Electromagnetic Induction — Part 2: Self-induction, mutual induction and the transformer: visual lesson, ISC Class 12 Physics

Lesson

Inductance and the transformer

Part 2 — a coil that resists changes in its own current, two coils that talk through their shared flux, and the transformer.

In this lesson you will

Define self-inductance, derive L for a long solenoid and explain back emf
Define mutual inductance and derive M for two coaxial solenoids
Explain the working, relations, efficiency and losses of a transformer and why power is transmitted at high voltage

Concept

Self-induction

The current $I$ in a coil sets up a flux through its own turns. The flux linkage is proportional to the current: $N\Phi = LI,\qquad \varepsilon = -L\,\dfrac{dI}{dt}$ $L$ is the self-inductance. Unit: henry (1 H = 1 V s A⁻¹ = 1 Wb A⁻¹).

A coil has 1 H if a current changing at 1 A s⁻¹ induces 1 V across it.
L depends on the number of turns, the size and shape of the coil and the core, not on the current.

Diagram

Self-inductance of a long solenoid

A long solenoid: $N$ turns, length $l$ , area $A$ , $n = N/l$ . Inside, $B = \mu_0 nI = \dfrac{\mu_0NI}{l}$ , so the flux linkage is $N\Phi = N\cdot\dfrac{\mu_0NI}{l}\cdot A = LI \;\Rightarrow\; L = \dfrac{\mu_0N^2A}{l} = \mu_0n^2Al$

L ∝ N²: twice the turns on the same length, four times the inductance.
With an iron core of relative permeability $\mu_r$ : $L = \mu_0\mu_rN^2A/l$ .

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