# How Is Terminal Velocity Achieved by a Ring Falling Through a Magnetic Field?

• FelaKuti
In summary: The current-carrying wire would be in the downward direction. So the current would be coming out of the page, something like this?Yes, the current-carrying wire would be in the downward direction. So the current would be coming out of the page, something like this?In summary, the terminal velocity of a ring with mass 2.66 x 10^-4 kg, radius 2.00 cm and R = 2.48 mOhms is found to be v = 9.81 m/s. The magnetic force is upward and the ring reaches terminal velocity by weight downward equal to the magnetic force.
FelaKuti

## Homework Statement

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

## Homework Equations

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

## The Attempt at a Solution

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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.

FelaKuti said:
I calculated the change in magnetic flux linkage as pir2 * v * dt * B.
Does your expression have the correct dimensions for magnetic flux?

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
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?

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?

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

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

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

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.

FelaKuti said:
The area would be the surface of the cylinder, so perhaps 2pi r2 + 2pi r v dt ?
What area or areas are represented by the first term, 2pi r2? Is there any magnetic flux through this area?

Would the length then be v * dt perhaps?
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?

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.
Yes. Good.

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?

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

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

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

FelaKuti said:
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
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.

## 1. What happens when a ring is dropped into a magnet?

When a ring is dropped into a magnet, it will experience a force due to the magnetic field. The direction and strength of this force will depend on the orientation and strength of the magnet, as well as the material and size of the ring.

## 2. How does the magnetic field affect the falling ring?

The magnetic field of the magnet will exert a force on the ring, causing it to accelerate and change its trajectory. The direction and magnitude of this force will depend on the relative positions and strengths of the magnet and the ring.

## 3. Will the ring be attracted to the magnet?

In most cases, the ring will be attracted to the magnet due to the magnetic force. However, the strength of this attraction will depend on the materials and sizes of both the magnet and the ring.

## 4. Can the ring get stuck inside the magnet?

It is possible for the ring to get stuck inside the magnet, especially if the magnet is strong and the ring is made of a highly magnetic material. In this case, the force of attraction between the two objects may be strong enough to hold the ring in place.

## 5. Is it safe to drop a ring into a magnet?

In general, it is safe to drop a ring into a magnet. However, it is important to be cautious and avoid dropping the ring from a great height, as it may cause damage or injury if it falls with a lot of force.

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