Consider a long air-cored solenoid of length Z, diameter d and N turns of wire. We assume that the length of the solenoid is much greater than its diameter so that the magnetic field inside the solenoid may considered to be uniform, that is, end effects in the solenoid can be ignored. With a steady current I in the solenoid, the magnetic field within the solenoid is
B = µ0nI ………….. (1)
where n = N/l is the number of turns per unit length. So the magnetic flux through one turn is
Φm = BA = µ0nIA ……….. (2)
Hence, the self inductance of the solenoid,
L = \(\cfrac{NΦ_m}l\) = (nl)µ0nA = µ0n2 lA = µ0n2 V …………..(3)
= µ0n2 l \(\cfrac{πd
^2}4\)…………. (4)
where V = lA is the interior volume of the solenoid. Equation (3) or (4) gives the required expression.
[Note: It is evident thatthe self inductance of a long solenoid depends only on its physical properties – such as the number of turns of wire per unit length and the volume, and not on the magnetic field or the current. This is true for inductors in general.] .
