# Homework Help: Induced emf in a square loop.

1. Dec 8, 2003

### discoverer02

One more problem that's causing me grief:

A square loop of wire, b meters on a side, moves with constant velocity v (m/sec) toward the right in the plane of a long straight wire carrying a steady current I amperes. Calculate the emf induced in the loop when the side of the loop nearest the wire is at distance x meters from the wire.

The treatment of the non-uniform magnetic field with the velocity is what troubles me here.

I know that the magnetic field coming from the wire = uI/(2pir) and that if I integrate [uI/(2pir)]b](dr) from x to x + b then I get the magnetic flux for the stationary loop but this, of course, induces no current. I'm having trouble seeing how to incorporate the velocity into the non-uniform magnetic field.

I tried to substitute r = vt into the equation, and then take the derivative of the magnetic flux with respect to time to get the induced emf, but that didn't agree with the answer in the book..

Any suggestions are be greatly appreciated.

Thanks.

2. Dec 8, 2003

### lethe

yes, as i recall, this calculation is a little messy

i think you should substitute x=vt, after you have finished integrating. remember, r is just a dummy variable of integration, so it shouldn t really appear in your final answer

that is essentially correct. if you show your calculation in a little more detail, i can be a little more specific in where you went wrong.

3. Dec 8, 2003

### discoverer02

The magnetic flux for the loop with no velocity is:

[buI/(2pi)]ln[(x+b)/x]

If I plug x = vt here and take the derivative with respect to time, I end up with:

emf = [buI/(2pi)][vt/(vt+b)? This doesn't agree with the answer in the book.

4. Dec 8, 2003

### lethe

i don t think you took the derivative correctly. remember the t appears twice, once in the numerator, and once in the denominator, and both are inside a logarithm.

5. Dec 8, 2003

### discoverer02

You're right. What was I thinking? I'll try it again and see what comes out.

Thanks.