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• Ramon Lopez Lapiña

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CONDUCTORS

CURRENT AND CURRENT DENSITY • Current - is defined as electric charges in motion or defined as a rate of movement of a charge passing in a given reference point (or crossing in a given reference plane) of one coulomb per second

CURRENT AND CURRENT DENSITY • Current - is defined as the motion of positive charges (even though conduction in metals takes place through the motion of electrons). The unit of current is ampere (A) and current is symbolized by I,

dQ I dt

( convention current)

CURRENT AND CURRENT DENSITY • Consider the field theory, “the events occurring at a point was the main concerned rather than within the large region”, and in this case the concept of current density were important matter. • Current density - is defined as a vector represented by J and measured in amperes per square meter (A/m2). • The increment of current ΔI crossing an incremental surface, ΔS, is given as;

CURRENT AND CURRENT DENSITY

I  J N S

( normal to the current density)

or,

I  J  S

( current density is not perpendicular to the surface)

CURRENT AND CURRENT DENSITY • And the total current is obtained by integrating,

I   J  dS S

( convention current)

CURRENT AND CURRENT DENSITY • Current density can be related to the velocity of volume charge density at a given point. Consider the element of charge ΔQ = ρvΔv = ρvΔSΔL, as shown in the figure .

CURRENT AND CURRENT DENSITY • Assume that the charge element is oriented within the edges and parallel to the coordinate axes, and it shall only posses an x component of velocity.

CURRENT AND CURRENT DENSITY • With respect to the time interval Δt, the element of charge has moved a distance Δx, as shown in the figure.

CURRENT AND CURRENT DENSITY • Therefore a charge ΔQ = ρvΔSΔx has moved through a reference plane perpendicular to the direction of motion in a time increment Δt, and the resultant current is Q x I    S t t

CURRENT AND CURRENT DENSITY • And taking the limit with respect to time,

I   S x where: vx represents the x component of the velocity, v. • And in terms of current density,

J x   vx

CURRENT AND CURRENT DENSITY and in general

J   v where: J or ρvv is the convention current density

Sample Problem 1 The vector current density is given as J = (4/r2)cos θ ar + 20 e-2r sin θ aθ – r sin θ cos  a A/m2. (a) Find J at r = 3, θ = 0,  = π. (b) Find the total current passing through the spherical cap r = 3, 0 < θ < 20O, 0 <  < 2π, in the ar direction.

Solution (a) J J J J

4 2 r  2 cosar  20e sin a  r sin  cosa r 4  2 cos(0)ar  20e 2(3) sin( 0)a  3 sin( 0) cosa 3 4  ar  0  0 9  0.444 ar A / m 2

Solution

I   J  dS

r  3;  20 0   9

S

(b)

4  2 2r cos  a  20 e sin  a  r sin  cos  a  2 r     r sin ddar 0 0 r  2  9 4 1 2 I   cos   r sin  d  d  ; sin  cos   sin 2 2 0 0 r 2 2  9  1  I    4  sin 2dd 0 0 2 I 

2

 9

2

 9

0

0

I  2 d 

sin 2d

 9  2 1  I  2 0  cos 2   2  0     2  I  2   cos  cos 0  9   

I  2 (0.234 )  1.470 A

CONTINUITY OF CURRENT • The principle of conservation of charge states simply that charges can be neither created nor destroyed, although equal amounts of positive and negative charge may be simultaneously created, obtained by separation, destroyed, or lost by recombination. • The continuity equation, follow this principle when considering of any region bounded by a closed surface. And the current through the closed surface is

CONTINUITY OF CURRENT

I   J  dS S

the outward flow of positive charge must be balanced by a decrease of positive charge (or perhaps an increase of negative charge) within the closed surface.

CONTINUITY OF CURRENT • The current at closed surface, however, is an outward-flowing current and it is the integral form of the continuity equation, and the differential, or point, form is obtained by using the divergence theorem to change the surface integral into a volume integral:

J  dS  (   J ) d    S

vol

CONTINUITY OF CURRENT • Next represent the enclosed charge Qi by the volume integral of the charge density,

d (   J ) d     d   vol  dt vol

CONTINUITY OF CURRENT • Keeping the surface constant, the derivative becomes a partial derivative and may appear within the integral,

 (   J ) d    d  vol vol t

• For an incremental volume,

 (  J )   d t

CONTINUITY OF CURRENT • And the point from the continuity equation,

 (the current, or charge (  J )   t per second, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point)

Sample Problem 2 • Assume that an electron beam carries a total current of – 500 μA in the az direction, and has a current density Jz that is not a function of ρ or ǿ in the region 0 < ρ < 10-4 m and is zero for ρ < 10-4 m. If the electron velocities are given by vz = 8 x 107 z m/s, calculate ρv at ρ = 0 and z =:(a) 1 mm; (b) 2 cm: (c) 1 m.

