (b) f c = ( m / a ) + ( n / b) = 402 + 502 = 1.5 41 GHz 2 2 2π f 2π × 109 β = ω με 1 − ( f c / f ) 2 = f 2 − fc 2 = 225 − 92.25 = 241.3 rad/m. c 3 × 108
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Principles of Electromagnetics, 6e
385 (c)
− jβ (40π )20cos 40π x sin 50π ye − j β z h2 − jβ E ys = 2 (50π )20sin 40π x cos50π ye− j β z h Ey = 1.25 tan 40π x cot 50π y Ex Exs =
P. E. 12.3 If TE13 mode is assumed, fc and β remain the same.
fc = 28.57 GHz, β = 1718.81 rad/m, γ = j β 377 / 2
ηTE13 =
= 229.69 Ω 1 − (28.57 / 50) 2 For m=1, n=3, the field components are: Ez= 0 H z = H o cos(π x / a )cos(3π y / b)cos(ωt − β z )
ωμ 3π
H o cos(π x / a )sin(3π y / b)sin(ωt − β z ) h2 b ωμ π E y = 2 H o sin(π x / a)cos(3π y / b)sin(ωt − β z ) h a β π H x = − 2 H o sin(π x / a)cos(3π y / b)sin(ωt − β z ) h a β 3π H y = − 2 H o cos(π x / a )sin(3π y / b)sin(ωt − β z ) h a Ex = −
Given that H ox = 2 = − H oy = −
β h2
β
h2
(π / a) H o ,
(3π / b) H o = 6a / b = 6(1.5) / 8 = 11.25
−2 × 14.51π 2 × 104 × 1.5 × 10−2 H oz = H o = − = = −7.96 βπ 1718.81π ωμ π 2ωμ Eoy = 2 H o = − = 2ηTE = −459.4 β h a 3a Eox = − Eoy = 459.4(4.5 / 0.8) = 2584.1 b 2h 2 a
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Principles of Electromagnetics, 6e
386 Ex = 2584.1cos(π x / a)sin(3π y / b)sin(ωt − β z ) V/m, E y = −459.4sin(π x / a)cos(3π y / b)sin(ωt − β z ) V/m, Ez= 0, H y = 11.25cos(π x / a)sin(3π y / b)sin(ωt − β z ) A/m, H z = −7.96cos(π x / a) cos(3π y / b)cos(ωt − β z ) A/m P. E. 12.4 f c11 =
3 × 108
up =
ug =
u' 1 1 3 × 108 × 102 + = 1 / 8.6362 + 1 / 4.3182 = 3.883 GHz 2 2 b 2 a 2
1 − (3.883 / 4)
2
= 12.5 × 108 m/s,
9 × 1016 = 7.2 × 107 m/s 8 12.5 × 10
P. E. 12.5 The dominant mode becomes TE01 mode
f c 01 =
c = 3.75 GHz, ηTE = 406.7Ω 2b
From Example 12.2, Ex = − Eo sin(3π y / b)sin(ωt − β z ) , Pave =
a
x =0
where Eo =
| Exs |2 Eo 2 ab dxdy = y = 0 2η 4η
ωμb Ho . π
b
Hence Eo = 63.77 V/m as in Example 12.5. Ho =
π Eo π × 63.77 = = 63.34 mA/m 10 ωμb 2π × 10 × 4π × 10−7 × 4 × 10−2
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Principles of Electromagnetics, 6e
387 P. E. 12.6 (a) For m=1, n=0, fc = u’/(2a)
σ 10−15 10−15 = = 11 GHz, only the dominant TE10 mode is propagated. u 1 1 = = 1.193 (a) p = 2 u 1 − ( fc / f ) 1 − (6 / 11) 2
(b)
ug u
= 1 − (6 / 11) 2 = 0.8381
Prob. 12.23 For the TE10 mode, π x − jβ z H zs = H o cos e a jβ a π x − jβ z H xs = H o sin e π a π x − jβ z H o sin e π a Exs = 0 = Ezs = H ys E ys = −
jωμ a
E s × H s* =
0 H
* xs
0
E ys 0
jωμ a
H
* zs
= E ys H zs* a x − E ys H xs* a z
ωμβ a 2 2 π x πx πx H cos H o sin =− sin ax + az π π2 a a a ωμβ a 2 2 2 π x 1 * H o sin Pave = Re E s × H s = az 2 2π 2 a 2
2 o
Prob. 12.24 Pave =
| Exs |2 + | E ys |2 2η
ω 2 μ 2π 2 2 2 H o sin π y / ba z az = 2η b 2 h 4
where η = ηTE10 . Pave = Pave .dS =
Pave =
ω 2 μ 2π 2 2 a b H o sin 2 π y / bdxdy 2 4 2η b h x =0 y =0
ω 2 μ 2π 2 2 H o ab / 2 2η b 2 h 4
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Principles of Electromagnetics, 6e
401
h 2 = (mπ / a ) 2 + (nπ / b) 2 =
But Pave =
π2 b2
,
ω 2 μ 2 ab3 H o 2 4π 2η
Prob. 12.25
πμ f π × 12 × 109 × 4π × 10−7 = = 2.858 × 10−2 Rs = 7 σc 5.8 × 10 f c10 =
u' 3 × 108 = = 4.651 GHz 2a 2 2.6 × 2 × 10−2 1/ 2
u' 1 1 f c11 = 2 + 2 = 10.4 GHz 2 a b μ 377 = = 233.81Ω η'= ε 2.6 (a) For TE10 mode, eq.(12.57) gives
α d + j β d = −ω 2 με + k x 2 + k y 2 + jωμσ d = −ω 2 / u 2 +
π2 a2
+ jωμσ d 2
2π × 12 × 109 π2 (2.6) = − + + j 2π × 12 × 109 × 4π × 10−7 × 10−4 −2 2 8 (2 × 10 ) 3 × 10 = 0.012682 + j373.57
α d = 0.012682 Np/m αc =
=
1 b fc 2 + ( ) bη ' 1 − ( f c / f ) 2 a f 2 Rs
2
2 x 2.858 x10−2
1 1 4.651 2 ) = 0.0153 Np/m + ( 10−2 (233.81) 1 − (4.651 / 12) 2 2 2 12
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Principles of Electromagnetics, 6e
402 (b) For TE11 mode,
α d + j β d = −ω 2 / u 2 + 1 / a 2 + 1 / b 2 + jωμσ d = −139556.21 +
