# Ampere's Law, magnitude of I?

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1. Mar 8, 2017

### PolarBee

1. The problem statement, all variables and given/known data
The value of the line integral around the closed path in the figure is 1.79×10−5 Tm .
https://session.masteringphysics.com/problemAsset/1385538/4/32-22.jpg

There is I2 and I3 inside the closed loop.
I2 = 12 A
What is I3?

2. Relevant equations
Amphere's Law: B = Uo I(enclosed)

3. The attempt at a solution
I rewrote the equation as B = Uo * (I2 + I3)
Solving for I3...
I3 = (B / Uo) - I2
Plugging in values,
1.79*10^-5 / (4*pi*10^-7) - 12 = 2.2 A
But this is not the right answer? I don't understand what is wrong.

Last edited by a moderator: Mar 8, 2017
2. Mar 8, 2017

### Staff: Mentor

Investigate how the direction taken around the closed path for the line integral affects the results; Check the definition of Ampere's Law.

3. Mar 8, 2017

### PolarBee

Using the right hand rule to check the direction of positive I, positive I points into the page, so the 12A is negative and, so I should add 12 instead of subtracting?

4. Mar 8, 2017

### Staff: Mentor

What you want to do is check the direction of the field lines produced by any currents inside the line integral's closed path. If they have the same sense (direction) as the Ampere Loop then they make a positive contribution to the sum. If they have the opposite sense then they make a negative contribution.

You have one given current, $I_2$, and it comes out of the page. Do its magnetic field lines have the same sense as the loop direction?

5. Mar 8, 2017

### PolarBee

Okay I see, so I2's field lines are opposite of the loop direction, so they make a negative contribution to the sum.

So it should be B = Uo (-I2 + I3)
Making I3 = (B/Uo) + I2

And I3 would be 26.2 A, and since it is positive it would be going into the page, correct?

6. Mar 8, 2017

Looks good!