# Induced Current

1. Apr 13, 2008

### Gear300

The circuit is shown below: Find the current through section PQ of length a = 65.0cm. The circuit is located in a magnetic field whose magnitude varies with time according to the expression B = (1.00 x 10^-3T/s)t. Assume the resistance per length of the wire is 0.100 Ohm/meter.

._________________P__________
.|xxxxxxxxxxxxxxxxxx|xxxxxxxxxxxx|
a|xxxxxxxxxxxxxxxxxx|xxxxxxxxxxxx|
.|xxxxxxxxxxxxxxxxxx|xxxxxxxxxxxx|
.|________________|__________|
---------2a--------Q-----a-----
The entire circuit is in a changing magnetic field that is going into the screen perpendicular to the plane of the circuit (nevermind the dashes between the 2a, Q, and a or the dots above and below the other a...they're just there for spacing).

I've tried several methods (actually, only 2 that I kept repeating, but they seemed right). The current along PQ should be found through finding the voltage between P and Q and using that (along with the resistance) to find the current. The actual answer is 283 x 10^-6A upward, but my answers dont match. Any help??

Last edited: Apr 13, 2008
2. Apr 13, 2008

### G01

Show me what you have tried so far. Remember the forum rules: You must show your attempt at a solution to get help here.

3. Apr 13, 2008

### Gear300

I thought that since there were 2 individual loops, I could take each one separately and find the emf the magnetic field would cause in each one. Since the magnetic field changes at a rate of 1.00x10^-3T/s and the area for each one is constant, the rate of change of magnetic flux, consequently the emf, through the larger loop would be .000845V through the larger loop and .000423V through the smaller loop. Since the 2 loops share a side PQ, I thought that the voltage along that side would be the subtraction of the 2 emfs. The result would be the voltage across PQ and since resistance is .1Ohm/m*.65m along PQ, I would use those two results to find the current. The answer I get from this is .0065A...which is not the answer.

Last edited: Apr 13, 2008
4. Apr 13, 2008

### G01

I also get that answer for the current. I also agree with your method to find the solution. The fact that we are 3 factors of ten away from the correct answer makes me think some number is incorrect. Also the larger loop will have a larger induced emf and will induce a downward current along PQ due to the right hand rule, so the net current should be downward, not upward. Are you sure your numbers are correct?

Last edited: Apr 13, 2008
5. Apr 13, 2008

### Gear300

pretty sure they are.

6. Apr 25, 2008

### pwelp

I just did some searching on google after working on this problem for a few hours now myself... and this is what I have found. Basically you have to treat it like a kirchoff rules problem.

physicsforums/archive/index.php/t-69984.html

(i am not allowed to post links apparently... but physicsforums.com goes where it says physicsforums)
Hope this helps

7. Apr 25, 2008

### Gear300

Last edited: Apr 25, 2008
8. Apr 25, 2008

### apelling

pwelp is right - use Kirchoff's circuit laws.

The emf s you have calculated are correct but they are spread around each loop.

I found it helped to model each loop as 4 resistors of either 65mOhm (milliOhm) for 65cm sides or 130mOhm resistance for 130cm sides. The two loops share a 65mOhm resistor ie side PQ. Insert into each loop a battery of the emf you have calculated previously. Then solve using Kirchoff. This gave me an answer of 280 microA (rounding errors probably)

9. Apr 25, 2008

### Gear300

Alright...to be sure, the emf generated in these loops is distributed across the loops, right?

10. Apr 28, 2008

### apelling

Yes there will be a potential difference across each resistor given by Ohm's law V=IR.

11. Apr 28, 2008

### Gear300

alright then...thanks, I think I got it.