In the figure a cross-section has been shown of a hollow cylindrical conductor of radii and . It carries a uniformly distributed current . (a) Using the circular Amperian loop shown in the figure, we have to show that

for the range is given by

(b) We have to test this formula for the special cases of , , and .

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We are considering a hollow cylindrical conductor of radii and , which carries a uniformly distributed current . Therefore, the current density

Because of the cylindrical symmetry the magnetic field inside the conductor will be circular and will depend on the distance from the axis of the cylinder.

We consider an Amperian loop of radius r as shown in the diagram.

We now apply the Ampere’s law and write

We will now test this formula for different special cases.

(a)

For ,

It is the magnetic field outside a wire of radius a carrying current .

(b)

For

There is zero magnetic field at the surface and within the hollow part of the cylindrical conductor.

(c)

For ,