# Homework Help: Lenz' Law, induced current in a ring falling

1. Sep 27, 2011

1. The problem statement, all variables and given/known data

A ring falls through a horseshoe magnet. Before reaching the magnet, is the current induced in the ring clockwise or counterclockwise? What about after the ring has passed through the magnet?

The answers are clockwise before the ring passes through, and counterclockwise after. I'd like to understand why this is so.

2. Relevant equations

Lenz' law, a magnetic field will be induced on the ring to oppose change in magnetic flux in the magnetic field of the magnet.

3. The attempt at a solution

Part A) Before reaching the magnet
When falling, the magnetic flux is increasing because more magnetic field lines of the magnet pass through the ring. By Lenz' law, a magnetic field will be induced on the ring to resist this, so the magnetic force would push it upwards, and the direction of the magnetic field would be opposite to the direction of the magnetic field of the horseshoe magnet.

Using the right-hand rule, I have B from right to left (Lenz' law), F upwards. Thus, v at the bottom of the ring would be going right, which is counterclockwise, which is incorrect.

Part B)
The ring is moving away, decreasing the flux, so by Lenz' law, a magnetic field is induced on the ring where the force is upwards. Since the field is attempting to maintain itself here, the induced magnetic field points in the same direction as the magnet's field. B is from left to right, F is upwards, so v at the top of the ring is going left, which is counterclockwise, which is correct.

How should I do part A? Is my explanation for part B correct? I don't know if my usage of the right-hand rule is correct, since it applies to particles and I'm for the direction of current ring, and current flowing right/left at any point in a ring is flowing in the opposite clock direction of at the opposite side of the ring.

2. Sep 27, 2011

### PeterO

This is not a Current-Field-Force problem, it is a right hand grip problem or right hand screw rule - which ever way you learnt it.

A. As you said, the induced field is right to left . Hold your right hand with fingers curled and thumb straight out.
Point your thumb from right to left and see which way your fingers curl [clockwise]

B. As you said, induced field is left to right. Right hand grip rule again.

3. Sep 27, 2011

As I understood it, the right hand screw (thumb and curl, not cross product) is only for the magnetic field produced by a current-carrying charged wire. How does that work here?

My instructor told me that I need the cross-product rule, is there any way to use that?

4. Sep 27, 2011

### PeterO

Alternate explanation:

A. as the bottom of the ring encounters the B field, the positive charges in that section of the ring are traveling down through a left to right field.
Right hand rule says charges down, Field right, force on charges from back of ring to front of ring. Since that is in the bottom of the ring, that means clockwise.
(we all know the positive charges don't actually move but that has never worried us)

B. Charges pushed in the same direction [back of ring to front of ring], but now it is the top of the ring that is the last to fall through the B-field, so anti-clockwise

5. Sep 27, 2011

I agree about the alternate version, but where does Lenz' law come in? If there is a magnetic eddy in the opposite direction of the magnetic field to try to counteract the increasing flux before the ring reaches the magnet, how would that affect the direction of induced current?

6. Sep 27, 2011

### PeterO

The thumb and curl rule is for one straight element and one circular element.

case A: a straight wire - fingers give circular field around the straight wire

case B: a circular wire [loop or solenoid] straight thumb gives the straight field inside the "circular current"

EDIT: It is Lenz's law on case B here that gives the answer - as in my first explanation. [or is that Faraday's law?]

Last edited: Sep 27, 2011
7. Sep 27, 2011