# Homework Help: Using ampere's law for a circular loop

1. Apr 18, 2015

### Essence

1. The problem statement, all variables and given/known data
Sorry to bother you guys,

I've heard that Ampere's Law is either ineffective for calculating the B field due to a circular loop or needs some modification (I wasn't sure which).

I'm trying to figure out why this doesn't work so I can get a better understanding of Ampere's Law. I have attached a picture of two loops (probably trivial). The inner loop is the one with a current the outer loop is the surface I chose for Ampere's Law.

I know you guys have to deal without a lot of **** so I hope I haven't imposed too much.

Thanks,

2. Relevant equations
(integral) Bdl = u * I

3. The attempt at a solution

To be more concrete about this whole thing imagine I had a looped wire with current I. I will now use Ampere's Law. First I will choose a surface which is another circular loop that contains the first loop. Now I will use:
(integral) B dl = u * I. Since each piece of current in the loop has another piece of current going in the opposite direction (it's a circle after all). I should get an I enclosed of 0. This would suggest the B field is zero (which I could justify using the Biot-Savart law isn't true). For space I will not do that here; that's a number plugging game.

#### Attached Files:

• ###### Two loops.png
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2. Apr 18, 2015

### BvU

Not so much a number game as a vector game.
It is not so that $\oint \vec B\cdot \vec{dl} = 0\quad \Rightarrow \quad |\vec B| = 0$.

3. Apr 18, 2015

### Essence

I think I get it. Since the B fields may be opposite at some points if I integrate the B fields over the surface I will get zero. But if I'm to find a specific B field at a location since B is not a constant function (it flips signs) I can't just divide by 2* pi * r and get 0. This is because the result of taking the integral is not B * 2pi * r (again because B is changing and so can't be treated like a constant).

Last edited: Apr 18, 2015
4. Apr 19, 2015

### BvU

I think you have the wrong picture of the situation

Re pictures: Perhaps google "magnetic field of current loop" and look at some of the pictures.

Re Ampere's law: the integral is not a surface integral but a line integral: $\vec {dl}$ is a little section of the wire, so to say.
(But, considering you do write "the result of taking the integral is not $B\ 2\pi r$", it appears you have the right intention but just use the term "integrate the B fields over the surface" a little sloppy ? -- "B field over the loop" would be better)

Combining the two: From the pictures and the circular symmetry you can see that at the location of the outer loop wire $\vec B$ is in fact a constant, and the direction is perpendicular to the plane of the two wire loops. No flipping sign, no "opposite at some points".

Since $\vec {dl}$ is in that plane, the inner product $\vec B\cdot \vec{dl} = 0$ and that is what makes the $\oint\$ come out as zero.

--

Last edited: Apr 19, 2015
5. Apr 19, 2015

### Essence

Ok.
(Just confirming)
$$\vec{B}\cdot\vec{dl}$$ will equal zero (because we're taking a dot product of perpendicular vectors) but this does not mean B equals zero. Ideally we want to choose a loop where B is in the same direction as dl such that we can just divide by the distance over which the line integral is taking place to get B.

Sorry about earlier; that was very very sloppy. I did mean line integral rather than surface integral, and I had temporarily assumed that the B field switched from into the page to out of the page depending on what side of the loop you were on. I didn't actually check this with greater care because I assumed that that was the only thing that could cause issues, but I hadn't considered the dot product.

Thanks,

Last edited: Apr 19, 2015
6. Apr 19, 2015

### Essence

(can't delete posts)

Last edited: Apr 19, 2015
7. Apr 19, 2015

### BvU

I think you are doing just fine. Ok you are !