Q. 48

A long, cylindrical tube of inner and outer radii a and b carries a current i distributed uniformly over its cross section. Find the magnitude of the magnetic field at a point

(a) just inside the tube

(b) just outside the tube

Answer :

Given:


Inner radii of tube = a


Outer radii = b


Current distributed over its cross section = i


Diagram:



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.


(a) Inside a conducting tube, I (current enclosed) is 0.


Hence, by Ampere’s circuital law, = 0 => B = 0.


Hence, the magnetic field just inside the tube is 0.


(b) Just outside the tube, the distance from the center is b.


Hence, we consider an Amperian loop of radius b from the center.


Hence, from Ampere’s circuital law:


,


Where B = magnetic field,


dl = line element, μ0 = magnetic permeability of vacuum, I = current enclosed.


In this case, dl = 2πb (circumference of the loop of radius b)


Therefore, B x 2πb = μoi (since the total current enclosed is i).


Hence, the magnetic field at a point just outside the tube = μoi/2πb(Ans)


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