EL 321 Electromagnetic Theory

Assignment No 3 - Solution

March 26, 2014

1 . Some arbitrary vector fields are given. Which of them represent amagnetic field? (a) A1 = −yy+zz; (b) A2 = 5 exp(−2z)(ρρ+z); (c)A3 =1r (2 cos θr + sin θθ)

2 . A thick slab extending from z = −a to Z = +a carries a uniform volumecurrent J = J x. Find the magnetic field, as a function of z, both insideand outside the slab.

Soln.:

The field points in the −y direction for z > 0, and in the +y direction forz < 0. At z = 0; B = 0.

Consider a rectangular Amperian loop of length L running perpendicularto current.∮B.dl = BL = µoIenc = µoJ(zL)⇒ B = −µoJzy for (−a < z < a).

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If z > a µoIenc = µoJaLB = −µoJay. for z > +a= +µoJay. for z > −a

3 . A steady current I flows down a long cylindrical wire of radius a. Thecurrent is distributed in such a way that J is proportional to ρ, the distancefrom the axis. Find the magnetic field, both inside and outside the wire.

Soln.:∮B.dl = µoIenc

J = ks⇒∫ a0Jda =

∫ a0ks(2πs)ds = 2πka3

3 k = 3I2πa3

Ienc =∫ a0Jda =

∫ a0ks(2πs)ds = 2πks3

3 = I s3

a3 for s < a.

For s > a Ienc = I∮B.dl = B2π = µoIenc

For s < a B = µoI2πs

s3

a3 φ For s > a B = µoI2πs φ

4 . A uniformly distributed current I flows down a long straight wire ofradius a. If the wire is made of linear material with susceptibility χm,what is the magnetic field a distance s from the axis? Find all the boundcurrents. What is the net bound current flowing down the wire?

Soln.: Ampere’s law:∮H.dl = µoIenc

For, s < a, H.2πs = Iπs2

πa2 ⇒ H = Is2πa2 φ

B = µo(1 + χm) Is2πa2 φ

For, s > a, H.2πs = I ⇒ H = I2πs φ

B = µo(1 + χm) I2πs φ

M = χmH; Jb = ∇×M = 1s∂∂s

(sχmIs2πa2

)z = χm

πa2 z

Kb = M× n |s=a= χmH× s = χmI2πa φ× s = −χmI

2πa z

Total current, Jbπa2 +Kb(2πa) = 0

5 . A small circular loop of radius 10 cm is centered at the origin and placedon the z = 0 plane. If the loop carries a current of 1 A along φ, calculate(a) The magnetic moment of the loop. (b) The auxiliary field at (2, 2, 2).(c) The magnetic field at (-6, 8, 10).

Soln.: Use spherical polar coordinates

(a) m = mz = IπR2z = 1.π(0.1m)2 = 0.01πz

(b) A(r) = µoI4π

m×rr2 ,

=µo4π

m sin θ

r2φ

m×r = (mz)×(sin θ cosφx+sin θ sinφy+cos θz) = m sin θ cosφy−m sin θ sinφx = m sin θ[− sinφx + cosφy] = m sin θφ

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B = ∇×A = 1r2 sin θ

∣∣∣∣∣∣r rθ r sin θφ∂∂r

∂∂θ

∂∂φ

Ar rAθ r sin θAφ

∣∣∣∣∣∣= 1

r2 sin θµom4π

[r ∂∂θ (r sin θ sin θ

r2 )− rθ ∂∂r (r sin θ sin θ

r2 )]

= µom4πr3

[2 cos θr + sin θθ

]Coordinate of P (2,2,2): r =

√22 + 22 + 22 = 2

√3

sin θ = 42√3

= 2√3; cos θ = 2

2√3

= 1√3

B = µo0.01π

4π24√3( 2√

3r + 2√

3θ) = µoH

(c) Magnetic field B at (-6,8,10)?r =√

62 + 82 + 102 = 10√

2sin θ = 4

2√3

= 2√3; cos θ = 2

2√3

= 1√3

6 . A finite line current flowing through −l ≤ z ≤ l along z direction. Findmagnetic vector potential at finite distance point (x,y,z). Determine Bfrom A.

Soln.:

r =√x2 + y2 + (z − z′)2; If z � z′ r =

√x2 + y2 + z2

For line current along z direction: A = µoI4π

∫ ll

1rdz′z

A =µoI

4π

1

r2lz =

µoIl

2π

1

rz

B = ∇×A = µoIl2π

∣∣∣∣∣∣x y z∂∂x

∂∂y

∂∂z

0 0 1r

∣∣∣∣∣∣∂∂y

1r = −y

r3 and ∂∂z

1r = −z

r3

B = µoIl2π

[−yr3 x + x

r3 y]

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