π2 (10−2 ) 2
+ j 9.4748 = 0.02344 + j 202.14
α d = 0.02344 Np/m αc =
(b / a )3 + 1 (1 / 8) + 1 2 × 2.858 × 10−2 = 2 bη ' 1 − ( f c / f ) 2 (b / a) + 1 10−2 (233.81) 1 − (10.4 / 12) 2 (1 / 4) + 1 2 Rs
α c = 0.0441 Np/m Prob. 12.26
ε c = ε '− jε '' = ε − j
Comparing this with
σ ω
ε c = 16ε o (1 − j10−4 ) = 16ε o − j16ε o × 10−4 σ = 16ε o x10−4 ε = 16ε o , ω
For TM21 mode,
1/ 2
u ' m2 n2 fc = 2 + 2 2 a b
= 2.0963 GHz,
f = 1.1 f c = 2.3059 GHz
σ = 16ε oω × 10−4 = 16 × 2π × 2.3059 × 109 × η'=
αd =
10−9 × 10−4 = 2.0525 × 10−4 36π
μ = 30π ε
ση ' 2 1 − ( fc / f )2
Eo e −α d z = 0.8 Eo
=
4.1 × 10−4 × 30π = 0.0231 Np/m 2 1 − 1 / 1.12 z=
1
αd
ln(1 / 0.8) = 9.66 m
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Principles of Electromagnetics, 6e
403 Prob. 12.27 For TM21 mode,
αc =
2 Rs bη ' 1 − ( f c / f ) 2
π fμ π × 2.3059 × 109 × 4π × 10−7 Rs = = = = 0.0246 σ cδ σc 1.5 × 107 1
αc =
2 × 0.0246 = 0.0314 Np/m 4π × 10−2 × 30π × 0.4166
Eo e − (αc +α d ) z = 0.7 Eo
z=
1 ln(1 / 0.7) = 6.5445 m αc + αd
Prob. 12.28 u' 3 × 108 = = 2.5 GHz 2a 2 × 6 × 10−2 377 η' = = = 483 2 2 f 2.5 1− 1− c 4 f
f c10 =
ηTE
From Example 12.5, Pave =
Eo2 ab (2.2) 2 × 106 × 6 × 3 × 10−4 = = 9.0196 mW 2η 2 × 483
Prob. 12.29 For TE10 mode, u' 3 × 108 fc = = = 2.151 GHz 2a 2 2.11 × 4.8 × 10−2
(a) loss tangent =
σ =d ωε
σ = dωε = 3 × 10−4 × 2π × 4 × 109 × 2.11 × η'=
120π = 259.53 2.11
10−9 = 1.4086 × 10−4 36π
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Principles of Electromagnetics, 6e
404
αd =
ση ' 2 1 − ( fc / f )
2
=
1.4067 × 10−4 × 259.53 2 1 − (2.151 / 4)
2
= 2.165 × 10−2 Np/m
μ fπ π × 4 × 109 × 4π × 10−7 (b) Rs = = = 1.9625 × 10−2 7 σc 4.1 × 10 αc =
3.925 × 10−2 (0.5 + 0.5 × 0.2892) 1 b 2 + = ( f / f ) c 2.4 × 10−2 × 259.53 × 0.8431 bη ' 1 − ( f c / f ) 2 2 a 2 Rs
= 4.818 × 10−3 Np/m Prob. 12.30
αc =
=
2 Rs bη ' 1 −
π fμ 2 1 b f 2 σc + c = 2 2 fc f c 2 a f bη ' 1 − f f 1 2 4π × 10−7 × π f × 2
0.5 × 10−2 × (120π / 2.25) 5.8 × 107 1 − =
10−5 f 30 (5.8 / 2.25) 1 −
1 1 f 2 + c 2 2 f
f 2 1 + c 2 f c f f
f 2 1 + c 2 f c f f
The MATLAB code is shown below k=10^(-5)/(30*sqrt(5.8/2.25)); fc=10^10; for n=1:1000 f(n)=fc*(n/100+1); fn=f(n); num=sqrt(fn)*(1 +(fc/fn)^2); den=sqrt(1- (fc/fn)^2); alpha(n) =k*num/den; end plot(f/10^9,alpha) xlabel('frequency (GHz)') ylabel('attenuation') grid
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Principles of Electromagnetics, 6e
405 The plot of attenuation versus frequency is shown below.
attenuation (Np/m)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
10
20
30
40
50
60
frequency (GHz) Prob. 12.31 (a) For TE10 mode,
fc =
u' , 2a
c 2.11
u' =
3 × 108 fc = = 4.589 GHz 2.11(2 × 2.25 × 10−2 ) (b)
α cTE10 =
1 b + ( f c / f )2 bη ' 1 − ( f c / f ) 2 2 a 2 Rs
π fμ π × 5 × 109 × 4π × 10−7 Rs = = = 3.796 × 10−2 7 σc 1.37 × 10 η' =
377 = 259.54 2.11
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70
80
90
100
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Principles of Electromagnetics, 6e
406 1.5 (4.589 / 5) 2 ] 2.25 = 0.05217 Np/m −4 1.5 × 10 (259.54) 1 − (4.589 / 5) 2
2 × 3.796 × 10−2 [0.5 +
αc =
Prob. 12.32
(a)
f c10
u' 3 × 108 = = = 3.947 GHz 2a 2 × 3.8 × 10−2 2
f u g = u ' 1 − c = 3 × 108 1 − (0.3947) 2 = 2.756 × 108 m/s f (b) α = α d + α c
α d = 0 since the guide is air-filled. π fμ π × 1010 × 4π × 10−7 = = 2.609 × 10−2 Ω σc 5.8 × 107
Rs =
2 b fc αc = 0.5 + 2 a f f c bη ' 1 − f
2 Rs
=
2 × 2.609 × 10−2 1.6 × 10−2 (377) 1 − ( 0.3947 )
2
1.6 5.218 × 0.5656 2 0.5 + 3.8 ( 0.3947 ) = 554.23
= 5.325 × 10−3 Np/m
α c (dB) = 8.686 × 5.325 × 10−3 = 0.04626 dB/m Prob. 12.33 f c10 =
β' =
ω u'
=
u' c 3 × 108 = = = 3.991 GHz 2a 2a ε r μr 2 × 2.5 × 10−2 2.26 2π f ε r c 2
2
f 3.991 F = 1− c = 1− = 0.8467 7.5 f
β = β 'F =
2π × 7.5 × 109 2.26 0.8467 = 199.94 rad/m 3 × 108
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Principles of Electromagnetics, 6e
