15D (Electromagnetism)

Translation symmetry in the plane and reflection symmetry across it force the magnetic field to be uniform on each side, parallel to

e_{y}

, and opposite on the two sides. Apply Ampère's law to a rectangular loop perpendicular to the sheet whose two long sides have length

L

. The enclosed current is

K L

, so

2 B L = μ_{0} K L .

The right-hand rule fixes the directions:

B (z > 0) = - \frac{μ _{0} K}{2} e_{y}, B (z < 0) = \frac{μ _{0} K}{2} e_{y} .

Consequently

e_{z} \times B (0^{+}) - e_{z} \times B (0^{-}) = \frac{μ _{0} K}{2} e_{x} - (- \frac{μ _{0} K}{2} e_{x}) = μ_{0} K,

which is the magnetic jump condition for a surface current density.

b

The accumulated point charge

Q (t)

produces

E (r, t) = \frac{Q ( t )}{4 π ϵ _{0} r ^{2}} e_{r} .

Since

\dot{Q} = I

ϵ_{0} \frac{\partial E}{\partial t} = \frac{I}{4 π r ^{2}} e_{r} .

r

and polar angle

θ < π /2

B = B_{ϕ} e_{ϕ}

is constant along the boundary, whose circumference is

2 π r sin θ

. The displacement-current flux through the cap is

\int_{cap} \frac{I}{4 π r ^{2}} d A = \frac{I}{4 π r ^{2}} 2 π r^{2} (1 - cos θ) = \frac{I}{2} (1 - cos θ) .

Thus

2 π r sin θ B_{ϕ} = \frac{μ _{0} I}{2} (1 - cos θ) .

Using

(1 - cos θ) / sin θ = tan (θ /2)

gives

B (r, θ) = \frac{μ _{0} I}{4 π r} tan (\frac{θ}{2}) e_{ϕ} .

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