Define self-inductance and its S.I. unit. Derive an expression for self-inductance of a long, air-cored solenoid of length, radius r, and having N number of turns.

Self-inductance is a property of an electrical circuit that quantifies how much voltage is induced in a coil due to a change in current flowing through it. It essentially measures the coil's ability to oppose changes in current. The S.I. unit of self-inductance is the henry (H).

Deriving the Expression for Self-Inductance

To derive the expression for the self-inductance of a long, air-cored solenoid, we consider the following parameters:

• L: Self-inductance
• N: Number of turns
• l: Length of the solenoid
• r: Radius of the solenoid
• B: Magnetic field inside the solenoid

Step 1: Magnetic Field Inside the Solenoid

The magnetic field B inside a long solenoid is given by the formula:

B = μ₀ * (N/l) * I

where μ₀ is the permeability of free space, I is the current flowing through the solenoid, and N/l is the number of turns per unit length.

Step 2: Magnetic Flux

The magnetic flux Φ through one turn of the solenoid is calculated as:

Φ = B * A

Here, A is the cross-sectional area of the solenoid, given by:

A = πr²

Substituting for B, we get:

Φ = μ₀ * (N/l) * I * πr²

Step 3: Induced EMF and Self-Inductance

The induced electromotive force (EMF) in the solenoid due to a change in current is given by Faraday's law:

ε = -dΦ/dt

Substituting the expression for magnetic flux:

ε = -d(μ₀ * (N/l) * I * πr²)/dt

This leads to:

ε = -μ₀ * (N/l) * πr² * (dI/dt)

By definition, self-inductance L is related to the induced EMF and the rate of change of current:

L = -ε / (dI/dt)

Substituting for ε, we find:

L = μ₀ * (N² * πr²) / l

Final Expression

The self-inductance of a long, air-cored solenoid is given by:

L = (μ₀ * N² * πr²) / l

This formula shows how the self-inductance depends on the number of turns, the radius, and the length of the solenoid, as well as the permeability of free space.