Solution I   J  dS S

I 

2

0

110 4



0

Jzdd

2

110 4

0

0

I  Jz  d 

d

 2  2 110 I  Jz  0 2 0     I  Jz 2  0       I  Jz (  10 8 )

4

   

110 

4 2

2

(0) 2    2  

I  500  10 6 2 Jz     15915 . 5 A / m   10 8   10 8

Solution (a)

z  1mm  1 10 3 m Jz  15915 .5 v   Vz 8  10 7 1 10 3

at





 v  0.1989 C / m (b)



3

z  2cm  0.02 m Jz  15915 .5 v   Vz 8 10 7 0.02 

at





 v  9.95 10 3 C / m3  9.95mC / m3

Solution (c) at z  1m Jz  15915 .5 v   Vz 8 10 7 1





 v  1.989 10  4 C / m3  198 .9C / m3

Seatwork 1. Given the vector current density J = 10ρ2zaρ - 4ρcos2ΦaΦ A/m2: (a) find the current density at P(ρ = 3, Φ = 30O, z = 2); (b) determine the total current flowing outward through the circular band ρ = 3, 0 < Φ < 2π, 2 < z < 2.8. Answer 180aρ – 9aΦ A/m2; 518 A

Seatwork 2. Current density is given in cylindrical coordinates as J = - 106z1.5az A/m2 in the region 0 ≤ ρ ≤ 20μm; for ρ ≥ 20 μm, J = 0. (a) Find the total current crossing the surface z = 0.1m in the az direction. (b) If the charge velocity is 2 x 106 m/s at z = 0.1 m, find ρv there. (c) If the volume charge density at z = 0.15 m is -2000 C/m3, find the charge velocity there. Answer -39.7 mA; - 15.81 kC/m3; -2900 m/s

METALLIC CONDUCTOR • Metallic Conductor permit a higher-energy level in the valence band to merges smoothly, to a conduction band by the help of kinetic energy produce by an external field that will result in an electron flow.

METALLIC CONDUCTOR • The valence electrons, or conduction, or free, electrons, having a charge Q = -e will move under the influence of an electric field, E, and will experience a force: F = - eE

METALLIC CONDUCTOR • And the valence electron velocity (drift velocity) is linearly related to the electric field intensity by the mobility of the electron in a given materials (i.e. free space or crystalline) vd = - μeE (electron velocity is opposite in direction to E) where μє is the mobility of an electron (positive)

METALLIC CONDUCTOR • In terms of current density, J:

J    e e E where:

    e e Therefore;

J  E

METALLIC CONDUCTOR

• Consider a uniform current density J and electric field intensity E in a cylindrical region of length L and cross-sectional area S.

METALLIC CONDUCTOR I   J  dS  JS S

a

L Therefore; V  I S a

Vab   E  dL   E   dL   E  Lba b

Vab  E  Lba

or

b

or

V  IR

V  EL but

I V J   E   S L

where:

L R S

METALLIC CONDUCTOR • Insulator did not permit any electron flow due to an existing gap between the valence band and the conduction band. In this case the electron cannot accept any additional amounts of energy.

METALLIC CONDUCTOR • Semiconductors had a small “forbidden region” that separates the valence band and the conduction band. A small amounts of energy in the form of heat, light, or an electric field may raise the energy of the electrons and provide a conduction.

Sample Problem 3 Find the magnitude of the electric field intensity in a sample of silver having σ = 6.17 x 107 mho/m and μe = 0.0056 m2/V.s if: (a) the drift velocity is 1 mm/s; (b) the current density is 107 A/m2; (c) the sample is a cube, 3 mm on a side, carrying a total current of 80 A; (d) the sample is a cube, 3 mm on a side, having a potential difference of 0.5 mV between opposite faces.

Solution (a)

Vd   e E 3

Vd  1 10 E   0.1786V / m  e 0.0056

(b)

J  E 1 10 7 E   0.1621V / m 7  6.17  10 J

Solution (c)

(d)

I 80 6 J   8 . 889  10 A 9  10 6 J 8.889  10 6 E   0.1441V / m 7  6.17  10 V 0.5 10 3 E   0.1667V / m 3 l 3 10

Sample Problem 4 An aluminum conductor is 1000 ft long and has a circular cross section with a diameter of 0.8 in. If there is a dc voltage of 1.2 V between the ends, find: (a) the current density; (b) the current; (c) the power dissipated, using your vast knowledge of circuit theory.

Solution (a)

(b)

3.82 10 7 1.2 J   1.50 105 A / m2 l 304.8

V





I  JA  1.50  10 5 3.243  10 4  48.8 A

(c)

P  VI  (1.2)(48.8)  58.5W

Conductivity of Metallic Conductor

Seatwork 1. Find the magnitude of the current density in a sample of silver for which σ = 6.17 x 107 S/m and μe = 0.0056 m2/V∙s if: (a) the drift velocity is 1.5 μm/s; (b) the electric field intensity is 1 mV/m; (c) the sample is a cube 2.5 mm on a side having a voltage of 0.4 mV between opposite faces; (d) the sample is a cube 2.5 mm on a side carrying a total current of 0.5A. Answer: 16.53 kA/m2; 61.7 kA/m2; 9.87MA/m2; 80.0 kA/m2

Seatwork 2. A copper conductor has a diameter of 0.6 in and it is 1200 ft long. Assume that it carries a total dc current of 50 A. (a) Find the total resistance of the conductor. (b) What current density exists in it? (c) What is the dc voltage between the conductor ends? (d) How much power is dissipated in the wire? Answer: 0.0346Ω; 2.74 x 105 A/m2; 1.729 V; 86.4 W

CONDUCTOR PROPERTIES AND BOUNDARY CONDITIONS Conductor Characteristics: 1. It has a surface charge density that resides on the exterior surface and within has zero charge density. 2. In static conditions, in which no current will flow, follows directly Ohm’s law: the electric field intensity within the conductor is zero.

