Self & Mutual Inductance: JEE Main Physics Guide

Inductance is a property of coils that opposes changes in current — the electromagnetic analogue of inertia. Self-inductance and mutual inductance generate questions in both the standalone Electromagnetic Induction section and in the AC Circuits section of JEE Main. The formulas are derivable, and the question types are predictable. This guide covers everything from the derivation of inductance for standard geometries to energy stored and the coupling coefficient.

Self-Inductance: Definition and Formula

The self-inductance L of a coil is defined by: EMF = −L (dI/dt). It relates flux linkage to current: NΦ = LI → L = NΦ/I. For a long solenoid of n turns per unit length, length l, cross-sectional area A: B = μ₀nI, flux per turn = μ₀nIA, total flux linkage = N × μ₀nIA = nL × μ₀nIA = μ₀n²lA. Therefore L = μ₀n²lA = μ₀N²A/l. Unit: henry (H) = V·s/A = Ω·s. A solenoid of 1000 turns, length 0.5 m, area 4 cm² has L = μ₀ × (1000)² × (4×10⁻⁴) / 0.5 ≈ 1 mH.

For a toroid of N turns, mean radius R, cross-sectional area A: L = μ₀N²A/(2πR). For a coaxial cable (inner radius a, outer radius b, length l): L = (μ₀l/2π) ln(b/a). JEE uses these directly in formula-application questions. Energy stored in an inductor: U = ½LI². This is the magnetic energy, analogous to ½CV² for a capacitor. Take a free EM induction mock. For the related chapter on LC oscillations see our LC oscillations guide.

Mutual Inductance: Two Coils

When two coils are near each other, current in coil 1 creates a changing magnetic flux through coil 2, inducing an EMF. Mutual inductance M: EMF₂ = −M (dI₁/dt) and M = N₂Φ₂₁/I₁. M is symmetric: M₁₂ = M₂₁ = M. For two coaxial solenoids (1 inside 2), area A, lengths l₁ and l₂, turns N₁ and N₂: M = μ₀N₁N₂A/l (approximating they are the same length). Neumann's formula gives M in general, but JEE uses the solenoid formula. Coefficient of coupling k = M/√(L₁L₂) where 0 ≤ k ≤ 1. k = 1 means all flux links both coils (perfect coupling); k = 0 means no mutual linkage. Ideal transformer has k = 1.

Series and Parallel Inductors

Inductors in series (no mutual coupling): L_eq = L₁ + L₂ + … Inductors in parallel: 1/L_eq = 1/L₁ + 1/L₂ + … (same rules as resistors, but mutual inductance complicates it). With mutual inductance: series aiding: L_eq = L₁ + L₂ + 2M; series opposing: L_eq = L₁ + L₂ − 2M. JEE tests the aiding/opposing distinction — the orientation of the coils determines the sign of M's contribution. The energy stored in two mutually coupled coils: U = ½L₁I₁² + ½L₂I₂² ± MI₁I₂ (± depending on whether the fluxes aid or oppose). For RL circuit growth of current: i(t) = (V/R)(1 − e^(−Rt/L)) with time constant τ = L/R. Decay: i(t) = I₀e^(−Rt/L).

Transformer Action and Energy Transfer

An ideal transformer (k = 1, no resistance) has V₁/V₂ = N₁/N₂ and I₁/I₂ = N₂/N₁ (power in = power out). A step-up transformer (N₂ > N₁) increases voltage and decreases current. For the exam: power transmitted = V₁I₁ = V₂I₂; power loss in transmission lines ∝ I² → high-voltage (low-current) transmission reduces power loss. JEE frequently pairs the transformer with power loss calculations. For the broader AC circuit strategy, see our alternating current guide and our AC phasors guide.