# Magnetic field between two long, straight wires

1. Oct 31, 2009

### exitwound

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

2. Relevant equations

$$B = \frac{\mu_o i}{2\pi d}$$

3. The attempt at a solution

To do this, you'd find the magnetic field magnitude due to the left wire, and add it to the magnetic field due to the right wire. Correct?

If so:

Wire 1:
$$B = \frac{1.26x10^-6 (8)}{2\pi(.02)} = -4x10^-6 \hat k$$

Wire 2:
$$B = \frac{1.26x10^-6 (12)}{2\pi(.02)} = 1.2x10^-5 \hat k$$

But this isn't right. The answer is B.

What's gone wrong?

2. Oct 31, 2009

### ApexOfDE

You just forget to add 2 B together.
$$B1 = \frac{\mu_o i1}{2\pi d} = \frac{20 \mu_0}{\pi}$$

$$B2 = \frac{\mu_o i2}{2\pi d} = \frac{30 \mu_0}{\pi}$$

Total B at midpoint is:

B = B1 + B2

Because direction of B1 is in opposite that of B2 and B2 > B1, so you should choose direction of B2 as positive direction.

==> B = B2 - B1

3. Oct 31, 2009

### exitwound

Hmm. I thought that's what I was doing. I must have put something into the calculator incorrectly because I got something else for B1. But when I just did it, following what you did, it works out.

Thanks.