Magnetic field of a thick cylindrical shell conductor using Ampere's Law.

In this video, we calculate the magnetic field in all three regions for a thick cylindrical shell conductor carrying a uniform current density. We are given a thick conducting shell with inner radius a, outer radius b and total current I, and we want to express the magnetic field in terms of I.

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We quickly review Ampere's Law and the definition of current density (current per unit cross sectional area), then we quickly compute the magnetic field inside the hollow center of the thick cylindrical shell wire (B=0 inside because there is no enclosed current) and the magnetic field outside the wire (B=mu_0*I/2pi*r -- the same result we found outside the long thin wire and the long thick wire).

The interesting part of the problem is the magnetic field inside the actual conducting material of the thick shell, because an Amperian loop with a radius between a and b is only going to capture part of the total current! The path integral on the left hand side of Ampere's Law reduces to B(2pi*r) as usual, but to get the enclosed current on the right hand side, we have to multiply current density by the conducting cross section within the Amperian loop. Since this cross section is an annulus, we take the area of the outer circle and subtract the area of the inner circle, and we are able to express the magnetic field in terms of the current density J.

Finally, we need to relate the current density and total current, and we do this by going back to the definition of current density as J=I/A. Using the entire wire, we have a current of I the area of an annulus with outer radius b and inner radius a, so we find a current density of J=I/(pi*b^2-pi*a^2). Subbing this expression into our previous expression for the magnetic field, we find the final expression for the magnetic field of a thick cylindrical shell conductor using Ampere's Law!