# Help clarifying an answer for magnetics question.

#### Kruum

1. Homework Statement

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

2. Homework 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:
Related Introductory Physics Homework Help News on Phys.org

#### Doc Al

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

#### Kruum

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

#### Doc Al

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?
$$B= \frac{ \mu _0}{2 \pi r}I$$
That's the field from a long straight wire, not half of a current loop.

#### 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:

#### Doc Al

Mentor
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?
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.

#### Kruum

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.
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.

#### 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:

#### Doc Al

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

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving