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Magnetic field on the axis along the plane of loop

  • Thread starter wytang
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1. Homework Statement
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. Homework 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
 

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:)
dB= μ0I/4∏ *(dl X r)/r2

But I don't know how to solve the problem when it is on the axis of the plane of loop.
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:)
 
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Re: 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:)
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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