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.

Answer :

Formula used:


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.


Diagram:



(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:



Where


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.


Hence,


,


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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