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.

Hint:

Self-inductance depends on geometry and material of the coil. For solenoids, use flux linkage \( \Phi = N\phi \) and relate it to current.

Answer 1

Self-Inductance: Self-inductance of a coil is the property by which it opposes any change in the current flowing through it, by inducing an emf in itself. The self-induced emf is given by: \[ \mathcal{E} = -L \frac{dI}{dt} \] Where \( L \) is the self-inductance. Derivation: Consider a solenoid of: - Cross-sectional area \( A \) - Length \( l \) - Turns per unit length \( n \) - Total number of turns \( N = n \cdot l \) Magnetic field inside a long solenoid: \[ B = \mu_0 n I \] Magnetic flux through each turn: \[ \phi = B \cdot A = \mu_0 n I A \] Total flux linkage for \( N \) turns: \[ \Phi = N \cdot \phi = n l \cdot \mu_0 n I A = \mu_0 n^2 A l I \] From the definition of self-inductance: \[ \Phi = L I \Rightarrow L = \mu_0 n^2 A l \] Final Expression: \[ \boxed{ L = \mu_0 n^2 A l } \]