# Why should there be an induced electric field if magnetic field is constant? [closed]

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?

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.