# Ampere's Law

1. Oct 28, 2009

### exitwound

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

2. Relevant equations

$$\int \vec B \cdot d\vecs = \mu i_{enc}$$

3. The attempt at a solution

This should be straight forward. I'm really starting to become frustrated with this textbook.

Amperian Loop 1 encircles both current loops.

$$\int \vec B \cdot d\vecs = \mu i_{enc}$$
$$\int \vec B \cdot d\vecs = (1.26x10^-6) i_{enc}$$

The enclosed current is *using the right hand rule* +6A and -15A = -9Amps.

1.26x10^-6(-9) = -1.1x10^-5 which is apparently wrong.

What the heck is wrong?

Last edited: Oct 28, 2009
2. Oct 28, 2009

### Troels

Did you remember the minus? :)

3. Oct 28, 2009

### exitwound

That's a typo. I actually DID submit -1.1e-5 which is wrong.

4. Oct 28, 2009

### Troels

5. Oct 28, 2009

### exitwound

Yup. I doublechecked. That's what I put in, the negative.

6. Oct 29, 2009

### exitwound

From a homework study group on Facebook. Apparently, this gives the right answer:
Why do I use both positive 15 and postive 6 in the equation??

7. Oct 29, 2009

### Troels

You don't - notice it says "(-1 * first red number)" which would give a -15

8. Oct 29, 2009

### exitwound

It's still not what we computed though. Either you use a negative 15A current, and a positive 6A current and get -9, or you get -21A as they compute. Why would you get -21? Why would you use both negative currents?

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