407
αd =
αc =
Rs =
ση ' 2F
=
σηo 10−4 (377) = = 1.481 × 10−2 Np/m 2 F ε r 2 × 0.8467 2.26
2 Rs bη ' 1 −
2 b fc 0.5 + 2 a f f c f
π fμ π × 7.5 × 109 × 4π × 10−7 = = 0.0519 σc 1.1 × 107
2 1.5 3.991 2 × 0.0519 0.5 + 2.5 7.5 0.1038 × 0.6698 αc = = 377 3.1848 −2 1.5 × 10 × × 0.8467 2.66 = 0.02183 Np/m
up =
u' c 3 × 108 = = = 2.357 × 108 m/s F F ε r 0.8467 2.26
3 × 108 × 0.8467 = 1.689 × 108 m/s 2.26 3 × 108 u' c λc = = = = 0.05 m = 5 cm(= 2a, as expected) f c f c ε r 3.991 × 109 2.26
ug = u ' F =
Prob. 12.34
The cutoff frequency of the dominant mode is u 3 × 108 f c10 = = = 6.56 GHz 2a 4.576 × 10−2 The surface resistance is Rs =
π fμ π × 8.4 × 109 × 4π × 10−7 = = 23.91 × 10−3 5.8 × 107 σc
For TE10 mode,
αc =
2 Rs
2 b fc 0.5 + 2 a f f c f
bη ' 1 − f c 6.56 = = 0.781, f 8.4
η ' = ηo = 377
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Principles of Electromagnetics, 6e
408 2 × 23.91 × 10−3
1.016 2 0.5 + 0.781) ( 2.286 1.016 × 10−2 × 377 1 − 0.7812 −3 47.82 × 10 (0.5 + 0.2711) = = 15.42 × 10−3 Np/m 3.83 × 0.6245 =15.42 × 10−3 × 8.686 dB/m = 0.1339 dB/m
αc =
Prob. 12.35 For TE10 mode,
αc =
1 b fc 2 + ( ) bη ' 1 − ( f c / f ) 2 a f 2 Rs
2
But a = b, Rs =
1
σ cδ
=
π fμ σc
π fμ σc
1 f k f + ( c )2 f 1 f 2 αc = + ( c )2 = 2 2 f aη ' 1 − ( f c / f ) 1 − ( fc / f )2 2
where k is a constant. f c 2 1/ 2 1 −1/ 2 3 2 −5/ 2 k 1 1/ 2 f ) ] [ f ] − [ f + f c 2 f −3/ 2 ](2 f c 2 f −3 )[1 − ( c ) 2 ]−1/ 2 − fc f dα c f 4 2 2 2 f = 2 df 1 − ( fc / f ) dα c For minimum value, = 0 . This leads to f = 2.962 fc. df k[1 − (
Prob. 12.36
For the TE mode to z,
Ezs = 0, H zs = H o cos(mπ x / a ) cos(nπ y / b)sin( pπ z / c) E ys = −
γ ∂ Ezs jωμ ∂ H zs jωμ + 2 = − 2 (mπ / a ) H o sin(mπ x / a )cos(nπ y / b)sin( pπ z / c) 2 ∂x h ∂y h h
as required. Exs = −
γ ∂ Ezs jωμ ∂ H zs jωμ − 2 = 2 (nπ / b) H o cos(mπ x / a )sin(nπ y / b)sin( pπ z / c) 2 h ∂x h ∂y h
From Maxwell’s equation,
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Principles of Electromagnetics, 6e
409
∂ − jωμ H s = ∇ × E s = ∂ x
∂ ∂y
∂ ∂z
Exs
E ys
0
H xs =
1 ∂ E ys 1 = − 2 (mπ / a )( pπ / c ) H o sin(mπ x / a ) cos(nπ y / b) cos( pπ z / c) jωμ ∂ z h
Prob. 12.37 Maxwell’s equation can be written as
jωε ∂ Ezs γ ∂ H zs − h2 ∂ y h2 ∂ x For a rectangular cavity, H xs =
h 2 = k x 2 + k y 2 = (mπ / a ) 2 + (nπ / b) 2
For TM mode, Hzs = 0 and Ezs = Eo sin(mπ x / a )sin(nπ y / b) cos( pπ z / c) Thus jωε ∂ Ezs jωε = 2 (nπ / b) Eo sin(mπ x / a)cos(nπ y / b) cos( pπ z / c) 2 h ∂y h as required. H xs =
H xs = −
jωε ∂ Ezs γ ∂ H zs − h2 ∂ x h2 ∂ y
jωε (mπ / a ) Eo cos(mπ x / a )sin(nπ y / b) cos( pπ z / c) h2 From Maxwell’s equation, =−
∂ jωε E s = ∇ × H s = ∂ x
∂ ∂y
∂ ∂z
H xs
H ys
0
E ys =
1 ∂ H xs 1 = 2 (nπ / b)( pπ / c) Eo sin(mπ x / a )cos(nπ y / b)cos( pπ z / c) jωε ∂ z h
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Principles of Electromagnetics, 6e
410 Prob. 12.38
fr =
u' ( m / a ) 2 + ( n / b) 2 + ( p / c ) 2 2
where for TM mode to z, m = 1, 2, 3,…, n=1, 2, 3, …., p = 0, 1, 2, …. and for TE mode to z, m = 0,1, 2, 3,…, n=0,1, 2, 3, …., p = 1, 2, 3, … , (m + n) ≠ 0 . (a) If a < b < c, 1/a > 1/b > 1/c, u' 1 1 + 2 2 2 a b u' 1 1 u' 1 1 The lowest TE mode is TE011 with f r = + 2 < + 2 2 2 2 b c 2 a b
The lowest TM mode is TM110 with f r =
Hence the dominant mode is TE011. (b) If a > b > c, 1/a < 1/b < 1/c, u' 1 1 + 2 2 2 a b u' 1 1 u' 1 1 The lowest TE mode is TE101 with f r = + 2 > + 2 2 2 2 a c 2 a b
The lowest TM mode is TM110 with f r =
Hence the dominant mode is TM110. (c) If a = c > b, 1/a = 1/c < 1/b, u' 1 1 + 2 2 2 a b u' 1 1 u' 1 1 The lowest TE mode is TE101 with f r = + 2 < + 2 2 2 2 a c 2 a b
The lowest TM mode is TM110 with f r =
Hence the dominant mode is TE101.
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Principles of Electromagnetics, 6e
