Question

Derive the relation for self inductance of a long solenoid

Answer 1

Answer:

Let us consider a solenoid of N turns with length l and area of cross section A. It carries a current I. Hence, the total magnetic flux (φ) linked with the solenoid is given by the product of flux through each turn and the total number of turns.

Answer 2

Answer:

In expression for the self-inductance of a long solenoid.

Consider a solenoid of N turns with length l and area of cross section A. It carries a current I. If B is the magnetic field at any point inside the solenoid, then Magnetic flux per turn =B × area of each turn

But, B= llμ0NI

Magnetic flux per turn =

lμ0NIA

Hence, the total magnetic flux (ϕ) linked with the solenoid is given by the product of flux through each turn and the total number of turns.

ϕ= lμ 0

NNI ×Ni.e. ϕ= lμ 0 N 2 II ....(1)

If L is the coefficient of self induction of the solenoid, then

ϕ=LI....(2)

From equation (1) and (2),

LI= μ0 N 2IIAL= lμ 0 N 2 AAIf the core is filled with a magnetic material of permeability μ,

Then, L= lμN2AASolve any question of Moving Charges and Magnetism with:-