A long hollow cylindrical conductor with inner radius a and outer radius b carries a current I uniformly distributed over the cross-sectional area of the conductor (see fig.).
Use Ampere's law to show that the magnitude of the magnetic field at a radius r from the axis of the conductor is zero if r < a and 
if b < r < a and equal to 
if r > b.
Text Solution
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Sol. Due to symmetry of the conductor we expect magnetic field lines to be circular. Therefore we choose a circular closed curve of radius r < a. From Ampere's law,
= B2 π r = µ0I encl
As I encl = 0, B (r < a) = 0
Current density j = 
inside the conductor. Now we take a circular closed curve in the region a < r < b. Then current enclosed by curve,
I incl = (j) [ π (r 2 – a 2 )]
= 
From Ampere's law,
The Magnetic Field
= B2 π r = µ 0 
B = 
For a circular closed curve outside the conductor (r > b),
= B2 π r = µ 0 I
B = 
As total current is enclosed in our circular loop.
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