411 Prob. 12.39 (a) Since a > b < c, the dominant mode is TE101.
c 1 1 3 × 108 fr = + = 2 a2 c2 2
δ=
(b)
1
π f r101μoσ
=
1 1 + = 9.014 GHz −2 2 (2 × 10 ) (3 × 10−2 ) 2 1
π × 9.014 × 10 × 4π × 10−7 × 5.8 × 107 9
1
=
4π 2 × 5.8 × 9.014 × 109 (4 + 9)6 × 10−2 0.78 (a 2 + c 2 )abc Q= = = δ 2b(a 3 + c 3 ) + ac(a 2 + c 2 ) δ [ 2(8 + 27) + 6(4 + 9)] 148δ =
0.78 4π 2 × 5.8 × 9.014 × 109 = 7571.5 148
Prob. 12.40 u' =
1
με
=
c 3 × 108 = = 1.897 × 108 2.5 2.5 2
2
2
2
2
u ' m n p 1.897 × 108 × 102 m n p fc = + + = + + 2 a b c 2 1 2 3 = 9.485 m 2 + 0.25n 2 + 0.111 p 2 GHz
f r101
= 9.485 1 + 0 + 0.111 = 10 GHz
f r 011
= 9.485 0 + 0.25 + 0.111 = 5.701 GHz
f r 012
= 9.485 0 + 0.25 + 0.444 = 7.906 GHz
f r 013
= 9.485 0 + 0.25 + 0.999 = 10.61 GHz
= 9.485 0 + 1 + 0.111 = 10 GHz f r 021 Thus, the first five resonant frequencies are:
5.701 GHz(TE 011 ) 7.906 GHz (TE 012 ) 10 GHz (TE101 and TE 021 ) 10.61 GHz (TE 013 or TM110 ) 11.07 GHz (TE111 or TM111 )
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2
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Principles of Electromagnetics, 6e
412 Prob. 12.41 Q=
(a 2 + c 2 )abc δ 2b(a 3 + c3 ) + ac(a 2 + c 2 )
When Q=
a = b = c,
a 2a 2 a 3 2a 5 = = 4 3 2 2 3δ δ 2a × 2a + a × 2a 6δ a
Prob. 12.42 (a) Since a > b < c, the dominant mode is TE101
f r101 =
u' 1 1 3 × 108 × 102 + + = 0 c2 2 a2 2
(b) QTE101 = = But
1 1 + = 16.77 GHz 22 12
(a 2 + c 2 )abc δ 2b(a 3 + c3 ) + ac(a 2 + c 2 )
(400 + 100)20 × 8 × 10 × 10−3 3.279 × 10−3 = δ [16(8000 + 1000) + 200(400 + 100)] δ
δ=
1
π f r101μoσ
QTE101 = 3.279 × 10−3
=
1
π 16.77 × 109 × 4π × 10−7 × 6.1 × 107
=
10−4 200.961
200.961 = 6589.51 10−4
Prob. 12.43
c m2 + n2 + p 2 2a The lowest possible modes are TE101, TE011, and TM110. Hence fr =
fr =
c 2 2a
a=
c fr 2
=
3 × 108 = 7.071 cm 2 × 3 × 109
a = b = c = 7.071 cm
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Principles of Electromagnetics, 6e
413 Prob. 12.44 (a) a = b = c u' fr = m2 + n2 + p 2 2a For the dominant mode TE101 ,
u' c 1+1 = 2 2a 2a c 2 3 × 108 2 a= = = 0.03788 m 2 f r 2 × 5.6 × 109 fr =
a = b = c = 3.788 cm (b) For ε r = 2.05, a=
c
u' =
εr
c 2 0.03788 = = 0.02646 2 fr ε r 2.05
a = b = c = 2.646 cm
Prob. 12.45
(a) This is a TM mode to z. From Maxwell’s equations, ∇ × E s = − jωμ H s
Hs = −
1 jωμ
∇ × Es =
j
ωμ
∂ ∂x
∂ ∂y
∂ ∂z
0
0
Ezs ( x, y )
=
j ∂ Ezs ∂ Ezs ax − ay ωμ ∂ y ∂x
But Ezs = 200sin 30π x sin 30π y,
1
ωμ
=
1 10−2 = 6 × 109 × 4π × 10−7 24π
j10−2 Hs = × 200 × 30π {sin 30π x cos30π ya x − cos30π x sin 30π ya y } 24π H = Re (Hs e jωt )
H = 2.5{− sin 30π x cos30π ya x + cos30π x sin 30π ya y } sin 6 × 109 π t A/m
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Principles of Electromagnetics, 6e
414
(b)
E = Ez a z ,
H = H xa x + H y a y
EH = 0
Prob. 12.46
(a) a = b = c a=
⎯⎯ →
f r101 =
3 × 108 = 12 × 109 a 2
3 × 108 = 1.77 cm 2 × 12 × 109
(b) QTE101 = =
a a π f r101μσ = 3δ 3 1.77 × 10−2 π × 12 × 109 × 4π × 10−7 × 5.8 × 107 = 9767.61 3
Prob. 12.47 2
2
u' m n p fr = + + 2 a b c f r101 =
3 × 108 2
2
1 1 + = 44.186 MHz 2 (10.2) (3.6) 2
f r 011 = 150
1 1 MHz = 45.093 MHz + 2 (8.7) (3.6) 2
f r111 = 150
1 1 1 MHz = 47.43 MHz + + 2 2 (10.2) (8.7) (3.6) 2
f r110 = 150
1 1 MHz = 22.66 MHz + 2 (10.2) (8.7) 2
f r102 = 150
1 4 MHz = 84.62 MHz + 2 (10.2) (3.6) 2
f r 201 = 150
4 1 MHz = 51 MHz + 2 (10.2) (3.6) 2
Thus, the resonant frequences below 50 MHz are f r110 , f r101, f r 011 , and f r111 Prob. 12.48
n = c/um =
3 × 108 = 1.4286 2.1 × 108
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Principles of Electromagnetics, 6e
415 Prob. 12.49 When an optical fiber is used as the transmission medium, cable radiation is eliminated. Thus, optical fibers offer total EMI isolation because they neither emit nor pick up EM waves. Prob. 12.50
(a) NA =
n12 − n2 2 =
1.622 − 1.6042 = 0.2271
(b) NA = sin θ a = 0.2271 or θ a = sin –1 0.2271 = 13.13o (c) V =
πd π × 50 × 10−6 × 0.2271 = 27.441 NA = 1300 × 10−9 λ
N = V2/2 6 modes Prob. 12.51 π d 2 3 π × 2 × 5 × 10−6 1.482 − 1.462 = 5.86 V= n1 − n2 = −9 1300 × 10 λ 2 V = 17.17 or 17 modes N= 2 Prob. 12.52
(a) NA = sin θ a =
n12 − n2 2 = 1.532 − 1.452 = 0.4883
θ a = sin –1 0.4883 = 29.23o (b) P(l)/P(0) = 10- α l / 10 = 10-0.4X5/10 = 0.631 i.e. 63.1 % Prob. 12.53
P () = P(0)10−α /10 = 10 × 10−0.5×0.85/10 = 9.0678 mW Prob. 12.54 As shown in Eq. (10.35), log10 P1/P2 = 0.434 ln P1/P2 ,
1 Np = 20 log10 e = 8.686 dB or 1 Np/km = 8.686 dB/km, or 1Np/m = 8686 dB/km. Thus,
α12 = 8686α10
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Principles of Electromagnetics, 6e
416 Prob. 12.55
α = 10log10
Pin 1.2 × 10−3 = 10log10 = 30.792 1 × 10−6 Pout
0.4 Np/km 8.686 30.792 30.392 dB = = 76.98 km = 0.4 dB/km α
α = 0.4 dB/km =
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Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
417 415 CHAPTER 13 P.E. 13.1 (a) For this case, r is at near field. I dl sin θ j β 1 − j β r 2π λ . = 72o + 2 e Hφ s = o , βr = λ 5 4π r r
2π c
2π × 3 × 108 λ= = = 6π , 108 ω
Hφ s =
(0.25)
β=
2π
λ
=
1 3
6π sin 30o − j 72o 1 j1 / 3 100 e + = 0.2119∠ − 20.511o mA/m 2 4π 6π / 5 (6π / 5)
H = Im ( Hφ s e jωt aφ )
Im is used since I = Io sin ωt
= 0.2119sin(108 − 20.5o )aφ mA/m
(b) For this case, r is at far field. β = Hφ s =
j (0.25)(
2π
)
λ
λ
Sin60o e − j 0
λ 100 4π (6π × 200)
H = Im ( H φ s aφ e jωt )
2π
× 200λ = 0o
o
= 0.2871e j 90 μ Am o
= 0.2871sin(108 + 90o )aφ
P. E. 13.2
(a) l =
λ
4
= 1.5m ,
(b) Io = 83.3mA 1 (c) Rrad = 36.56 Ω , Prad = (0.0833)2 36.56 2 = 126.8 mW. (d) ZL = 36.5 + j21.25,
Γ=
36.5 + j 21.25 − 75 = 0.3874∠140.3o 36.5 + j 21.25 + 75
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μ Am .
