# Current Carry Wire With A Loop

1. Mar 10, 2010

### Charanjit

1. The problem statement, all variables and given/known data:
A. Determine the magnetic flux through a square loop of side a (see the figure) if one side is parallel to, and a distance b from, a straight wire that carries a current I.

Express your answer in terms of the variables I, a, b, and appropriate constants.

B. If the loop is pulled away from the wire at speed v, what emf is induced in it?
Express your answer in terms of the variables I, a, b, v, and appropriate constants.

2. Relevant equations:
Flux=integral(B*da)
E=BLv

I think those are the two equations we need, but not 100% sure.

3. The attempt at a solution:
I have already figured out part A, and the solution is posted above, just having issue with part B.

So all I did was replaced L=length with "a", one side of the area. I am totally lost here. So please help me.

2. Mar 10, 2010

### collinsmark

Have you tried using Faraday's law?

3. Mar 10, 2010

### Charanjit

Well yes, but how to relate velocity?

4. Mar 10, 2010

### collinsmark

The rate of change per unit time of $$\Phi_B$$ is a function of the loop's velocity.

Hint: Distance = Velocity times Time. Distance here, as your problem describes it, is your variable 'b'. Next find the negative of the time derivative of $$\Phi_B$$.

5. Mar 10, 2010

### Charanjit

Which b are you talking about? The equation E=BLv? That "B" is the magnetic field.

6. Mar 10, 2010

### collinsmark

In the problem statement, "...and a distance b from ..."

7. Mar 10, 2010

### Charanjit

Ok... so how do I import that into the equation? We don't have the "time" variable.

8. Mar 10, 2010

### collinsmark

Distance is equal to Time times Velocity, or b = vt.

Make the substitution as an interim step. You can always make the reverse substitution (t = b/v) later to get rid of the 't' variable if you need to, once you have your final answer.

9. Mar 10, 2010

### Charanjit

Ok, but in which equation? E=BLv?

10. Mar 11, 2010

### collinsmark

try the answer to the first part:
After you make the substitution, you'll still need to use Faraday's law as a step to the final answer.

11. Mar 11, 2010

### Charanjit

Ahh ok. I got it. Thanks.