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

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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?

J7b5L.png



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.
 

Answers and Replies

  • #2
collinsmark
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I used RHR.I am confused about when should it is positive and when is negative.
Ampère's law states that:

[tex] \oint _P \vec B \cdot \vec{dl} = \mu_0 I_{enc} [/tex]
The left is a closed path integral. In other words, assuming that the path is closed, the total area under the curve of [itex] \vec B \cdot \vec{dl} [/itex] 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 [itex] \vec{dl} [/itex] (the corresponding arrow in the figure will show you the direction of [itex] \vec{dl} [/itex]). Use the right hand rule to determine the direction of [itex] \vec B [/itex]. For that path, is the direction of [itex] \vec B [/itex] in the same general direction of [itex] \vec{dl} [/itex] (making [itex] \vec B \cdot \vec{dl} [/itex] positive)? Or are they generally in the opposite direction (making [itex] \vec B \cdot \vec{dl} [/itex] negative)? As a sanity check, repeat for the other path components.
 
  • #3
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Ampère's law states that:

[tex] \oint _P \vec B \cdot \vec{dl} = \mu_0 I_{enc} [/tex]
The left is a closed path integral. In other words, assuming that the path is closed, the total area under the curve of [itex] \vec B \cdot \vec{dl} [/itex] 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 [itex] \vec{dl} [/itex] (the corresponding arrow in the figure will show you the direction of [itex] \vec{dl} [/itex]). Use the right hand rule to determine the direction of [itex] \vec B [/itex]. For that path, is the direction of [itex] \vec B [/itex] in the same general direction of [itex] \vec{dl} [/itex] (making [itex] \vec B \cdot \vec{dl} [/itex] positive)? Or are they generally in the opposite direction (making [itex] \vec B \cdot \vec{dl} [/itex] negative)? As a sanity check, repeat for the other path components.
So the direction of path is same as direction of current?
Thanks!
 
  • #4
collinsmark
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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 [itex] \vec{dl} [/itex] 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 [itex] \vec{B} [/itex] by using the right hand rule. (Hint: the direction of [itex] \vec{B} [/itex] is perpendicular to the current in the infinitely long wires. But you need to think in three dimensions.)
 

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