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Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
418 416
s=
1 + 0.3874 = 2.265 1 − 0.3874
P.E. 13.3
D=
4π U max Prad
(a) For the Hertzian monopole U (θ ,φ ) = sin 2 θ , 0 < θ < π / 2, π
Prad =
2 2π
sin θ φ
2
sin θ dθ dφ =
=0 =0
D= (b) For the
λ 4
0 < φ < 2π , Umax = 1
4π 3
4π (1) =3 4π 3
monopole,
π
cos 2 ( cosθ ) 2 U (θ ,φ ) = , Umax = 1 sin 2 θ
π
Prad
cos 2 ( cosθ ) 2 = sin θ dθ dφ = 2π (0.609) sin 2 θ θ =0 φ =0
D=
4π (1) = 3.28 2π (0.609)
π
2 2π
P. E. 13.4
(a) Prad = η r Pin = 0.95(0.4) 4π U max 4π (0.5) = = 16.53 D= Prad 0.4 × 0.95 4π (0.5) (b) D = = 20.94 0.3
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Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
419 417 P. E. 13.5 π
Prad =
2 2π
sin θ sin θ dθ dφ =
θ =0 φ =0
D=
4π (1)
π2
π2 2
, Umax = 1
= 2.546
2
P. E. 13.6
1 (a) f (θ ) = cosθ cos ( β d cosθ + α ) 2 2π λ . =π where α = π , β d = λ 2 1 f (θ ) = cosθ cos (π cosθ + π ) 2 unit pattern
group pattern
For the group pattern, we have nulls at
π (cosθ + 1) = ± π 2 2
θ = ±π 2
and maxima at
π (cosθ + 1) = 0, π 2
cosθ = −1,1
⎯⎯ → θ = 0, π
Thus the group pattern and the resultant patterns are as shown in Fig.13.15(a)
1 (b) f (θ ) = cosθ cos ( β d cosθ + α ) 2 − π , βd = π / 2 where α = 2 1 π f (θ ) = cosθ cos cosθ − π 2 2 2 unit pattern
group pattern
For the group pattern, the nulls are at
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Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
420 418
π (cosθ − 1) = − π 4 2
θ = 180o
cos θ − 1 = 0
θ =0
and maxima at
Thus the group pattern and the resultant patterns are as shown in Fig.13.15(b) P. E. 13.7 (a)
●
●
●
λ
●
●
● ● ● ● λ 2 ● 2 ●
λ ●● 2
1: 2 :1
1: 2 :1
x
x
λ
2 Thus, we take a pair at a time and multiply the patterns as shown below.
●
×
●
x
≡
x
(b) The group pattern is the normalized array factor, i.e. 1 N ( N − 1) i 2ψ N ( N − 1)( N − 2) i 3ψ ( AF ) n = 1 + Neiψ + e + e + ............ + ei ( N −1)ψ 2! 3! where
N −1
N
i −1
= i = 1+ N +
N − 1 N ( N − 1)( N − 2) + + ........... 2! 3!
= (1 + 1) N −1 = 2 N −1 ( AF ) n =
1 2 N −1
1 + e jψ
N −1
=
1 2 N −1
e
jψ
2
e
− jψ
2
+e
jψ
N −1 2
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Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
421 419
=
1 2 N −1
2cos
ψ
N −1
= cos
2
N −1
ψ 2
P. E. 13.8
λ2 c 3 × 108 Ae = Gd , λ = = =3m f 108 4π For the Hertzian dipole, Gd = 1.5sin 2 θ
λ2 (1.5sin 2 θ ) 4π 1.5λ 2 1.5 × 9 = = 1.074 m 2 Ae,max = 4π 4π Ae =
By definition, Pr 3 × 10−6 Pave = = Ae 1.074
Pr = Ae Pave
= 2.793 μ W / m 2
P. E. 13.9
(a) Gd =
4π r 2 Pave = Prad
4π r 2
1 E2 2 η
Prad
=
2π r 2 E 2 η Prad
2π × 400 × 106 × 144 × 10−6 = = 0.0096 120π × 100 × 103 G = 10log10 Gd = -20.18 dB
(b) G = η r Gd = 0.98 × 0.0096 = 9.408 × 10−3 P. E. 13.10 1
λ 2Gd 2σ Prad 4 r= 3 (4π ) Pr c 3 × 108 where λ = = = 0.05m f 6 × 109
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Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
422 420
Ae = 0.7π a 2 = 0.7π (1.8) 2 = 7.125m 2 Gd =
4π Ae
λ
4π (7.125) = 3.581 × 104 −4 25 × 10
=
2
25 × 10−4 × (3.581) 2 × 108 × 5 × 60 × 103 r= (4π )3 × 0.26 × 10−3
1
4
= 1168.4m = 0.631 nm
At r = P=
rmax
2
= 584.2m,
Gd Prad 3.581 × 104 × 60 × 103 = = 501 W/m 2 4π r 2 4π (584.2) 2
Prob. 13.1
Using vector transformation, Ars = Axs sin θ cos φ , Aθ s = Axs cosθ cos φ , Aφ s = − Axs sin φ As =
50e− j β r (sin θ cos φ ar + cosθ cos φ aθ − sin φ aφ ) r
∇ × As
μ
−
= Hs =
100cosθ sin φ − j β r 50 e ar − 2 (1 − j β r )sin φ e − j β r aθ 2 μ r sin θ μr
50 cosθ cos φ (1 + j β r )e − j β r aφ 2 μr
At far field, only
1 term remains. Hence r
j 50 − j β r β e (sin φ aθ − cosθ cos φ aφ ) μr − j 50 βη e − j β r (sin φ aφ + cosθ cos φ aθ ) E s = −η ar × H s = μr Hs =
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H Sadiku Sadiku & & Kulkarni Kulkarni
= Re H s e jωt =
Prob. 13.2
(a)
λ=
−50 β sin(ωt − β r )(sin φ aθ − cosθ cos φ aφ ) A/m μr
3 × 108 c = = 0.75 m f 400 × 106
