# Calculating an expression for work in a magnetic field

1. Jun 17, 2013

### QuickSkope

As shown, I must derive a formula to find work.

http://i.imgur.com/0U0eLqm.jpg

I proceeded as follows.

W= F * D
= (B * I * L) * D

V = BA/T

I = V/R
= (BA/t)/R
= BA/tR

F = B * (BA/tR) * A
W = ((B^2*A^2)/tR) * A
W = (B^2 * A^3)/tR

Does that seem right? IIRC, the B*I*L formula is only for solonoids?

Thanks

2. Jun 17, 2013

### barryj

I don't understand this

"F = B * (BA/tR) * A
W = ((B^2*A^2)/tR) * A
W = (B^2 * A^3)/tR

However consider Power = VI and work = power X time
I think you will get a similar ansawer

3. Jun 17, 2013

### QuickSkope

I solved L in terms of B, A, R and T, as the question doesn't allow me to use I. Only thing I'm confused about is if I can use V = B * I * L in this situation, as I thought that was for solonoids only.

4. Jun 17, 2013

### barryj

What is L? I recall that Force = B X current X length of wire.
Think about it this way. V = dphi/dt and this is what you have.
Power = V^2/r

Work = power X Time, yes?

5. Jun 17, 2013

### QuickSkope

((BA/t)^2 / r) * t gives me

(BA)^2 / tR, which is a different equation then the other one I got. So is mine wrong?

6. Jun 18, 2013

### barryj

I got (BA)^2 / tR. In your post #1 you have an extra A. There might be a math error. Do you have the correct answer?

7. Jun 18, 2013

### QuickSkope

Unfortuantetly, I don't have an answer. I'll check it today. Thanks :)