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    What does integrating this give?
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    (Original post by JokesOnYoo)
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    What does integrating this give?
    This is from a standard result for the B field on the axis of a current loop, at a distance z from the centre of the loop, radius R. You should be able to "integrate" this by symmetry arguments (Hint: the only thing that can vary is dL - how much of this is there, in total?)
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    (Original post by atsruser)
    This is from a standard result for the B field on the axis of a current loop, at a distance z from the centre of the loop, radius R. You should be able to "integrate" this by symmetry arguments (Hint: the only thing that can vary is dL - how much of this is there, in total?)
    As L varies doesn't R also vary?
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    (Original post by JokesOnYoo)
    As L varies doesn't R also vary?
    You mean "as dL varies" - L doesn't vary - L is the the total length of the current loop i.e. its circumference.

    But no, R is the fixed radius of the loop so that doesn't change at all. By the circular symmetry of the situation, the point at which you are finding the B field, namely P(0,0,z), is always at the same distance from any infinitesimal section dL of the loop. By the Biot-Savart law, that means that each dL contributes the same dB to the total field at P. So you can simply multiply up this contribution over the whole length of the loop i.e. it is proportional to the circumference of the loop.

    To think of this by integration, we want to find:

    B = \int dB = \int_\text{loop} f(I, z,R) dL = f(I, z,R)  \int_\text{loop} dL

    since f(I, z,R) is independent of the variable of integration, L. But what is  \int_\text{loop} dL?
 
 
 
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