Question

Derive an expression for the self-inductance of an air-filled long solenoid of length l and cross-sectional area A having N turns.

Answer 1

The self-inductance (L) of an air-filled long solenoid is derived by calculating the total magnetic flux linkage through the coil when a current (I) flows through it. For a long solenoid, the magnetic field is considered uniform inside and negligible outside.

Using Ampere’s Law, the magnetic field inside a long, air-cored solenoid is:

B = μ₀nI    ...(i)

The flux through one turn of area (A) is:

Φ = B × A    ...(ii)

= (μ₀nI)A    ...[From equation (i)]

The total flux linkage, denoted as λ, is the number of turns multiplied by the flux per turn:

λ = N × Φ    ...(iii)

= N(μ₀nIA)

= (nl)(μ₀nIA)    ...[N = nl]

By definition, self-inductance is the ratio of total flux linkage to the current:

L = `lambda/I`

= `(mu_0 n^2 l A I)/I`

= μ₀n²lA

The self-inductance of an air-filled long solenoid is:

L = μ₀n²lA

Or, in terms of total turns (N):

L = `(mu_0 N^2 A)/l`

Answer 2

Magnetic field inside solenoid:

B = `mu_0 N/l I`

Flux through one turn:

Φ = BA

= `mu_0 N/l IA`

Total flux linkage:

NΦ = `mu_0 (N^2 A)/l I`

Self-inductance:

L = `(N Phi)/I`

= `mu_0 (N^2 A)/l`