Magnetic field of a circular coil.
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In the problem it is explained that the magnetic field only has a phi-hat component that depends only on rho and z. To find H we can use the equation:
nabla x H = J
Integrating both sides over some surface O gives by Stokes' theorem:
Integral over boundary of O of H dot ds = Integral over O of J dot dO
The boundary of O is some curve to which we assign some orientation. Applying the right hand rule to that orientation then gives the direction of the normal of the surface O. These orientations fix the way you have to assign the correct directions to ds and dO in the integrals.
To compute H we take for the surface O a disk at height z with center at rho = 0 (i.e. on the z-axis) and radius r. The boundary of this disk is thus located at rho = r and is thus a circle with radius r and center on the z-axis. Let's choose the orientation of this circle to be counterclockwise, i.e. in the minus phi-hat direction. Then the surface of the disk has a normal that is oriented in the z-direction. Let's write:
H(r,z) = ...
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