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What is principle of solution behind this induction problem (problem 29)? The problem can be find in here (problem 29).

Problem is: The five separate figures below involve a cylindrical magnet and a tiny light bulb connected to the ends of a loop of copper wire. These figures are to be used in following question. The plane of the wire loop is perpendicular to the reference axis. The states of motion of the magnet and of the loop of wire are indicated in the diagram. Speed will be represented by v and CCV represent counter clockwise. (look at the picture in here( problem 29))

29) In which of the above figures will the light bulb be glowing?

(a) I, III, IV (b) I, IV (c) I, II, IV (d) IV (e) None of these.

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Let me ask that how do you post pictures to the questions in this website? –  laovultai Nov 2 '12 at 14:13
    
If you click on the little square on the question tool bar it asks you for the url of the image to upload either from your computer or from the web. –  anna v Nov 2 '12 at 14:28
    
I tried but it failed. It takes wrong link like; [3]: i.stack.imgur.com/2cO6y.png –  laovultai Nov 2 '12 at 14:49
    
Maybe you haven't noticed. It displays the image right now. And, that's not a wrong link. That's where you've uploaded the image through the link you've provided :-) –  Waffle's Crazy Peanut Nov 2 '12 at 15:20

1 Answer 1

up vote 4 down vote accepted

In situations I,II,IV, since the magnetic flux in the loop is changing in these cases, not in case III, since there is no change in magnetic flux in the loop, and the path of conduction is fixed in the loop.

For case I, the magnet is moving toward the loop, so the flux is decreasing, so there is an electric field which produces an EMF in the loop by Faraday's law. For case IV, the loop is moving, so the flux is increasing, and so there is an EMF in the loop by the Lorentz force law (it is relativistically related to the time-reverse of case I by relativity, although time reverse with resistors still means the light bulb is glowing, because that's the irreversible part).

Case II has a decreasing flux because the loop is collapsing, so the Lorentz force law gives an EMF in the loop.

Case III has no change in the flux through the loop, the loop is rotating. The Lorentz force is perpendicular to the direction of motion, so the integral around the loop is zero.

This looks like a physics GRE problem, I don't know if that counts as homework.

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Do you use formula " Faraday's law of induction states the induced electromotive force (EMF) in the wire is $E= - \frac{d \Phi_B}{dt}$ "? –  laovultai Nov 3 '12 at 11:10
    
@alvoutila: Yes, but carefully, because while the formula is true, it is due to two different effects (which are ultimately related by relativity)--- the Lorentz force law and the Faraday induction law, depending on whether it's the magnet or the wire that is moving. –  Ron Maimon Nov 3 '12 at 13:45

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