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 quantifies a coil's ability to oppose changes in current by generating an opposing electromotive force (emf) within itself.The self-induced emf is described by the equation:\[\mathcal{E} = -L \frac{dI}{dt}\]Here, \( L \) represents the self-inductance.Derivation: Consider a solenoid with the following parameters:- Cross-sectional area \( A \)- Length \( l \)- Turns per unit length \( n \)- Total number of turns \( N = n \cdot l \)The magnetic field within a long solenoid is given by:\[B = \mu_0 n I\]The magnetic flux through a single turn is:\[\phi = B \cdot A = \mu_0 n I A\]For \( N \) turns, the total flux linkage is:\[\Phi = N \cdot \phi = n l \cdot \mu_0 n I A = \mu_0 n^2 A l I\]Based on the definition of self-inductance, \( \Phi = L I \), we can derive \( L \) as:\[L = \mu_0 n^2 A l\]Final Expression:\[\boxed{ L = \mu_0 n^2 A l }\]