# Induced current in circuit

1. Apr 4, 2005

### jordanl122

This one is really getting to me.

Find the current through section PQ (the middle column of the circuit) of length a, a =.65 m. The circuit is located in a magnetic field whose magnitude varies with time according to the expression B(t) = .001t. Assume the resistance per length of wire is .1 ohm/m.
Diagram:

key: I: piece of circuit
X: direction of B field

-----------P
IIIIIIIIIIIIIIIIIIIIIIIII
I X X X X I X X X I
I X X X X I X X X I a
I X X X X I X X X I
IIIIIIIIIIIIIIIIIIIIIIIII
-----2a---Q----a---

I started by using Faraday's law of inductance,

Emf = -d Phi_m/dt
but Im not exactly sure how this will help me. I foresee a kirchoff analysis, but I'll be damned if I can figure out how I arrive there

Last edited: Apr 4, 2005
2. Apr 4, 2005

### jdavel

Jordan,

So using Faraday's law, what did you get for the emf around the left and right loops?

3. Apr 4, 2005

### jordanl122

for the bigger section I got..

Emf = -2(a^2)dB/dt
= -2(a^2)*.001 d/dt (t)
= -.001*2*.65^2
= -8.45*10^-4 V

and smaller

Emf = 4.225*10^-4 V

4. Apr 4, 2005

### jdavel

jordan,

The Emf magnitudes look right. Did you mean to say one was positive and the other negative? What would that mean?

Now think about currents. How many different currents will there be in the circuit?

5. Apr 4, 2005

### jordanl122

I'm thinking three, since it splits at some point. And I meant to make the second Emf negative as well. Now this definitely seems like a kirchoff set up, except that I'm not exactly sure how to play the resistance.

Im thinking if we treat each section like loops, then the current is going to be counterclockwise through each section. Since the initial flux is going into the page. The B field generated in response to maintain constant flux over the system is going to be out of the plane of the page. And hopefully, I am applying the right hand rule correctly here.

To deal with the resistance, Im guessing I just find the total resistance in each segment of the wire, and treat it as if it were a regular resistor in the circuit. And along that same path of logic, I can treat the emf as a battery somewhere in each loop.

Feel free to continue steering me in the right direction if I'm heading way off course.

6. Apr 4, 2005

### jdavel

jordan,

Doin't worry, you're not far off course. But hold off on replacing the Emf with a battery. That only works if the current everywhere in the loop is the same. We're not going to be that lucky.

You said there are three different currents. That's right. Call them I1, I2 and I3. You need to decide how to define them. So, where is each one, and which direction is each one going in?

7. Apr 4, 2005

### jordanl122

damn, I was almost positive it was going to be two batteries, one for each corresponding loop. As for the currents, would it be I1 traces out the bigger rectangle and splits off into I2 and I3. I2 running up the section we are looking for, and I3 running around the smaller rectangle

8. Apr 4, 2005

### jdavel

That will work fine for I1, I2 and I3. Now you need to find some (namely three) equations for them. Any ideas?

9. Apr 4, 2005

### jordanl122

well I1 = I2 +I3 is in the bag. So I guess I'll follow each individual loop to generate the two other equations. And scuttling along with the resistance length and two battery idea, seeing as it will all come out in the kirchoff loop rule wash anyway... I run into some trouble because I don't know what direction my batteries are, or should I just directly sub in my emfs and hope for the best. Anywho my second equation is:
(emf1 is the emf from the bigger section, and emf2 is the emf from the smaller one)
emf1 -.325I1 -.065I2 = 0
and my third:
emf2 + .065I2 - .195I3 = 0

to get those three resistor values I multiplied the length of wire by the resistance per unit length
so .1(5a) = .1(5*.65) = .325 ohms
and .1(a) = .065 ohms
and .1(3a) = .195 ohms
since those are the three lengths upon which the three currents run

10. Apr 4, 2005

### jdavel

jordan,

Lookin' good! You just go around the loop, and if you hit a current going the other way, you make the voltage contribution from that link negative.

So, since you know the emfs, you have 3 equations in 3 unknowns. I'd use the first one to eliminate I1 in the second, then solve the second and third for I2 and I3.

Go for it!

11. Apr 4, 2005

### jordanl122

I thank you oodles friend.

-Jordan

12. Apr 4, 2005

### jdavel

jordan,

You're welcome, but you did most of it.

By the way, after looking at what you did, I retract what I said about not being able to treat the Emf as a battery in the loop. Your second and third equations said that the sum of the voltage drops around each loop had to be the same as the Emf, but that's the same as saying they have to be the same as the voltage across a battery. I don't know what I was thinking!