CONDUCTOR PROPERTIES AND BOUNDARY CONDITIONS The principles applied to conductors in electrostatic fields: 1. The static electric field intensity inside a conductor is zero. 2. The static electric field intensity at the surface of a conductor is everywhere directed normal to the surface. 3. The conductor surface is an equipotential surface.

CONDUCTOR PROPERTIES AND BOUNDARY CONDITIONS

Prove: Consider a closed surface conductor in a free space boundary ( In static conditions, tangential E  dL  0 electric field intensity and electric flux density are zero.)



CONDUCTOR PROPERTIES AND BOUNDARY CONDITIONS • For the tangential field around the small closed path at the surface:



b

a

c

d

b

c

 



a

d

0

and E = 0,

1 1 Et w  EN , atb h  EN , ata h  0 2 2 where: a to b = c to d = Δw and b to c = d to a = Δh

CONDUCTOR PROPERTIES AND BOUNDARY CONDITIONS

Et w  0 Et  0

( Δh = 0 and Δw ≈ 0(finite))

Dt  Et  0

CONDUCTOR PROPERTIES AND BOUNDARY CONDITIONS • Consider the normal field in a small cylinder as the Gaussian surface; D  dS  Q (Using Gauss Law)



S



top



bottom



sides

 Q(Cylinder Sides)

DN S  Q   S S where: bottom and sides integral were equal to zero

CONDUCTOR PROPERTIES AND BOUNDARY CONDITIONS or

DN   S

DN   0 EN   S

SAMPLE PROBLEM 5 A potential field is given as V = 100e-5x sin 3y cos 4z V. Let point P(0.1,π/12,π/24) be located at a conductor-free space boundary. At point P, find the magnitude of: (a) V; (b) E; (c) EN; (d) Et; (e) ρS

Solution (a) find the magnitude of V at point P V  100 e 5 x sin 3 y cos 4 z V  100 e

5 ( 0.1)

V  100 e

0.5

    sin 3  cos 4   12   24 

sin



cos



4 6 0.5 0 0 V  100 e sin 45 cos30  37.1V

Solution (b) find the magnitude of E E  V   5 x   x e sin 3 y cos 4 z a x     5 x  E  100  e sin 3 y cos 4 z a y  y     e 5 x sin 3 y cos 4 z a  z   z 













Solution (b) find the magnitude of E

  

 

 5 e 5 x sin 3 y cos 4 z a x    5 x E  100  3 e cos 3 y cos 4 z a y    5 x  4 e sin 3 y ( sin 4 z ) a z  E  500 e 5 x sin 3 y cos 4 za x  300 e 5 x cos 3 y cos 4 za y  400 e 5 x sin 3 y sin 4 za z



Solution (b) find the magnitude of E

    Ep  500 e sin 3  cos 4 a x  12   24      5 ( 0.1)  300 e cos3  cos 4 a y  12   24      5 ( 0.1)  400 e sin 3  sin 4 a z  12   24  Ep  185.7a x  111 .43a y  85.78a z 5 ( 0.1)

| E | 185 .7  111.43  85.78 | E | 233V / m 2

2

2

Solution (c) find the magnitude of En Dp   0 E





Dp  8.854  10 12 185 .7a x  111 .43a y  85.78a z  Dp  1.644  10 9 a x  0.987  10 9 a y  0.760  10 9 a z Dn | Dp |

1.644 10   0.987 10   0.760 10  9 2

Dn | Dp | 2.06  10 9 2.06  10 9 En    233V / m 12  0 8.854  10 Dn

9 2

9 2

Solution (d)

Et  0

(e) s  DN   0 En  2.06 109 C / m2  2.06nC / m2

Methods of Images • Dipole Characteristics – infinite plane at zero potential that exists midway between the two charges. • Consider a vanishingly thin conducting plane that is infinite in extent and have an equipotential surface at a potential V = 0, and electric field intensity normal to the surface.

Methods of Images • This can be represented by a single charge (image) above the plane and maintain the same fields, removing the plane and locating a negative charge at a symmetrical location below the plane.

SAMPLE PROBLEM 6 A point charge of 25 nC is located in free space at P(2,-3,5), and a perfectly conducting plane is at z = 0. Find: (a) V at (3,2,4); (b) E at (3,2,4); (c) ρS at (3,2,0).

Solution to Problem 6 Given: z=0 V = 0 at z = 0 at (3,2,4) R = (3,2,4) – (2,-3,5) = (1,5,-1) R’ = (3,2,4) – (2,-3,-5) = (1,5,9)

(a)

(b)

Solution to Problem 6 (b)

(c)