- #1

spsch

- 111

- 21

- 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.

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.