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Question

A long cylinder of uniform cross section and radius R is carrying current i along its length and current density is uniform. There is a cylindrical cavity or uniform cross section and radius r in the cylinder parallel to its length. The axis of the cylindrical cavity is separated by a distance d from the axis of the cylinder. Calculate the magnetic field at the axis of the cylinder.

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Solution

Current density, j=iπ(R2r2)
Current through the smaller cross section: i=jπr2 (in absence of cavity)
Magnetic field due to this current at a point on the axis of bigger cylinder:
B1=μ0i12πd=μ0i2πd×jπr2=μ0r22d×iπ(R2r2)
Let B2 be the magnetic field due to remaining portion. Then
B1+B2=0 or B2=B1 or |B2|=|B1|
B2=μ0i2πd×r2(R2r2)

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