Question

Assuming that the length of the solenoid is large when compared to its diameter, find the equation for its inductance.

Answer 1

Consider a long solenoid of length l and cross-sectional area A. Let n be the number of turns per unit length (or turn density) of the solenoid. When an electric current i is passed through the solenoid, a magnetic field is produced by it which is almost uniform and is directed along the axis of the solenoid. The magnetic field at any point inside the solenoid is given by

B = μ₀ni

As this magnetic field passes through the solenoid, the windings of the solenoid are linked by the field lines. The magnetic flux passing through each turn is

`Φ_"B" = int_"A" vec"B"*"d"vec"A" = "BA" cos theta = "BA"` ..(since θ = 0°)

`Φ_"B" = (mu_0 "ni")"A"`

The total magnetic flux linked or flux linkage of the solenoid with N turns (the total number of turns N is given by N = nl) is

Self-inductance of a long solenoid

NΦ_B = n (nl) (μ₀ni)A

NΦ_B = (μ₀n²Al)i ….. (1)

From the self-induction

NΦ_B = LI ….. (2)

Comparing equations (1) and (2), we have L = μ₀n²Al

From the above equation, it is clear that inductance depends on the geometry of the solenoid (turn density n, cross-sectional area A, length l) and the medium present inside the solenoid. If the solenoid is filled with a dielectric medium of relative permeability μ_r, then

L = μ₀

L = μn₀μ_rn²Al