# Finding The Induced Current in This Loop

1. Apr 24, 2014

### Shakenbake158

1. The problem statement, all variables and given/known data

Hey guys, I have a physics II test tomorrow on Electricity and Magnetism, and I cannot seem to figure out this question.
The rectangular loop in the figure has 2.2×10−2Ω resistance.
What is the induced current in the loop at this instant?
Picture:

2. Relevant equations

E = d(flux m)/dt

B = (mu_0)(I)/2(pi)(r)

flux m = integral (B * DA)

3. The attempt at a solution

However, the area is not changing, so I can pull that out of the integral.
Then I have to integrate:

(mu_not)(I)/2(pi)(r)

Everything is constant except 1/r, so I can pull everything out and be left with:

integral (1/r)dr = ln(r)

So now we have:

E = d/dt(A)*(mu_not*I)*(ln(r))/(2*pi)

How do I take the derivative of this with respect to time?
Did I do the other steps correctly?

2. Apr 24, 2014

### Staff: Mentor

The whole area of the loop is being displaced with respect to time (it's moving away from the wire). So if you have a formula for the flux at a given distance then you can find how it changes with distance. And, since know how the distance changes with time, what rule from calculus springs to mind?

3. Apr 24, 2014

### Shakenbake158

Wait, I thought that the area was constant, since the area of the rectangular loop is not changing.

However, the B field is changing because the rectangular loop is getting further away.

4. Apr 25, 2014

### rude man

That's right. So what's Farady say about the emf induced around a loop with changing flux?

5. Apr 25, 2014

### Shakenbake158

Faraday's Law says that the induced EMF is equal to the changing flux. So do I take the derivtive of B?

6. Apr 25, 2014

### rude man

Sure! Flux - area x B field. If the B field is non-uniform you have to intgerate B over the area. And if B changes with time you have to integrate AND consider how that integral changes with time. But area is always the same constant.

In any case emf = - N d(flux)/dt. Your N is of course 1.

(Exception: under certain moving-media circumstances that will not get you the induced emf but let's leave that for later unless you're really interested).