423 421
2 λ50)ηβ 2d 2 2(0.02 jω t rEmin==Re E=s e = sin( r ωt − β r )(sin φ aφ + cosθ cos φ aθ ) V/m μr
λ
λ
λ
λ
i.e. r is in the far field. −50 jω t = − j β r β sin(ωt − β r )(sin φ aθ − cosθ cos φ aφ ) A/m H = RejIoH β dl se Hφ s = sin θ e μ r 4π r Prob. 13.2 2π 3 ×8 × 0.02λ × sin 90o 3 × 10 c β I dl λ= 0.75 m = θ= |(a) H φ s |= λ o= sin = 5 × 10−4 = 0.5 mA/m 6 4π rf 400 × 10 4π (60) 2 2 λ ) V/m |Eθ s |= 2ηdo | H φ2(0.02 |= 0.1885 rmin = = s r i.e. inφ sthe far field. (b) r is| H |= 0.5 mA/m| jI o β dl 2 H = e − j β r sinθdl 2 (c)φ s R rad4π= r80π 2 = 80π 2 ( 0.02 ) = 0.3158 Ω λ 2π 3 × 1 × 0.02λ × sin 90o 1 2 β I dl | I oθ| =R rad =λ (9)(0.3158) = 1.421 |(d) H φ s |=Prado = sin = 5W × 10−4 = 0.5 mA/m 2 4π (60) 4π r2 |Eθ s |= ηo | H φ s |= 0.1885 V/m Prob. 13.3 2 2 2 dl 2 0.024λ (a) (b) R|rad H φ=s 80 |= π 0.5mA/m| = 80π = 0.4548 Ω λ 2 λ dl − 502 2 2 0.4548 L − Zo (c) Γ = RZ ) = 0.3158 Ω −0.982 (b) rad = 80π = = 80π (=0.02 λ + 50 ZL + Z o 0.4548 1 2 1+ | Γ | 1 1.982 (d) = (9)(0.3158) = 1.421 W s = Prad == 2 | I o | R rad= 110.11 2 1− | Γ | 1 − 0.982 Prob. 13.3 Prob. 13.4 2 2 8 dl 210 2 0.024λ c 3 × (a) λR=rad ==80π = 6= m 80π = 0.4548 Ω λ f 50 × 10λ6 − Z 2− Z o 2(5 ×0.4548 10−3 ) 2 50 = −0.982 (b)Γ = 2dL −6 rmin = Z +=Z = 0.4548 =+ 8.333 50 × 10 m r = 15 cm o 6 λL 1+ | Γ | far 1.982 i.e. field. = 110.11 = s = r is in the 1− | Γ | 1 − 0.982
Prob. 13.4 c 3 × 108 =6m λ= = f 50 × 106
2(5 × 10−3 ) 2 Copyright © 2015−6by Oxford University Press = 8.333 × 10 m r = 15 cm 6 λ i.e. r is in the far field. rmin =
2d 2
=
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Sadiku Sadiku & & Kulkarni Kulkarni
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
424 422 Hφ s =
jI o β dl sin θ e − j β r 4π r
2π × 5 × 10−3 × sin 30o 5 I o β dl 6 | H φ s |= sin θ = = × 10−2 = 2.778 mA/m −2 4π r 4π × 15 × 10 18 −3 |Eθ s |= ηo | H φ s |= 377 × 2.778 × 10 = 1.047 V/m 2×
Prob. 13.5
e− jβ r (a) Azs = 4π r
l
2
I
−l
o
(1 −
2z l
2
)e j β z cos θ ∂ z
l 2 2z 2z e − j β r 2 = Io (1 − ) cos( β z cosθ )dz + j (1 − )sin( β z cos θ ) dz l l 4π r −l −l 2 2 l
l
2 e− j β r 2z = 2 I o (1 − ) cos( β z cosθ )dz l 4π r 0
I oe− jβ r 2 . 1 − cos( β l cosθ ) = 2 2 2 2π r β cos θ l
E s = − jωμ As
Eθ s = jωμ sin θ Azs = j βη sin θ Azs
sin θ 1 − cos( β l cos θ ) 2 2 β cos θ βl ( cosθ ) 2 If β l l , cos( β l cosθ ) = 1 − 2 . 2 2 2! Hence jη I o e − j β r Eθ s = π rl
jη I o β le− j β r sin θ , H φ s = Eθ s / η 8π r 2 Eθ s Pave = , Prad = Pave dS 2η
Eθ s =
Prad
2
2π π
n I βl 1 = o 2 sin 2 θ r 2 sin θ dθ dφ 0 0 2 8π r 2
l = 10π I o = 1 I o 2 Rrad 2 λ 2
2
l or Rrad = 20π 2 λ
2
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423 Sadiku Sadiku & & Kulkarni Kulkarni
l (b) 0.5 = 20π 2 λ
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
2
l = 0.05λ
425 423
Prob. 13.6 c 3 × 108 l 2 m λ = 0.5== 20π 26= 100 l = 0.05λ (b) f 3 × 10 λ = 4 m λ . Hence the antenna is electrically short. 2 2 Prob. 13.6 1 2 π I o dl π × 3× 4 8 Prad =c I 3o R×rad 10 = 40 λ = 40 100 = 5.685 W m = 100 λ = 2= f 3 × 106 =4m Prob. 13.7 λ . Hence the antenna is electrically short. Let us model this as a short Hertzian dipole. 2 2 1 2 π I o dl π2 × 3 × 4 Prad = I o Rrad = 40 dl =40 = 5.685 W 2 Rrad λ= 80π 2 =100 80π 2 (1 / 8) 2 = 12.34 Ω λ Prob. 13.7 Prob. Let us 13.8 model this as a short Hertzian dipole. 2 40Ω dl Rrad = 80π 2 = 80π 2 (1 / 8)I2 = 12.34 Ω λ Prob. 13.8 24 V
+ -
40Ω
I
Zin = 73+j42
+ -
24 V Zin = 73+j42 V 24 I= = = 0.1866 − j 0.0694 Rs + Z in 40 + 73 + j 42 1 | I |2 Rrad , Rrad = 73 2 1V 24 I rad = = (0.1991) = 2 × 73 = 1.447= W 0.1866 − j 0.0694 P Rs 2+ Z in 40 + 73 + j 42 Prad =
1 | I |2 Rrad , Rrad = 73 2 1 = (0.1991)2 × 73 = 1.447 W 2
Prad = Prad
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Copyright © 2015 by Oxford University Press
Sadiku Sadiku & & Kulkarni Kulkarni
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
426 424 Prob. 13.9 Change the limits in Eq. (13.16) to ± l As =
2
i.e.
