Plots of B•dl as a function of position along the closed path

[syhpui2]

Homework Statement

Two infinitely long current carrying wires run into the page as indicated. Consider a closed triangular path that runs from point 1 to point 2 to point 3 and back to point 1 as shown.
Which of the following plots best shows B•dl as a function of position along the closed path?

http://i.imgur.com/J7b5L.png

Homework Equations

miu*I= B•dl

The Attempt at a Solution

I used RHR.I am confused about when should it is positive and when is negative.

 

[collinsmark]

I used RHR.I am confused about when should it is positive and when is negative.

Ampère's law states that:

\oint _P \vec B \cdot \vec{dl} = \mu_0 I_{enc}
The left is a closed path integral. In other words, assuming that the path is closed, the total area under the curve of \vec B \cdot \vec{dl} is proportional to the current enclosed within the path. How much current is enclosed within the path? That should rule out one of the choices right there.

Pick a path component and note the direction of \vec{dl} (the corresponding arrow in the figure will show you the direction of \vec{dl}). Use the right hand rule to determine the direction of \vec B. For that path, is the direction of \vec B in the same general direction of \vec{dl} (making \vec B \cdot \vec{dl} positive)? Or are they generally in the opposite direction (making \vec B \cdot \vec{dl} negative)? As a sanity check, repeat for the other path components.

 

[syhpui2]

Ampère's law states that:

\oint _P \vec B \cdot \vec{dl} = \mu_0 I_{enc}
The left is a closed path integral. In other words, assuming that the path is closed, the total area under the curve of \vec B \cdot \vec{dl} is proportional to the current enclosed within the path. How much current is enclosed within the path? That should rule out one of the choices right there.

Pick a path component and note the direction of \vec{dl} (the corresponding arrow in the figure will show you the direction of \vec{dl}). Use the right hand rule to determine the direction of \vec B. For that path, is the direction of \vec B in the same general direction of \vec{dl} (making \vec B \cdot \vec{dl} positive)? Or are they generally in the opposite direction (making \vec B \cdot \vec{dl} negative)? As a sanity check, repeat for the other path components.

So the direction of path is same as direction of current?
Thanks!

 

[collinsmark]

So the direction of path is same as direction of current?
Thanks!

Um no. The direction of the current is into the board/paper, as indicated by the 'x's on the wires. The direction of \vec{dl} of each path segment is shown by the arrows on each path segment. (All of that is given to you in the problem statement.) You can determine the direction of \vec{B} by using the right hand rule. (Hint: the direction of \vec{B} is perpendicular to the current in the infinitely long wires. But you need to think in three dimensions.)