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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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  • $\begingroup$ Would it not be akin to the friction of the cylinder (resistance to rotation)? In an electrical circuit with a given current, the resistance sets the rate work is done, so in the cylinder case there needs to be some sink of energy. Of course the magnetic field depends only on the current not the resistance of the circuit (the resistance just sets the work required to move that current). $\endgroup$
    – Joe Iddon
    Aug 29 at 14:17
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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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  • $\begingroup$ This answer does not address the OP's question about resistance. $\endgroup$ Mar 22 '14 at 17:58

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