Solving Ampere's Law Homework Statement

[exitwound]

Homework Statement

Homework Equations

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

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?

 

[Troels]

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

What the heck is wrong?

Did you remember the minus? :)

 

[exitwound]

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

 

[Troels]

Hmm I seem to get the same

 

[exitwound]

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

 

[exitwound]

From a homework study group on Facebook. Apparently, this gives the right answer:

PART a
find current enclosed: use right hand rule to determine sign. thumb should point in direction of positive current and your other 4 fingers point in direction of the loop.
answer is 4 * pi * 10^-7 * [(-1 * first red number) + (second red number)]

PART b
follow the same right hand rule as in part a
answer is 4 * pi * 10^-7 * [(-2 * first red number) + (-1 * second red number)]

Why do I use both positive 15 and postive 6 in the equation??

 

[Troels]

Why do I use both positive 15 and postive 6 in the equation??

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

 

[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?