Question

Obtain an expression for the self inductance of a solenoid.

Answer 1

Imagine an air-cored solenoid with a length of l, a diameter of d, and N wire turns. We take it for granted that the solenoid's length will be far longer than its diameter, allowing for the uniformity of the magnetic field inside and the disregard of any end effects. The magnetic field inside the solenoid is when the current I in the solenoid is constant.

B = μ₀nI   ... (1)

where n = N/l is the number of turns per unit length. So the magnetic flux through one turn is

Φ_m = BA = μ₀nIA   ... (2)

Hence, the self inductance of the solenoid,

L = `(NΦ_m)/I` = (nl) μ₀nA = μ₀n²lA = μ₀n²V   ... (3)

= `μ_0n^2l (πd^2)/4`   ... (4)

where V = lA is the interior volume of the solenoid. Equations (3) or (4) gives the required expression.