# Homework Help: Faraday's Law, EMF, Finding Current in Loop

1. Oct 29, 2008

### ncm2

1. The problem statement, all variables and given/known data
A 26.4 cm diameter coil consists of 23 turns of circular copper wire 1.90 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.00E-3 T/s. Determine the current in the loop.

Part Two: Determine the rate at which thermal energy is produced.

2. Relevant equations
EMF=(deltaB*A)/delta T
EMF Total= -N*EMF
I=EMF Total/R
R=(Cu Resistivity*L)/A
L= N*2(Pie)R
A = TR^2

3. The attempt at a solution
EMF= 8E-3*(Pie*.132^2) = 4.379E-4
EMF Total = -23*4.379E-4 = -0.010072
I = -.010072 / ( 1.72E-8 * (23*2*Pie*.123) / (Pie*.95E-3^2) ) = -.093406 A

Part 2:
E THermal=I^2*R = -.093406^2 * .10783 = 9.40E-4 m^2 kg/s^3

Both answers are wrong and I have no idea where I went wrong, and neither do my TAs.

Last edited: Oct 29, 2008
2. Oct 29, 2008

### alphysicist

Hi ncm2,

The number in bold is incorrect; it should be 0.132m (like you have two lines before).

(If correcting that still does not work, you might check to see if they really just want the magnitude of the current; if so the minus sign might be enough to make it count as being wrong.)

3. Oct 29, 2008

### ncm2

Sorry that was a mistake in me writing it on here. I did calculate it with the proper .132m value, and tried positive and negative and it is still wrong, so something in my equation is wrong?

4. Oct 29, 2008

### alphysicist

So what answer did you get for the current? The answer in your post (I=0.093406 A) comes from using the 0.123m value; if you use the 0.132m value you get something different.

5. Oct 29, 2008

### ncm2

For the .132m value, I got 0.0870 A. This is incorrect as is -.0870 A.

I still don't know what I did wrong in my formula

Last edited: Oct 29, 2008
6. Oct 30, 2008

### alphysicist

That looks like the correct answer to me.

Here are two things, though.

First, just to check: assuming this in an online homework, are you sure they are asking for the current in amps and not milliamps?

Second, I would suggest trying $\rho=1.68\times 10^{-8}$ for the resistivity of copper.

7. Oct 30, 2008

### ncm2

I tried the change in units and the different resistivity of copper and they are both wrong.
It is an online assignment, but I also have to hand it in for my work to be marked.

Any other ideas?

Also, is my thermal energy equation correct?
E THermal=I^2*R

8. Oct 30, 2008

### alphysicist

Your work looks all correct to me.

There is a webpage with this problem (a publisher's page for a book by Giancoli); look here at problem #4:

http://cwx.prenhall.com/giancoli/chapter21/medialib/frameset.html [Broken]

Try out your formula for that problem (it uses the smaller resistivity) and you'll see it says it is correct.

Last edited by a moderator: May 3, 2017
9. Oct 30, 2008

### ncm2

Thanks for all your help. I used that website and my equation worked there, so maybe it is a problem with submitting the answer, I'm not sure.