In the wing frame the unperturbed plasma moves with velocity $-u_0\mathbf e_x$. The <ideal magnetohydrodynamics> condition $\mathbf E+\mathbf u\times\mathbf B/c=0$ therefore produces the <motional electric field>
$$
\mathbf E=-\frac{(-u_0\mathbf e_x)\times(B_0\mathbf e_z)}c
=-\frac{u_0B_0}{c}\mathbf e_y.
$$
The wing's <surface conductivity> then gives the <surface current density>
$$
\boxed{\mathbf J=\Sigma\mathbf E
=-\frac{\Sigma u_0B_0}{c}\mathbf e_y}.
$$
Its direction also follows directly from the <Lorentz force> on the wing's mobile charges.

Solution

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

{title2=$\Sigma$}

Surface conductivity is the conductivity of a thin layer integrated through its thickness. The sheet form of <Ohm's law> is $\mathbf K=\Sigma\mathbf E_\parallel$.

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