# Electromagnetic Induction (EMI)

A non conducting ring of mass m and radius R has a charge Q uniformly distributed over its circumference. This ring is placed on a rough horizontal surface such that the plane of the ring is parallel to the surface. A vertical magnetic field B=B0t2 τ is switched on. After 2 second from switching on the magnetic field, the ring is just about to rotate around vertical axis through its centre. Then

( A ) The induced electric field is quadratic in time "t"
( B ) The force tangential to the ring is B0QRt
( C ) Until 2 seconds, the friction force does not come into play
( D ) The friction coefficient between the ring and the surface is {2B0RQ}/{mg}

Can anyone please make me clear about WHY the ring is about to rotate and solve the question properly? I can't understand the problem clearly.

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I can't see why the ring would rotate either. Perhaps there is a mistake in the question. Sorry i can't be of more help.

TSny
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Gold Member
Hi purple prize. Welcome! If you click on the "Rules" tab at the top, you can read the regulations for posting questions at this forum. Please try to provide some attempt at solving the problem or at least some information regarding how you are thinking about the problem.

To help get started, maybe you could try answering one or more of the following.

What do you know about induced electric fields produced by changing magnetic fields? Have you seen any example problems (in class or in your text) that you might be able to relate to this problem? Do you know something about the geometrical pattern of the induced electric field lines produced by the changing magnetic field?

Alright. Firstly, I'm sorry for posting in Advanced Physics forum. This should go to Intro Physics. I didn't know the rules. My apology.

Yes, I know changing magnetic fields produce electric field. Here, in this problem, as the magnetic field is changing with time, the flux linkage with the non conducting loop also changes. However, as the ring is non conducting, there will NOT be any flow of current. There will be an induced electric field set up.

ε = - d $\phi$ / dt
= - d (B0t2 . A ) / dt
= - 2B0t . π R2

Now let the magnitude of the electric field is E.

∫E. dl = ε [Sorry I couldn't write surface integral and vector sign]
E. 2πR = |ε| = 2B0t. πR2
E = B0Rt

Force, F = q|E| = B0QRt

I honestly don't know direction of the induced electric fields. So I can't even properly tell the direction of the Force F too. In the book, it's written that "The lines of induced electric field are closed curves. There are no starting and terminating points of the field lines."

My question is how is this force tangential? How does it make the ring rotate about the vertical axis through its center?

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TSny
Homework Helper
Gold Member
OK, good. You're right, there won't be any induced current in the non-conducting ring. But there will be an induced electric field that will exert a force on the net charge of the ring. If the magnetic field is confined to a region of circular cross section, then the induced electric field lines will be circular as shown in the figure (B is assumed to be increasing in the figure). So, if the ring is placed so that the center of the ring is at the center of the cross section of the region of B, then you can see that E would be tangent to the ring at each point of the ring.

If the ring is not placed at the center of the region of magnetic field or if the region of B is not circular, then E would not be tangent to the ring at each point of the ring. But, you can still show that the net force on the ring due to the induced E field will be the same as long as the B field is uniform inside the ring.

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Thank you for the diagram. :)

Now, how do we generalize the direction of induced electric field? Is there a rule we should follow?

I think it is Lenz's law?

TSny
Homework Helper
Gold Member
As Basic Physics says, use Lenz's law. The direction of E is such that if it did set up a current in a conducting ring, the current's B field would oppose the change in flux of the external B field.