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

Given:

Inner radii of tube = a

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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