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. :uhh: 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.)

 

[asteriatic]

The plot that best represents the B•dl as a function of position along the closed path would be the one with a triangular shape, where the value of B•dl is highest at point 2 and decreases towards points 1 and 3. This is because the magnitude of the magnetic field, B, is directly proportional to the current, I, and the distance, dl, from the wire. As the closed path moves closer to the wire at point 2, the value of B•dl will be highest. As the path moves away from the wire at points 1 and 3, the value of B•dl will decrease. The direction of B•dl will also depend on the direction of the current and the orientation of the path, following the right-hand rule. It is important to note that the value of B•dl can be positive or negative, depending on the orientation of the path and the direction of the current.