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Suppose we have a cylindrical shell of radius $r$ with surface charge density $\sigma$. Then we start rotating the cylinder at an angular speed $\Omega$. You can show that in this case the surface current density on the cylinder is $\sigma r \Omega$.

Similarly, in a solenoid with loop density $n$ and current $I$, the surface current density can be thought to be $nI$, so you can use this to "convert" between a rotating cylinder and a solenoid. For example, the magnetic field inside the cylindrical shell must be $\mu_0 \sigma r \Omega$, because for the solenoid it's $\mu_0 n I$.

I was wondering if you could establish another relation between the solenoid and the cylinder if the solenoid has a resistance $R$. Based on the analogy, I imagine the resistance must be connected to the cylinder's moment of inertia, but I can't quite figure it out.

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If you use ampere's law, you will find that the magnetic field inside the rotating cylinder is uniform. With some fiddling, you will get that the magnetic field is equivalent to a solenoid with an infinite number of loops per length with a $\sigma r \Omega \mathrm{d}l$ current in each loop (a differential amount of current in each loop).

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This answer does not address the OP's question about resistance. –  ZachMcDargh Mar 22 at 17:58

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