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Magnetic field on the axis along the plane of loop

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A circular loop in the XY plane of radius R and with current I moving in the counterclockwise direction. Compute B at (L,0,0) for L >> R0.


    2. Relevant equations
    Biot–Savart law
    dB= μ0I/4∏ *(dl X r)/r^2


    3. The attempt at a solution
    I know that the B on the axis perpendicular to the plane of loop is μ0IR^2/ [2*(z^2/+R^2)^(3/2)].
    But I don't know how to solve the problem when it is on the axis of the plane of loop. Thanks for any helps
     
  2. jcsd
  3. Apr 9, 2012 #2

    tiny-tim

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    welcome to pf!

    hi wytang! welcome to pf! :smile:

    (try using the X2 button just above the Reply box :wink:)
    set up the integral of the biot-savart law, integrating wrt θ …

    what do you get? :smile:

    (L >> R means that at some stage it'll be ok to ignore the y components :wink:)
     
  4. Apr 9, 2012 #3
    Re: welcome to pf!

    tiny-tim, I tried to work out the problem myself out of curiosity. I know that usually, the z component is all that is asked.

    I may be wrong, but it seems that the only thing to change was the denominator. I found that dB = μ0I/4∏ *(dl X r)/r^2 = μ0I/4∏ *(dl)r^2, and the x component of dB is μ0I/4∏ *(dl * R)/r^2 where R is the radius of the loop and r is the distance from it (the original poster uses L).

    So then is seems that r = (R^2 + (r + R)^2). Integration of dl gives us 2piR. So the final answer that I have is μIR^2/2((r+R)^2 + R^2)^3/2). Is this work correct, or am I missing something?
     
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