A cylinder of radius $a$ has a uniform magnetisation along its axis, which results in zero bound volume current density and a bound surface current density $M\hat{\phi}$. A coil with $N$ turns and wire resistance $R$ and a given wire radius is wrapped around the magnetisation in the form of a solenoid with radius $b>a$. When the magnetised cylinder is flipped a charge $q$ flows through the wire in the coil. What happens in this situation, ignoring the self-inductance of the wire? (It was a question in my exam today asking to compute the new magnetisation of the cylinder that I couldn't do and am intrigued - before this was asked I had calculated the $\vec{H}$ and $\vec{B}$ fields everywhere.)

My initial thoughts were that if there is a charge flowing into the wire then there is then a current flowing through it since the flow of charge is what current is. This can then be modelled as a solenoid with a magnetic field inside of $\vec{B}=\mu_0nI\hat{z}$.

While the charge is flowing into the wire, would there then be a time-dependent current increasing with time? I was thinking of computing change in flux and thus induced emf inside the solenoid, and then go on to somehow combine this with what I had already discovered about the magnetised cylinder.

Could someone please take me through the physics of what is going on in this situation?

  • $\begingroup$ Your thoughts are correct, looking at the induced emf by considering the change in magnetic flux is one way to handle such a problem. You can think of it as a non-driven RL circuit. Technically, there would be some self-inductance and a little bit of capacitive reactance during the initial switch, but the non-driven (i.e., no external power supply) RL circuit should work... $\endgroup$ Jan 14, 2016 at 13:57
  • $\begingroup$ The coil is fixed in space, the magnetized cylinder is rotated along with the coil, the total resistance I would assume is R, although the question did not specify. $\endgroup$
    – ODP
    Jan 20, 2016 at 14:44

1 Answer 1


Well you are right in expecting an effect on the wire every time there is a change in the field. The magnetic induction phenomenon will occur with changes in the magnetic flux through the solenoid, which depends both the intensity of the magnetic field and the time in which the change occurs.

Now when you say "flip the magnetised wire" if you mean sudden change, it would be the unrealistic case in which infinite forces in the wire move the charges and rip it apart.

If you mean change with time in a known way, then the forces inside the coil will change proportional to the derivative of B (since the area of the solenoid remains constant) and the current in the coil behaves exactly like in certain simplified circuit with the corresponding time dependent voltage, R representing the coil resistance and L representing the inductance of the coil. Also if the ends of the coil are not connected, charge will accumulate in them, and they will behave further like a capacitor C, in which case the system would be analogous to a simplified RLC circuit.


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