# Ring Dropped into a Magnet

1. Mar 24, 2017

### FelaKuti

1. The problem statement, all variables and given/known data
A cylindrical magnet with its axis vertical provides a radial magnetic field. A thin, circular aluminium ring that is co-axial with the magnet falls through the magnetic field.

By finding an expression for the current in the ring when it is falling at the speed v, or otherwise, determine the terminal velocity of the ring if the magnetic flux density at the circumference of the ring is 2.00T, and the ring has mass 2.66 x 10^-4 kg, radius 2.00 cm and R = 2.48 mOhms

Q from here: https://isaacphysics.org/questions/ring_drop?board=6e219139-c944-4b22-9a12-9a560fd199ce

2. Relevant equations

induced emf = NdΦ/dt
W = mg
V = IR
Φ = BA
F = BIL

3. The attempt at a solution

If it reaches terminal velocity then weight downwards must equal magnetic force upwards. To work out magnetic force I need an equation for the induced emf so V = NdΦ/dt. I calculated the change in magnetic flux linkage as pir2 * v * dt * B.

Dividing by dt gives the rate of change of flux linkage which is pi * r2 * v * B which equals induced emf V.

Divide the LHS by R to give an equation for current, which I put into BIL and equate to mg. But I have a length l which I don't know how to eliminate whilst still being able to work out v, so I think my method here is wrong.

2. Mar 24, 2017

### TSny

Does your expression have the correct dimensions for magnetic flux?

What particular length does L represent in the expression BIL? Does the expression BIL apply to this problem? If so, can you explain why?

Did you work out why the magnetic force would be upward?

3. Mar 24, 2017

### FelaKuti

So it should be magnetic flux = B * pi r2 for the right dimensions.

L is the length of a current-carrying wire, I thought it would apply to the problem because an emf is being induced in the ring that opposes the flux, and since it's a closed loop there must be current..

I think the magnetic force has to oppose weight for a terminal velocity to occur?

4. Mar 24, 2017

### TSny

This equation is dimensionally correct, but it is not a correct equation for this problem. Magnetic flux is always associated with a surface area. Can you describe the area that you are working with?

If you've studied the concept of "motional emf", you might find it more appropriate for this problem than the concept of magnetic flux through a surface.

OK. You mentioned in your first post that you don't know how to "eliminate" L. Could you calculate the value of L?

Yes, that's right. But that doesn't explain why the magnetic force on the falling loop of wire is upward. You should be able to give an explanation based on your knowledge of the right-hand rules used to determine the direction of an induced current and the direction of the magnetic force on a current in a magnetic field.

5. Mar 25, 2017

### FelaKuti

The area would be the surface of the cylinder, so perhaps 2pi r2 + 2pi r v dt ?

Would the length then be v * dt perhaps?

For direction of the current, I have using the fleming's right hand rule the motion going downwards, the field going rightwards and thus the current would be coming out of the page, something like this?

If I use fleming's left hand rule from here I can see that the force acts upwards.

6. Mar 25, 2017

### TSny

What area or areas are represented by the first term, 2pi r2? Is there any magnetic flux through this area?

I'm not sure what length you are referring to here. Is it the L in the expression BIL? If so, then the answer would be no. Previously, you stated correctly that L is the length of a current carrying wire. In this problem, identify the current-carrying wire. How long is it?

Yes. Good.

7. Mar 25, 2017

### FelaKuti

The cross-sectional areas I guess don't have flux running through them. So my area would just be 2 pi r * v * dt

Would the length of the current-carrying wire be the circumference of the ring?

8. Mar 25, 2017

### TSny

Yes.

Yes.

9. Mar 25, 2017

### FelaKuti

now I've got V = B 2 pi r v dt / dt (change in flux over change in time, I guess I can cancel out the dts?)

I = 2 pi r v / R

using F = BIL with L = 2 pi r I get:

mg = B2 4 pi r2 v / R

still I've gone wrong somewhere

10. Mar 25, 2017

### TSny

Looks like you have the right approach. In your equation for the current, I, you left out a quantity (but I think you put it in later). Your expression for mg looks almost right, but you missed a square of one of the factors on the right.

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