μ I oe j β zCosθ ( j β cosθ cos β t + β sin β t ) − β 2 cos 2 θ + β 2 4π r
μ I oe jβ r 1 = 2π r β sin 2 θ
l 2 −l 2
βl βl βl βl sin 2 cos 2 cosθ − cosθ cos 2 sin 2 cosθ
But B = μ H = ∇ × A Hφ s =
1 ∂ ∂A ( rAθ ) − r , μr ∂ r ∂θ
where Ao = − Az sin θ , Ar = Az cosθ Hφ s =
I o e− j β r j β β l βl βl Io − jβ r βl e (......) sin cos cosθ − cosθ cos sin cosθ + 2 2π r β sinθ 2 2 2 2 2π r
For far field, only the
Hφ s =
jI o − j β r e 2π r
1 -term remains. Hence r
βl βl βl βl sin 2 cos 2 cos θ − cos θ cos 2 sin 2 cos θ sin θ
βl βl cos cosθ − cos 2 2 (b) f (θ ) = sin θ
For l = λ , f (θ ) =
cos (π cosθ ) + 1 sin θ
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Sadiku Sadiku & & Kulkarni Kulkarni
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
427 425
3π cos cosθ 3λ 2 For l = , f (θ ) = 2 sin θ
For l = 2λ , f (θ ) =
cosθ sin ( 2π cosθ ) sin θ
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Sadiku Sadiku & & Kulkarni Kulkarni
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
428 426
Prob. 13.10 (a) c 3 × 108 = 0.6667 m λ= = f 450 × 106 =
λ 2
= 0.333 m
(b)
σ = ωε
4 10−9 2π × 450 × 10 × 81 × 36π
= 1.975
6
με
2 2π × 460 × 106 81 σ 1 + (1.975 )2 + 1 1 + + 1 = c 2 2 ωε
β =ω
2π × 460 × 106 × 11.4086 = 109.91 3 × 108 2π λ= = 0.0572 =
=
β λ 2
= 28.58 mm
Prob. 13.11 (a) c 3 × 108 = 260.8 m λ= = f 1.150 × 106 =
λ 4
= 65.22 m
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Sadiku Sadiku & & Kulkarni Kulkarni
λ= =
c 3 × 108 = = 0.5 m f 600 × 106
λ 4
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
= 0.125 m
429 427
Prob. 13.12 (b) This cis a monopole 3 × 108 antenna λ = c = 3 × 1086 = 3.333 m λ = f = 90 × 10 6 = 200 λf 1.5 × 10 = = 0.8333 m 4 l λ ,hence it is a Hertzian monopole.
(c)
2
2
c 31× 1082 dl 2 1 λ =Rrad == 80π 6 = 3.75 =m40π = 9.87 mΩ 2 × 10 λ f 80 200 1 2 λ Pt = m I o Rrad = Prad==0.9375 2 4 (d) 8 c2 23P×t 108 = 810.54 −3 = 0.5 m λ = I o == R = 9.87 6 10 × rad f 600 × 10
λ
I o== 28.47 = 0.125 A m 4 Sadiku & Kulkarni
Principles of Electromagnetics, 6e
Prob. Prob. 13.13 13.12 8 c 3 × 10antenna This is a monopole 428 = 30 m (a) λ = = 8 c f3 × 10 10 × 106 = 200 λ= = f 1.5 × 106 ηπ I o S Emax rλ 2 Emax = I = o 2 ηπ S l λ ,hencerλit is a Hertzian monopole.
50 × 10−3 × 32× 302 2 2 =2 9.071 Io =1 dl 2 1 mA (S=Nπ r ) Rrad = 120 80π 22π (0.2) 40 9.87 m Ω =100 π = 2 λ Copyright©200 2015by Oxford University Press 1 Prad = Pt 320 = πI o42SR2rad 320π 4π 2 (0.2)4 × 104 (b) Rrad = 2 4 = = 6.077Ω 304 λ 2P 8 I o 2 = t1 = 2 1 3 = 810.54 2 PradR=rad I o 9.87 Rrad ×=10−(9.071) × 10−6 × 6.077 2 2 I o = 28.47 A = 0.25 mW Prob. 13.13 Prob. 13.14 c 3 × 108 = 30 m (a) λ = = 6 f c 10 ×310 × 108 = 3.75 m λ= = f 80 × 106
S = Nπρ o2 Rrad = N2 =
320π 4 S 2
λ4
=
320π 4 N 2π 2 ρo4
λ4
⎯⎯ → N2 =
λ 4 Rrad
320π 6 (1.2 × 10−2 )4
(3.75) 4 × 8 =2015 248006 ⎯⎯ → Press N 498 Copyright © by Oxford University 320π 6 (1.2 × 10−2 )4
1 2 1 I o Rrad = (9.071) 2 × 10−6 × 6.077 2 2
Prad =
Sadiku & Kulkarni
= 0.25 mW
Principles of Electromagnetics, 6e
430
Prob. 13.14
c 3 × 108 = 3.75 m λ= = f 80 × 106 S = Nπρ o2 Rrad = N2 =
320π 4 S 2
320π 4 N 2π 2 ρo4
λ4
(3.75) 4 × 8 = 248006 320π 6 (1.2 × 10−2 )4
Prob. 13.15
Rrad =
(a)
λ4
=
⎯⎯ → N2 =
λ 4 Rrad
320π 6 (1.2 × 10−2 )4
⎯⎯ → N 498
320π 4 S 2
λ4
S = πρ o2 = π (0.4)2 = 0.5027 m 2
λ= Sadiku & Kulkarni
c 3 × 108 = = 50 m f 6 × 106
Rrad = (b) (c)
320π 4 (0.5027)2 = 1.26 mΩ (50) 4 1 1 Prad = I o 2 Rrad = (50)2 × 1.26 × 10−3429 = 1.575 W 2 2 R =
2π R μ f π a a a R dc = = π fμσ = 2 2δ 2δ σ S 2σπ a 2π a σ
0.4 4π × 10−7 × 6 × 106 × π R μ fπ = = 63.91 mΩ R = σ 4 × 10−3 5.8 × 107 a R rad 1.26 = × 100% = 1.933% η= R rad + R 1.26 + 63.91 Prob. 13.16
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π cos cosθ 2 (a) f (θ ) = sin θ
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Principles of Electromagnetics, 6e
2δ
2δ σ S
2σπ a
2π a
σ
0.4 4π × 10 × 6 × 10 × π R μ fπ = = 63.91 mΩ −3 σ 4 × 10 5.8 × 107 a R rad 1.26 = × 100% = 1.933% η= 431 R rad + R 1.26 + 63.91
Sadiku & Kulkarni R
−7
=
6
Principles of Electromagnetics, 6e
Prob. 13.16 π cos cosθ 2 (a) f (θ ) = sin θ
(b) The same as for
λ 2
dipole except that the fields are zero for θ
Sadiku & Kulkarni
π 2
as shown.
Principles of Electromagnetics, 6e
430 Prob. 13.17 Let Prad1 and Prad2 be the old and new radiated powers respectively.
Let Pohm1 and Pohm2 be the old and new ohmic powers respectively.
