MHT-CET Physics · Electromagnetic Induction

Mutual Inductance, Coupling, Transformers and Generators

When the current in one coil changes, the flux it sends through a neighbouring coil changes and induces an e.m.f. there; the constant M linking the two (Nφ₂ = MI₁, e₂ = M dI₁/dt) is their mutual inductance, it is at most √(L₁L₂), and the transformer and the a.c. generator are both built on it.

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Why this matters

35 PYQs, 3 HARD. Twenty are mutual inductance — M from an e.m.f. and a rate of change, the flux one coil sends through another, the peak e.m.f. when the current is I₀ sin ωt, the time for a current change, and M for two concentric coplanar rings (asked seven times); three are the coefficient of coupling; twelve are transformers and generators — turns, voltage and current ratios, power lost, laminated cores, flux against e.m.f. in a rotating coil, and displacement current. Three cards.

Concept 1 of 3: Mutual Inductance

A current I₁ in coil 1 sends flux through coil 2; the flux linkage is proportional to I₁, N₂φ₂ = MI₁, and the same M works in both directions. So measure M one way and use it the other: an e.m.f. of 15 mV in P when Q's current rises at 10 A/s means M = 1.5 mH, and then 1.8 A in P links 2.7 mWb through Q. When the current is I₀ sin ωt, the e.m.f. in the other coil peaks at MI₀ω. For two concentric coplanar rings with r₂ ≪ r₁, the big ring's centre field μ₀I/(2r₁) is taken as uniform over the small one: M = μ₀πr₂²/(2r₁), which is proportional to r₂²/r₁.

Definition

$N_2\phi_2 = MI_1$ , $e_2 = M\dfrac{dI_1}{dt}$ ; M is the same whichever coil carries the current.
Peak e.m.f. for $I = I_0\sin\omega t$ : $e_{\max} = MI_0\omega$ ( $M = 1$ H, $I_0 = 2/\pi$ A, 50 Hz ⇒ 200 V).
Time for a change: $\Delta t = \dfrac{M\,\Delta I}{e}$ (2 H, 6 A → 3 A, 2 kV ⇒ $3 \times 10^{-3}$ s). From flux: $M = \dfrac{\Delta\phi}{\Delta I}$ .
e.m.f. per turn of a coil of N turns: $\dfrac{MI}{Nt}$ .
Concentric coplanar rings ( $r_2 \ll r_1$ ): $M = \dfrac{\mu_0\pi r_2^2}{2r_1} \propto \dfrac{r_2^2}{r_1}$ .
Toroid of N turns, major radius R, cross-section radius r: $\dfrac{\mu_0N^2r^2}{2R}$ .

Mutual inductance

e_2 = M\frac{dI_1}{dt}, \qquad M_{\text{rings}} = \frac{\mu_0\pi r_2^2}{2r_1}

Worked example

Current in coil A rising at 20 A/s induces 30 mV in coil B. What flux linkage does a steady 4 A in B produce in A?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift I · Q8Moderate

Example 1 · Electromagnetic Induction · Mutual Inductance, Coupling, and Transformer/Generator

Two coils P and Q are kept near each other. When no current flows through coil P and current increases in coil Q at the rate of

10\text{ }A/s

, the e.m.f. in coil

P

15mV

. When coil

Q

carries no current and current of 1.8 A flows through coil P, the magnetic flux linked with coil Q is

Putting the big ring's radius on top

The field comes from the LARGE ring (∝ 1/r₁) and passes through the SMALL ring's area (∝ r₂²): M ∝ r₂²/r₁. The options swap these.

Drill 19 more on mutual inductance

Concept 2 of 3: Coefficient of Coupling

Not all of one coil's flux reaches the other. The coefficient of coupling K = M/√(L₁L₂) measures the fraction: 1 when every line of one threads the other, less otherwise. So two coils whose flux is completely shared have M = √(L₁L₂).

Definition

$K = \dfrac{M}{\sqrt{L_1L_2}}$ , $0 \le K \le 1$ .
Perfect coupling (K = 1): $M = \sqrt{L_1L_2}$ — 25 mH and 9 mH give 15 mH.

Coupling

K = \frac{M}{\sqrt{L_1L_2}}

Worked example

Coils of 16 mH and 25 mH have M = 12 mH. K?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q18Easy

Example 2 · Electromagnetic Induction · Mutual Inductance, Coupling, and Transformer/Generator

Two coils of self-inductance 25 mH and 9 mH are placed close together such that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is

Adding the inductances

Perfectly coupled 25 mH and 9 mH give M = √(25 × 9) = 15 mH, a geometric mean — not 34 or 16.

Drill 2 more on coefficient of coupling

Concept 3 of 3: Transformers and Generators

A transformer is two coils on one iron core: the alternating flux through both induces the same e.m.f. PER TURN in each, so V_s/V_p = N_s/N_p. An ideal one wastes no power, so the current goes the other way, I_s/I_p = N_p/N_s; a real one loses some, and the secondary current comes from the power that remains. The core is laminated to cut eddy currents. An a.c. generator is a coil turning in a field: its flux is φ₀cos ωt and its e.m.f. ωφ₀ sin ωt, a quarter-cycle apart — flux largest when the coil's plane faces the field squarely, e.m.f. zero at that instant.

Definition

$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$ ; e.m.f. per turn is the same in both coils (1000 : 3000 turns, 80 V ⇒ 0.08 V per turn).
Ideal: $V_pI_p = V_sI_s$ (4.4 kW at 3.3 kV ⇒ $\tfrac{4}{3}$ A). With loss: $I_s = \dfrac{\eta V_pI_p}{V_s}$ (1100 W, 50% lost, 2200 V ⇒ 0.25 A). Input power fixes $V_p = P/I_p$ .
Impedance matching: $\dfrac{N_p}{N_s} = \sqrt{\dfrac{Z_p}{Z_s}}$ (8000 Ω to 8 Ω ⇒ about 32 : 1).
Laminated core: reduces eddy current losses. The secondary e.m.f. comes from the varying magnetic field.
Generator: flux and e.m.f. differ in phase by $\dfrac{\pi}{2}$ ; plane perpendicular to B ⇒ flux maximum, e.m.f. zero.
Displacement current in a capacitor: $i_d = C\dfrac{dV}{dt}$ (1 μF at 4 V/s ⇒ 4 μA).

Transformer

\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}\ \text{(ideal)}

Worked example

A transformer steps 240 V down to 12 V for a 60 W lamp. Turns ratio and primary current (ideal)?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q34Moderate

Example 3 · Electromagnetic Induction · Mutual Inductance, Coupling, and Transformer/Generator

A current of 5 A is flowing at 220 V in a primary coil of a transformer. If the voltage produced in the secondary coil is 2200 V and

50\%

of power is lost, then the current in the secondary coil will be

Scaling current the same way as voltage

A step-up transformer raises the voltage and LOWERS the current: power in ≈ power out. More secondary turns means less secondary current.

Drill 11 more on transformers and generators

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Mutual Inductance
Mutual inductance
$e_2 = M\frac{dI_1}{dt}, \qquad M_{\text{rings}} = \frac{\mu_0\pi r_2^2}{2r_1}$
Coefficient of Coupling
Coupling
$K = \frac{M}{\sqrt{L_1L_2}}$
Transformers and Generators
Transformer
$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}\ \text{(ideal)}$

Watch out for (3)

Putting the big ring's radius on top
→ Mutual Inductance
Adding the inductances
→ Coefficient of Coupling
Scaling current the same way as voltage
→ Transformers and Generators

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