Question

Define self-inductance and write its S.I. unit. Find the self-inductance for a solenoid of \( N \) turns, length \( l \), and radius \( r \).

Hint:

Self-inductance depends on geometry of the coil and the magnetic permeability of the medium.

Answer 1

Step 1: Self-inductance is defined as the property of a coil (or an electrical circuit) by which a changing current in the coil induces an electromotive force (emf) in the same coil itself. This induced emf opposes the change in current according to Lenz's law. The self-inductance \( L \) quantifies the ratio of the induced emf to the rate of change of current: \[ \mathcal{E} = -L \frac{dI}{dt}. \]
Step 2: Consider a solenoid of length \( l \), cross-sectional area \( A \), and \( N \) turns:

The magnetic field inside a long solenoid carrying current \( I \) is uniform and given by: \[ B = \mu_0 \frac{N}{l} I, \] where \( \mu_0 \) is the permeability of free space.
The magnetic flux \( \phi \) through one turn of the solenoid is: \[ \phi = B \times A = \mu_0 \frac{N}{l} I \times A. \]
Since the solenoid has \( N \) turns, the total magnetic flux linked with the coil is: \[ \Phi = N \phi = N \times \left( \mu_0 \frac{N}{l} I A \right) = \mu_0 \frac{N^2 A}{l} I. \]

Step 3: By definition, the self-inductance \( L \) is the ratio of the total magnetic flux linked to the current: \[ L = \frac{\Phi}{I} = \mu_0 \frac{N^2 A}{l}. \] This expression shows that the self-inductance of a solenoid depends on the number of turns squared, the cross-sectional area, and inversely on its length.