Board Formulas

Electromagnetic Induction

Faraday's law, Lenz's law, motional EMF, self and mutual inductance, AC generator — NCERT Class 12 Physics Ch 6

📐 12 formulas✏️ 3 examples🎯 6 practice⚖️ 5 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus
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Board Exam Tips

  • Faraday's law + Lenz's law (statement + direction of induced current) is nearly guaranteed as a 2–3 mark question.
  • Motional EMF derivation (\varepsilon = BLv) is a repeat 3-mark board favourite — memorise the force/work chain.
  • Self-inductance of a solenoid (L = \mu_0 n²Al) derivation appears in ~40% of past papers.
  • Never drop the minus sign in \varepsilon = -d\Phi/dt without stating you're finding magnitude.
  • For AC generator, know peak EMF \varepsilon_0 = NBA\omega and its instantaneous form.

📊 Diagram

C

Coil with changing magnetic flux inducing EMF in a closed loop

📐 Formulas(12)

1

Magnetic Flux

SymbolMeaning
Magnetic flux (Wb = T·m²)
Angle between B and area vector
2

Faraday's Law of Induction★ Board fav

SymbolMeaning
Induced EMF (V)
Number of turns
Rate of change of flux (Wb/s)
3

Lenz's Law★ Board fav

4

Motional EMF★ Board fav

5

Self-Induced EMF

SymbolMeaning
Self-inductance (H = Wb/A)
Rate of change of current (A/s)
6

Self-Inductance of a Solenoid★ Board fav

7

Mutual Inductance

SymbolMeaning
Mutual inductance (H)
Current in primary coil (A)
8

Mutual Inductance of Two Coaxial Solenoids

9

Energy Stored in an Inductor

10

Magnetic Energy Density

11

AC Generator EMF

12

Series/Parallel Inductors (no mutual coupling)

✏️ Solved Examples

1Solved Exampleeasy4 steps

A rectangular coil of 100 turns and area 200 cm² is placed with its plane perpendicular to a magnetic field that changes uniformly from 0.2 T to 0.6 T in 0.5 s. Find the induced EMF.

1

Convert area to SI, note field ⊥ plane means cos\theta = 1

2Solved Exampleboard5 steps

A metal rod of length 50 cm slides on horizontal rails at 4 m/s in a vertical uniform field of 0.5 T. The rails and a resistor of 2 \Omega form a closed loop. Find (a) motional EMF, (b) current in the loop, (c) mechanical power supplied by the external agent.

1

Motional EMF

3Solved ExampleHOTS6 steps

A long solenoid of length 60 cm and radius 2 cm has 1200 turns. A small coil of 40 turns and area 4 cm² is placed at its centre, with axis parallel. If the current in the solenoid changes from 0 to 3 A in 0.2 s, find the mutual inductance and the EMF induced in the small coil.

1

Turns per metre of solenoid

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6
  • 1

    Confusing 'flux' with 'rate of change of flux' — the flux itself need not induce an EMF

    EMF appears only when \Phi *changes* with time. A steady flux gives zero EMF.

  • 2

    Ignoring Lenz's law sign and getting wrong current direction

    Induced current opposes the change in flux. If flux is increasing, induced B inside loop opposes it — use right-hand rule to fix current direction.

  • 3

    Confusing self-inductance L with mutual inductance M

    L relates EMF in a coil to its *own* current change. M relates EMF in coil 2 to current change in *coil 1*.

  • 4

    Forgetting that inductance depends only on geometry (and medium), not on I

    L and M are constants for a fixed coil. They do NOT change when I changes — the *EMF* does.

  • 5

    Not converting cm² to m² in flux/inductance calculations

    1 cm² = 10⁻⁴ m². Missing this factor gives an answer off by 10⁴.

  • 6

    Applying \varepsilon = BLv when v is parallel to L or B

    \varepsilon = BLv only when B, L, v are mutually perpendicular. Otherwise use \varepsilon = (\vec{v}\times\vec{B})\cdot\vec{L}.

🎯 Practice Yourself

🎯Practice Yourself6 questions
  1. Q1

    A coil of 500 turns and area 10 cm² is in a field 0.4 T. It is pulled out completely in 0.1 s. Average induced EMF?

  2. Q2

    A solenoid of length 1 m has 2000 turns and radius 2 cm. Compute its self-inductance.

  3. Q3

    Current in an inductor of 0.5 H changes at 20 A/s. Self-induced EMF?

  4. Q4

    Energy stored when 4 A flows through a 2 H inductor?

  5. Q5

    An AC generator has 200 turns, coil area 100 cm², rotates at 50 Hz in field 0.5 T. Peak EMF?

  6. Q6

    A rod 40 cm long moves at 5 m/s perpendicular to a 0.2 T field. Motional EMF?

📝 Notes

Electromagnetic Induction — key concepts

The single phenomenon: a changing magnetic flux through a circuit induces an EMF around that circuit. This is the basis of the electric generator, transformer, and induction motor.

Faraday, in three sentences

  • Whenever the magnetic flux \Phi through a closed loop changes, an EMF \varepsilon = −d\Phi/dt is induced.
  • The change can come from a changing B, a moving area (motional EMF), or a rotating coil (generator).
  • With N turns, replace \Phi with N\Phi (flux linkage).

Lenz's law — the "why" of the minus sign

The induced current always flows in a direction that opposes the change producing it. If flux is increasing, the induced B opposes it; if decreasing, the induced B supports it. This is a statement of energy conservation: an induced current cannot make its own source of energy.

Motional EMF — two equivalent derivations

  • Force picture: free electrons in the moving rod feel qv × B, get pushed along the rod, produce EMF = BLv.
  • Flux picture: area swept per second = Lv, so d\Phi/dt = BLv. Same answer.

Self and mutual inductance

  • Self-inductance L (Henry) — property of a coil. \varepsilon_{self} = −L(dI/dt). Opposes changes in its own current — "electrical inertia".
  • Mutual inductance M — property of a pair of coils. \varepsilon_2 = −M(dI_1/dt). Basis of the transformer.

Quick sanity checks

  • If flux is not changing, EMF is zero — full stop.
  • Magnetic energy density B²/(2\mu_0) and electric energy density \tfrac{1}{2}\varepsilon_0 E² look alike — remember which \mu goes where.
  • L is measured in henry (H) = V·s/A = Wb/A.
  • Doubling turns quadruples L (since L ∝ N²) but only doubles M for coupled coils (M ∝ N_1 N_2).
  • Any solution must satisfy energy conservation: mechanical work done = electrical energy dissipated (motional EMF problems).

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