# Help clarifying an answer for magnetics question.

1. Feb 26, 2009

### Kruum

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

http://www.aijaa.com/img/b/00334/3685081.jpg [Broken]

2. Relevant equations

$$B= \frac{ \mu _0}{2 \pi r}I$$

3. The attempt at a solution

I got it wrong the two first times (I got $$\frac{ \mu _0I}{2 \pi a}(1- \frac{a}{b})$$), then the assistant teacher showed how it was done, but wasn't able to tell me why. This is the answer I was showed: $$B=B_a-B_b= \frac{ \mu _0I}{4}( \frac{1}{a}- \frac{1}{b})= \frac{ \mu _0I}{4a}(1- \frac{a}{b})$$ Can anybody clarify what the heck happened between steps 2 and 3, everything else I can understand.

Last edited by a moderator: May 4, 2017
2. Feb 26, 2009

### Staff: Mentor

Just factor out 1/a from within the parentheses:
(blah) = 1/a(a*blah)

3. Feb 26, 2009

### Kruum

Uh, if you compare my answer and the correct answer, I'm missing $$* \frac { \pi}{2}$$. If you misunderstood my original text $$B_a-B_b= \frac {\mu _0I}{4}( \frac{1}{a}- \frac{1}{b})$$ is the part where, the black magic happens.

4. Feb 26, 2009

### Staff: Mentor

Sorry about that. I wasn't clear on what you meant by "between steps 2 and 3", so I had to guess. (You meant between expressions 2 and 3 in your equation.)

In any case, why are you using this formula?
That's the field from a long straight wire, not half of a current loop.

5. Feb 26, 2009

### Kruum

That's the thing, I don't yet quite get the idea behind line integrals and I tried to derive the expression from Amperé's law. But isn't the magnetic field in the center of half current loop the sum of infinitely small straight wires?

Last edited: Feb 26, 2009
6. Feb 26, 2009

### Staff: Mentor

Yes, infinitely small straight wires. You cannot use the result for an infinitely long straight wire. Instead, integrate the magnetic field from a small current element as given by the Biot-Savart law. (The integration is trivial, by the way.) Give it a shot.

7. Feb 26, 2009

### Kruum

Ah, okay. I've never heard of that law so we haven't gone through it in our course. I'm going to look it up. I'll make sure the lecturer gets a few praises.

8. Feb 26, 2009

### Kruum

Now it makes sense. So $$B= \frac{ \mu_0I}{4 \pi r^2} \int dl$$, from 0 to $$\pi r$$. It gives me $$B= \frac{ \mu_0I}{4r}$$. Thanks for the help and expertise our teachers obviously lack, Doc Al! :tongue2:

9. Feb 26, 2009

### Staff: Mentor

Now you're cooking. (And you're welcome!)