CBSE 2025, Set 2

Question

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

Edvaya · Accepted Answer

The correct answer is Subjective. **Definition of Self-Inductance:**
Self-inductance of a coil is defined as the flux linked with the coil when unit current flows through it. (Alternatively: It is the ratio of the total magnetic flux linkage to the current flowing in the coil, $L = \frac{N\Phi}{I}$.)

**Expression for self-inductance of a long solenoid:**
Consider a long solenoid of length $l$, cross-sectional area $A$, and $n$ turns per unit length. Let a current $I$ flow through it.

The magnetic field $B$ inside the long solenoid is uniform and given by:
$$B = \mu_0 n I$$

The magnetic flux $\Phi$ linked with each turn of the solenoid is:
$$\Phi = B 	imes A = (\mu_0 n I)A$$

The total number of turns in the solenoid is $N = nl$.
The total flux linkage $N\Phi$ is:
$$N\Phi = (nl) 	imes (\mu_0 n I A)$$
$$N\Phi = \mu_0 n^2 A l I$$

By definition, the self-inductance $L$ is given by $N\Phi = LI$. Therefore:
$$L = \frac{N\Phi}{I}$$
$$L = \frac{\mu_0 n^2 A l I}{I}$$
$$\boxed{L = \mu_0 n^2 A l}$$

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Physics > Magnetism Induction And Alternating Current > Electromagnetic Induction

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