# Ampere's Law (I think). Induced current from current carrying wire

1. Jun 16, 2009

### in10s3gamer

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

A 1.60 cm x 1.60 cm square loop of wire with resistance $$1.10 \times 10^{-2} \Omega$$ is parallel to a long straight wire. The near edge of the loop is 1.10 cm from the wire. The current in the wire is increasing at the rate of 100 A/s

2. Relevant equations
I think I have to use Ampere's law to enclose the parallel wires.

$$\oint B*dl = \mu_0I_{enclosed}$$

3. The attempt at a solution
This is all I have so far:

$$B*2\pi r=\mu_0(100-I_1+I_2)t$$

I_1 would be the part of the loop closest to the long wire, and I_2 would be farthest. I'm not sure how to proceed further because I get the feeling that I'm just plain doing this problem incorrectly.

2. Jun 16, 2009

### dx

If the current in the wire is I, what is the flux of the magnetic field generated by it through the square loop?

Now if the current I is changing, the flux through the square loop also changes, and by Faraday's law there will be an induced emf in the square loop. From the induced emf and the resistance you can calculate the induced current.

3. Jun 16, 2009

### in10s3gamer

The flux's sign is arbitrary, but it's equal to BA, which is B(0.016)^2 in this case. And B can be determined by Biot-Savart's law? Integrating from 0 to 0.016:

$$\frac{\mu_0}{4\pi}\times \frac{I}{r^3}\int dl$$

Is that correct?

From that I would get B, then I could plug in the function for the current in the wire I which is changing, which would result in a changing flux.

4. Jun 16, 2009

### dx

It is easier to calculate B from Ampere's law, because of the symmetry. You should get

$$B = \frac{\mu_{0}I}{2\pi r}$$

Since B is not constant over the square loop (it changes with r), you will have to perform an integration to find the flux.

5. Jun 16, 2009

### in10s3gamer

That's where I was stuck earlier. I'm not exactly sure what I is equal to. I am thinking it is a function of time. Is it 100t, where t equals time in seconds? Or do I have to include the current inside of the square loop as well?

6. Jun 16, 2009

### dx

I is the current in the wire, I = C + 100t, where C is some constant. It doesn't matter what the constant is because only the derivative of I is required here.

7. Jun 16, 2009

### in10s3gamer

Thank you, it is very clear now.

edit:
r is in the denominator, so do i end up with [100mu/(2pi)]ln(r), with r evaluated from 0.01 - 0.026?

Last edited: Jun 16, 2009
8. Jun 16, 2009

### dx

You should get

$$\Phi_{B} = \frac{0.016\mu_{0}(C + 100t)}{2 \pi}\ln\frac{0.027}{0.011}$$

9. Jun 16, 2009

### in10s3gamer

Where does the 0.016 come from?

edit: i think 0.016 should be 0.016^2

Last edited: Jun 16, 2009
10. Jun 16, 2009

### in10s3gamer

If I take the derivative of the flux with respect to t i get your equation for flux, but instead of (C+100t) it is just (100).

After calculating the derivative of the flux with respect to t, this equals the EMF. I then set EMF = IR, and so the current = EMF/R. Since R = 1.1*10^-2, I get 1630 microAmps, but the answer is 26.1. Is there something I'm missing?

Last edited: Jun 16, 2009
11. Jun 16, 2009

### in10s3gamer

Something was wrong with my calculator, and now I get the right answer. However, why is there only one factor of 0.016 in the calculation for flux?

Isn't flux = BA, and A is equal to 0.016^2.

12. Jun 16, 2009

### dx

Flux is BA only when B is constant over the area. Here, B depends on r.

A small element of the area over which B is constant is therefore the part of the square between r and r + dr, whose area is 0.016⋅dr (draw a picture). The flux through this small element is 0.016⋅dr⋅B(r). The total flux is then the integral of this: ∫0.016⋅dr⋅B(r).

13. Jun 16, 2009

### in10s3gamer

Okay, 100% clear now, ready to be locked