Problem statement:

The curved surface of a very long hollow non conducting cylinder of radius $R$ is uniformly charged with surface charge density $\sigma$. A non conducting small circular ring of radius $a$ and mass $m$ having charge $q$ uniformly distributed over its length is placed coaxially inside the hollow cylinder at its centre. The arrangement is located in a gravity free space. If the cylinder is rotated with a constant angular velocity $\omega$ about its axis as shown, what is the angular velocity acquired by the ring?

It's solution is given as:

My question is, since the calculated magnetic field is $\mu_0 \sigma \omega R$ and as given in problem statement, $\omega$ is constant, magnetic field is constant as rest all parameters are constant. So, why should there be an induced electric field which would rotate the ring? As far as I know, electric field is induced only when there is a time varying magnetic field.

Secondly, since they have not shown how magnetic field is $\dfrac{\mu_0 I}{l}$, I reasoned it as:

We consider an Amperian loop which is rectangular of length $x$ and breadth $b$ (similar to how we do in solenoid), integral along the breadths would vanish as field is perpendicular to the path, and integral along length would be $Bx$ (inside portion) and for outside, since field would be in radially outward direction, that integral would also vanish as field would be perpendicular to the length, hence we have $Bx=\dfrac{\mu_0 I x}{l}$. Is this proof correct?


1 Answer 1


For the first point - the ring actually acquires its angular velocity while angular velocity of cylinder changes from $0$ to $w$. During this process, magnetic field inside changes from $0$ to $\mu_0 I/l$ and there is induced electric field inside.

For the second point - yes, the proof is mostly as you said, apart from that the field outside is not radially directed. It is, however, indeed negligible compared with the field inside.