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.