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Circular loop magnetic field

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data

    As shown in the figure below, a circular loop of radius a = 0.1 m lies in the horizontal x-y plane with its center located at the origin, with steady current I = 2.0 A circulating in a counter-clockwise direction in the loop. The magnetic field strength B(0,0,z) at the observation point (x, y, z) = (0, 0, 0.1 m) and its direction, due to the current I flowing in the circular loop is:
    (a) B(0,0,z=0.1m) = 0.628 × 10-7 T, -z direction
    (b) B(0,0,z=0.1m) = 2.323 × 10-6 T, +z direction
    (c) B(0,0,z=0.1m) = 3.678 × 10-6 T, -z direction
    (d) B(0,0,z=0.1m) = 4.443 × 10-6 T, +z direction

    Answer is D

    http://i.imgur.com/i6Zxf.png

    i6Zxf.png

    2. Relevant equations

    Ampere's Law

    3. The attempt at a solution

    I tried Ampere's law, but I am not sure what is B dot dl (especially angle) and what is the path in this case...

    Thx
     
  2. jcsd
  3. Oct 25, 2011 #2

    ehild

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  4. Oct 25, 2011 #3

    dynamicsolo

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    I believe you want the Biot-Savart Law, since we are dealing with a current loop, rather than a current-carrying straight wire.

    At any point on the loop, which way does the magnetic field circulate? At the observation point, what is going to happen to the "horizontal components" (the parts parallel to the xy-plane) of all the bits of magnetic field coming from the loop? What will that mean as far as how the bits of magnetic field add up at ( 0, 0, 0.1 m.)?
     
  5. Oct 25, 2011 #4
  6. Oct 25, 2011 #5
    At any point on the loop, which way does the magnetic field circulate?

    Up, by RHR



    At the observation point, what is going to happen to the "horizontal components" (the parts parallel to the xy-plane) of all the bits of magnetic field coming from the loop?

    Bcos(45)?

    What will that mean as far as how the bits of magnetic field add up at ( 0, 0, 0.1 m.)?


    integral Bcos(45)dl?
     
  7. Oct 26, 2011 #6

    ehild

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    If you scroll down the page you arrive at the title "Magnetic field on the axis".

    ehild
     
  8. Oct 26, 2011 #7
    Okay. Thanks
     
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