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Magnetic flux of rectangular loop w/ wire

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the magnetic flux of the rectangular loop. Show that [tex]\int B x dl[/tex] is independent of path by calculating [tex] \oint B x dl[/tex]


    2. Relevant equations

    3. The attempt at a solution

    First step was to solve for B from the wire, I did that using Biot-Savart Law for a straight wire.

    [tex]B = \frac{ \mu_{o}I}{2 \pi s}[/tex]

    To get flux use

    [tex] \Phi = \int B \bullet da = \int ^{b}_{a} \frac{ \mu_{o} I}{2 \pi s} w ds = \frac{ \mu_{o} I w}{2 \pi} ln ( \frac{b}{a})[/tex]

    So far so good?

    Okay, now I can show that it's independent of path by doing a line integral.

    Here's where I'm sketchy.

    I need to use [tex] \oint B \! x \! dl[/tex]

    So I need to evaluate an integral over each side of the loop. The value is going to be dependent on distance from the straight wire. Now, B is coming out of the board. The magnitude of this will be the same on the left as it is on the right, but because it's a cross product, they'll cancel each other out. Then, right hand rule shows that evaluating the top segment's integral yields something that points toward the wire, and the bottom one points up. We'll say up is in the [tex]\widehat{s}[/tex] direction. So they won't cancel each other out, but I'm forgetting how to do this...

    [tex]B_{top} = \frac{ \mu_{o}I}{2 \pi b}[/tex]

    [tex]B_{bottom} = \frac{ \mu_{o}I}{2 \pi a}[/tex]

    So now I'm confused, do I have to integrate these? I would think that the only things that I had to integrate were the sides, because B changes for them as they get further from the straight wire.
  2. jcsd
  3. Apr 25, 2010 #2
    Since B is constant along the top of the wire, you will just integrate along dx. Since the length is 'w', then the integral will just be 'w'.

    For cross product please use the "\times" ([itex]\times[/itex]) latex command.
    Last edited: Apr 25, 2010
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