Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 36 / 1 / b

Solution

B_{wire} (x, y) = \frac{μ _{0} J}{2 π} \frac{- ( y + b ) e _{x} + x e _{y}}{x ^{2} + ( y + b ) ^{2}} .

An oppositely directed image current

- J

(0, b)

gives

B_{image} (x, y) = - \frac{μ _{0} J}{2 π} \frac{- ( y - b ) e _{x} + x e _{y}}{x ^{2} + ( y - b ) ^{2}} .

The method of images makes the normal oscillating magnetic field vanish at

y = 0

, as required when the alternating component cannot penetrate a perfect conductor. The two tangential components reinforce one another there. Consequently

B (x, 0^{-}) = C (x) e_{x}, C (x) = - \frac{μ _{0} J b}{π ( x ^{2} + b ^{2} )} .

Only the oscillating field is being excluded. A pre-existing steady magnetic field is not erased by the perfect conductor limit.

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