Q. 603.7( 3 Votes )

A tightly-wound, long solenoid is kept with its axis parallel to a large metal sheet carrying a surface current. The surface current through a width dl of the sheet is Kdl and the number of turns per unit length of the solenoid is n. The magnetic field near the center of the solenoid is found to be zero.

(a) Find the current in the solenoid.

(b) If the solenoid is rotated to make its axis perpendicular to the metal sheet, what would be the magnitude of the magnetic field near its center?

Answer :

Given:


Number of turns per unit length = n


Surface current of sheet = K


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 in the circuit.


(a) Since the magnetic field near the center of the solenoid is 0, we infer that the magnetic field due to the solenoid = magnetic field due to the sheet


For the sheet, using Ampere’s circuital law, we get:


B x 2l = μ0 x Kl, where Kl = current enclosed by sheet and we take 2l because of the two surfaces of the sheet.


=> Bsolenoid = μ0K/2


Now, magnetic field due to a solenoid Bsolenoid = μ0ni, where μ0 = magnetic permeability of vacuum, i = current, n = number of turns per unit length


Hence, μ0K/2 = μ0ni => i (current in solenoid) = K/2n (Ans)


(b) Now, if the axes of the solenoid and the plate are perpendicular to each other, their magnetic fields are also perpendicular to each other.


Hence, net magnetic field at center,



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