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Let's say we have an infinite current-carrying wire parallel to an infinite conducting plate. Is the magnetic field behind the plate zero? I would say yes, because the field in the plate is zero. Also is the field in front of the plate, so the region containing the wire, given by the field produced by the wire only or is the field altered by the presence of the plate?

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If the plate has a finite thickness, then there will be a magnetic field on the other side. This is because the magnetic field inside a superconductor does not vanish discontinuously, but instead decays exponentially over a distance the order of the London penetration depth $\lambda_\mathrm{L}$:

$$B(x) = B_0 \exp\left(- \frac{x}{\lambda_\mathrm{L}}\right),$$

where $B_0$ is the field at the surface. $\lambda_\mathrm{L}$ can be determined from the properties of the conductor. For a plate of thickness $d$ the field on the other side would therefore be

$$B(d) = B_0 \exp\left(- \frac{d}{\lambda_\mathrm{L}}\right).$$

If $d \gg \lambda_\mathrm{L}$ the field on the other side of the plate would be negligibly small. But for $d \sim l $ there will be an appreciable magnetic field making it through to the other side.

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  • $\begingroup$ Thank you. In case of infinitely thin plate: how does the magnetic field look like in the region containing the current-carrying wire? $\endgroup$
    – 88888888
    Jun 28 '20 at 15:48

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