Expression For Self Inductance of a Solenoid

Derive an expression for self inductance of a solenoid. What happens to the self inductance of the coil if it is wound on a rod of magnetic material. State the factors on which the self inductance of a coil depends.

Consider a long solenoid of length $l$, area of cross section $A$ and number of turns per unit length $n$. Let $I$ be the current flowing through the solenoid. The magnetic field inside this solenoid is uniform and given by

$ B = \mu_0 n I $

Total number of turns in the solenoid, $N = nl$.

Now the magnetic flux linked with each turn of the solenoid, $d\phi_B = B \times A = \mu_0 n I A$.

Total magnetic flux linked with the whole solenoid,

$ \phi_B = \text{magnetic flux linked with each turn} \times \text{number of turns in the solenoid} $

$ \phi_B = (\mu_0 n I A) \times (nl) = \mu_0 n^2 I Al \quad \dots (1) $

or

$ \phi_B = LI \quad \dots (2) $

Also,

From (1) and (2), we get

$ LI = \mu_0 n^2 I Al $

or

$ L = \mu_0 n^2 Al \quad \dots (3) $

Since $n = \frac{N}{l}$. Hence eqn. (3) becomes

$ L = \mu_0 \left(\frac{N}{l}\right)^2 Al = \mu_0 \frac{N^2}{l^2} Al $

$ L = \mu_0 \frac{N^2 A}{l} $

Thus, self inductance of an air cored solenoid ($L$) depends on (i) the total number of turns ($N$) of the solenoid and (ii) the length ($l$) of the solenoid and (iii) the area of cross-section ($A$) of the solenoid.

Solenoid Wound on Magnetic Material : 

When a solenoid is wound on a rod of magnetic material of permeability $\mu_r$ (Figure 25), the self inductance of the solenoid is given by

$ L' = \mu_r \frac{\mu_0 N^2 A}{l} = \mu_r L $

Thus, self inductance of a coil increases if air core of the coil is replaced by an iron core.

Factors on which self inductance of a solenoid or a coil depends :

Self inductance of a solenoid depends on

(i) length of solenoid.

(ii) area of cross-section of the solenoid.

(iii) number of turns of the solenoid.

(iv) nature of the material of the core of the solenoid.