η r1 = 20% =
Prad 1 1 = Prad 1 + Pohm1 5
But
⎯⎯ →
4 Prad 1 = Pohm1
(1)
1 2 I Rs Δz 2 1 = I 2 Rs 2Δz = 2Pohm1 2
Pohm1 = Pohm2
(2) 2
1 2 1 2 Δz Prad 1 = I o Rrad = byI o2Oxford × 80πUniversity Copyright © 2015 Press 2 2 λ 2
1 1 2Δz Prad 2 = I o2 Rrad = I o2 × 80π 2 = 4 Prad 1 2 2 λ From (1) to (3),
η r2 =
4 Prad 1 Prad 2 P = = ohm1 = 33.3% Prad 2 + Pohm 2 4 Prad 1 + 2 Pohm1 3Pohm1 Copyright © 2015 by Oxford University Press
Prob. 13.18
(3)
Pohm2 = Sadiku & Kulkarni
1 2 I Rs 2Δz = 2Pohm1 2
(2) 2
1 1 Δz Prad 1 = I o2 Rrad = I o2 × 80π 2 2 2 λ 2 432 1 2 1 2 2 2 Δz Prad 2 = I o Rrad = I o × 80π = 4 Prad 1 2 2 λ From (1) to (3),
η r2 =
(3)
4 Prad 1 Prad 2 P = = ohm1 = 33.3% Prad 2 + Pohm 2 4 Prad 1 + 2 Pohm1 3Pohm1
Prob. 13.18
(a) Let H s =
cos 2θ − j β r e aH ηo r
a E × a H = ak
Hs =
Principles of Electromagnetics, 6e
⎯⎯ →
aθ × a H = ar
⎯⎯ → a H = aφ
cos 2θ − j β r e aφ 120π r
| Es |2 cos 2 (2θ ) ar = ar (b) Pave = 2η 2η r 2 Sadiku & Kulkarni Principles of Electromagnetics, 6e π 1 cos 2 2θ 2 1 Prad = r sin θ dθ dφ = (2π ) cos 2 2θ sin θ dθ 2 r 2η 240π 0 431 2 2 2 But cos 2θ = cos θ − sin θ = 2cos θ − 1
Prad
π
1 (2cos 2 θ − 1) 2 d (cos θ ) =− 120 0 =−
π
1 (4cos 4 θ − 4cos 2 θ + 1) d (cos θ ) 120 0
π 1 4cos5 θ 4cos3 θ =− − + cos θ 120 5 3 0 1 4 4 4 4 1 14 [− + − 1 − + − 1] = ( ) 120 5 3 5 3 120 15 = 7.778 mW =−
(c) Prad
1 =− 120 =−
120o
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(2cos 2 θ − 1) 2 d (cos θ )
60o
120o 1 4cos5 θ 4cos3 θ − + θ cos o 120 5 3 60
1 4 1 4 1 1 4 1 4 1 1 1 1 1 1 [ (− ) − (− ) − − ( ) + ( ) − ] = [ + − ] 120 5 32 3 8 2 5 32 3 8 2 60 40 2 6 = 5.972 mW 5.972 = 0.7678 or 76.78% 7.778 =−
which is Prob. 13.19 1 2
1 2
Pave = Re( E s × H s* ) = η | Hφ s |2 ar © 2015 by Oxford University Press 2 2 β Copyright Io 1 2 Prad = Pave dS = η θ cos 2 φ r 2 sin θ dθ dφ sin 2 16π 2 r 2
432
Sadiku & Kulkarni
Rrad
377 × 2502 × 1012 = = 87.27Ω 3 × 9 × 1016
Principles of Electromagnetics, 6e
433
Prob. 13.20 (a) Prad = Prad ⋅ dS = Pave .2π r 2 (hemisphere)
Prad 200 × 103 = = 12.73μW / m 2 2 6 2π r 2π (2500 × 10 )
Pave =
Pave = 12.73ar μ W/m 2 . (b)
(E ) Pave = max 2η
2
Emax = 2η Pave = 240π × 12.73 × 10−6
= 0.098 V/m
Prob. 13.21
U 4π r 2 Pave 8π sin θ cos φ = = Gd = U ave Pave .∂ s Pave .dS But Pave .dS =
π
π
2
2sin θ cosφ sin θ dθ dφ
θ =0 φ =0 π
2
π
π
= 2 cos φ dφ sin θ dθ = 2sin φ 0
0
2
2
0
π =π 2
Gd = 8sin θ cos φ D = Gd ,max = 8
Copyright © 2015 by Oxford University Press
Copyright © 2015 by Oxford University Press
Sadiku Sadiku & & Kulkarni Kulkarni
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
434 433 Prob. 13.22
3λ 2π 3λ β = = 3π , λ 2 2 2π 1 3λ 2π 1 3λ cos cos θ − cos −β r jI e λ 2 2 λ 2 2 Hφ s = o 2π r sin θ 3π 3π cos cosθ − cos −β r −β r jI e 2 2 jI o e cos (1.5π cos θ ) = o = 2π r sin θ 2π r sin θ Hence, the normalized radiated field pattern is From Prob. 13.11, set =
f (θ ) =
cos (1.5π cosθ )
sin θ which is plotted below.
Copyright © 2015 by Oxford University Press
Sadiku Sadiku & & Kulkarni Kulkarni
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
435 434 Prob. 13.23 The MATLAB code is shown below N=20; del= 2*pi/N; sum=0; for k=1:N theta = del*k; term = (1 – cos(theta))/theta; sum = sum + term; end int = del*sum
When the program is run, it gives the value of 2.4335. The accuracy may be increased by increasing N. Prob. 13.24
jη I o β dl sin θ e − j β r 4π r 2 dl Rrad = 80π 2 λ 1 4π r 2 . Eθ s 2 4π r Pave 2η Gd = = 1 2 Prad I o Rrad 2
(a) Eθ s =
2
2 2 2 2 4π r 2 1 λ 1 η I o β ( dl ) sin θ = . . I o 2 80π 2 dl η 16π 2 r 2 2
2
Gd = 1.5sin 2 θ (b) D = Gd ,max = 1.5 (c) Ae =
λ2 1.5λ 2 sin 2 θ Gd = 4π 4π
(d) Rrad
1 = 80π = 3.084 Ω 16
2
2
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Sadiku Sadiku & & Kulkarni Kulkarni
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
436 435 Prob. 13.25
Eφ s =
(a)
120π 2 I o S sin θ e − j β r 2 λ r 320π 4 S 2 Rrad = 4
λ
2
4π U (θ , φ ) 4π r 2 Pave 8π r 2 1 Eφ s Gd = = = 2 . 1 2 Prad I o 2η Rrad I o Rrad 2 8π r 2 1 I 2 S2 λ2 = 2 . .14400π 4 o2 4 sin 2 θ I o 2η r λ 320π 4 S 2
Gd = 1.5sin 2 θ (b)
D = 1.5
λ 2Gd λ 2 = 1.5sin 2 θ 4π 4π
(c)
Ae =
(d)
S = π a2 =
πd2 4
=
320π 6 (576) 2
Rrad = 0.927Ω
Prob. 13.26 (a) c 3 × 108 = 250 λ= = f 1.2 × 106 =
λ
= 62.5 m 4 (b) From eq. (13.30), Rrad = 36.5 Ω
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Sadiku Sadiku & & Kulkarni Kulkarni
Principlesof ofElectromagnetics, Electromagnetics,6e 6e Principles
437 436
(c) For λ /4-monopole,
π
cos( cosθ ) 2 f (θ ) = , sin θ
0