Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 3 / c

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- Solution c

Solution

Integrating the

y

component of Ampère's law,

\partial_{z} B_{x} - \partial_{x} B_{z} = \frac{4 π}{c} j_{y},

through the current sheet gives the magnetic-field jump across a surface current

δ B_{x} (0^{+}) - δ B_{x} (0^{-}) = \frac{4 π}{c} J_{y} = - \frac{4 π Σ u _{0} B _{0}}{c ^{2}} .

Reflection in the wing plane reverses the tangential perturbation

δ B_{x}

. If

b = \frac{2 π Σ u _{0} B _{0}}{c ^{2}},

the upper and lower boundary values on the wing are consequently

δ B_{x} (x, 0^{+}) = - b, δ B_{x} (x, 0^{-}) = b, ∣ x ∣ < \frac{L}{2} .

Outside the wing, the corresponding boundary value is zero.

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