# Find the speed of a copper loop falling in a magnetic field

Homework Statement:
A square copper loop with side length a is falling straight down at constant speed perpendicular to a magnetic field B. The cross-sectional area of the loop is A. What is its speed v?
Relevant Equations:
V = B*dA = Blv (or Bav here) F = ILB and V = IR
Hi all, so I had this problem and on the exam and I got a solution but I had an mass-term in there which wasn't given.

I used Farraday's Law of Induction to get the Voltage induced.
Then I used ##rho* \frac{A}{4a} ## for the resistance and divided the Voltage by that to get the current.
I then inserted this for the current I in F=IlB.
From the drawing on the exam it looked like only half the loop was in the field so I the forces up and down wouldn't cancel each other only the ones on the side.

I then set that equal to mg because ## ma = mg-IlB ## since the speed is constant ## mg = IlB ##.
This question really threw me off because m wasn't given and I was sure it had to cancel somehow.
Could someone show me where I made a mistake in thinking?

Thanks so much!
I don't remember exactly but I think B was 1.5T, a = 25cm, A was 10^-6m.

TSny
Homework Helper
Gold Member
It appears to me that you have the correct approach. The terminal speed of the loop will depend on the mass, as you found. So, a value of the mass should have been given in order to obtain a numerical value for the speed.

• spsch
Hello TSny.

Thank you very much. I lost quite a bit of time there. (trying to find a formula to make the masses cancel)
And I thought about it for the past two days without figuring it out.
I thought about it and with the density I probably could have figured it out. (Volume being 4a times A). There was a two page tableau with resistances and thermal data for the other problems.
I'm afraid I may have missed that densities were included as well.
(it also didn't occur to me then that m=density * volume)

But I wrote out the whole term without the numeric value. I hope that I may at least get a few points for that.

TSny
Homework Helper
Gold Member
OK. I hope that you receive almost full credit. The question is mostly about testing your understanding of some electromagnetism and I think you succeeded in doing that.

• spsch
kuruman
Homework Helper
Gold Member
Then I used ##\rho \frac{A}{4a}## for the resistance ...
So that you don't make the same mistake again, the correct expression for the resistance here is ##R=\dfrac{\rho(4a)}{A}##.

• spsch and TSny
rude man
Homework Helper
Gold Member
Yeah, very poorly posed problem.
You further need wire resistance per unit length (or total resistance) plus m.

kuruman
Homework Helper
Gold Member
Yeah, very poorly posed problem.
You further need wire resistance per unit length (or total resistance) plus m.
Tables were provided. See post #3.

Hi all, I'm sorry I just thought about this problem a little more and have another question.

Though it is a square, can I not see this wire as a loop and thus, wouldn't it also create a magnetic dipole that could end up feeling a torque?

Because by lenz's law the current induced should be in such a direction that opposes the change in magnetic flux? (Making the created dipole opposite)
So the wire would turn 180 degrees back and forth?

I'm sorry I hope I'm making sense, I could draw what I mean if it helps.

• kuruman
kuruman
Homework Helper
Gold Member
I'm sorry I hope I'm making sense, I could draw what I mean if it helps.
You are making perfect sense and it is to your credit that you should ask about this. Indeed there will be a magnetic dipole moment generated by the induced current in which case there will be a torque tending to align the magnetic moment with the field. However, unlike a compass needle oscillating back and forth in the Earth's magnetic field because there is a restoring torque, this is not the case here because the torque is not restoring. If the loop tries to rotate because of the external torque, Lenz's law says that there will be an induced current, in addition to the one already there, that will oppose the proposed change of angle. The counterpart of the magnetic "friction" force that opposes the linear motion of the loop no matter in what direction the loop is moving is a magnetic "friction" torque that opposes the rotational motion of the loop no matter which way the loop is turning. Without having written and solved the equations, my guess is that if the loop is initially oriented so that the magnetic field is not perpendicular to the plane of the loop, the loop will settle at some "terminal" angle by the time it reaches terminal velocity. If I solve the equations, I will post here.

• spsch
If the loop tries to rotate because of the external torque, Lenz's law says that there will be an induced current, in addition to the one already there, that will oppose the proposed change of angle. The counterpart of the magnetic "friction" force that opposes the linear motion of the loop no matter in what direction the loop is moving is a magnetic "friction" torque that opposes the rotational motion of the loop no matter which way the loop is turning.

Hi Kuruman,

thank you for answering me again. That makes sense.

The loop is originally falling vertically down perpendicular to the magnetic field. It's why when I thought about it I imagined the dipole being created inside the loop, directly opposite the direction of the magnetic field which it is falling in, and hence I imagined it should feel a torque until it balances again, maybe oscillate back and forth.

But it makes sense that the rotation would also create a change in flux and those a counter current and dipole.
It's hard to wrap my head around it but i'm having quite a bit of fun thinking about this and am grateful for the patience you guys show me.

OK. I hope that you receive almost full credit. The question is mostly about testing your understanding of some electromagnetism and I think you succeeded in doing that.
Update, got the results yesterday and I got half the points for this problem! :-) Thanks for helping!!

• TSny