Question

Derive the expression for the self-inductance of a long solenoid of cross sectional area A and length l, having n turns per unit length.

Obtain an expression for self-inductance of a long solenoid of length l, area of cross-section A having N turns.

Answer 1

The self-inductance of a solenoid is defined as the magnetic flux to current ratio via the solenoid. It is given by,

L = `phi/I`

Total number of turns in the solenoid,

N = nl

The magnetic field inside the long solenoid,

B = μ₀ni

Flux through one turn,

`phi_1 = BA = mu_0niA`

Thus, total flux through N turns,

`phi_t = Nphi_1 = nl xx mu_0niA = mu_0n^2lAi`

Using `phi_t = Li`

where L is the self-inductance of the coil.

∴ `mu_0n^2lAi = Li`

L = `mu_0n^2Al`