Q. 475.0( 2 Votes )
A thin but long, hollow, cylindrical tube of radius r carries a current i along its length. Find the magnitude of the magnetic field at a distance r/2 from the surface
(a) inside the tube
(b) outside the tube.
Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ0 times the current enclosed by the surface.
where B = magnetic field, dl = line element, μ0 = magnetic permeability of vacuum, I = current enclosed.
(a) Now, since inside a conductor, the current enclosed is 0, in Ampere’s circuital law, we put I = 0.
Hence, the magnetic field at a distance r/2 from the surface inside the tube is 0. (Ans).
(b) Now, for any point outside the tube, the current enclosed will be given by Ampere’s circuital law:
B = magnetic field,
dl = line element,
μ0 = magnetic permeability of vacuum,
I = current enclosed.
Now, since we have to find the magnetic field at a distance r/2 from the surface,
we consider an Amperian loop of radius from the centre of the loop.
The total current enclosed will be i, since we are considering a loop outside the tube.
since = circumference of the Amperian loop.
Hence, we get, B(at a distance r/2 from the surface outside the tube) =(Ans)
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