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Magnetic field homework

  1. May 8, 2006 #1
    Can anyone point me in the right direction to do this question (i.e the correct law to use, or correct method). I have no idea where to even start with this.

    It says:
    A current path is shown below, with an arc radius R =1.2m, and carries a current of i = 10 amps. Find the magnetic field at P if the arc subtends an angle of pi/6 at P.

    [​IMG]

    (Sorry about the quality of my diagram)
     
  2. jcsd
  3. May 8, 2006 #2

    Chi Meson

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    The straight portions of the wire do not contribute to a magnetic field at point P. Only a "portion of a loop" (where the number of turns, N, is less than 1.0) will contribute to this field.
     
  4. May 8, 2006 #3
    Thank you. It makes sense now and (I think) I know where to look in my notes.
     
  5. May 8, 2006 #4

    berkeman

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    So you can either do the integration of the Biot-Savart law, or use the trick that Chi is hinting at to do it quicker. Or do both to check your work.
     
  6. May 8, 2006 #5
    I think you need Ampere's law for this: nabla x B = J/mu (or the integral version of it)
     
    Last edited: May 8, 2006
  7. May 8, 2006 #6
    I was going to use the Biot-Savart law (but its very confusing). Whats this other trick that is being hinted at then?
     
  8. May 8, 2006 #7

    berkeman

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    Probably in the textbook they do the example problem of finding the magnetic field at the center of a loop of current....
     
  9. May 8, 2006 #8
    This is my worked answer. Does it make sense to you?

    Loop makes 1/12 the field of a full loop
    [tex]B = \frac{\mu_{0} I}{4\pi R^2}*(\frac{2\pi}{12}*R)\\ = \frac{4\pi*10^-^7 * 10}{4\pi * 1.2^2} * (1.2*\frac{2\pi}{12})\\ = 4.36 * 10^-^7 T[/tex]

    Using the right hand rule, the field will be pointing out of the screen.

    (Sorry about my graphics - I hope they make sense. I've never used them before)
     
    Last edited: May 8, 2006
  10. May 9, 2006 #9

    Chi Meson

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    This is what I get. The special case equation for the B at the center of a loop is (mu I N)/2r , where N is the number of turns of wire inthe loop. As you can see, N can be a fraction of a single turn as well.
     
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