Electromagnetic induction for JEE Main

Electromagnetic induction has one law and many disguises. Every question, whether a sliding rod, a spinning wheel, a magnet dropped through a coil or a transformer, comes down to asking how fast the magnetic flux through a circuit is changing. Lenz's law then tells you the direction. Once you see it that way, the chapter becomes short and very scoring.

The official 2026 syllabus lists Faraday's law, induced emf and current, Lenz's law, eddy currents, and self and mutual inductance. The AC generator and transformer are in the same unit and are covered in alternating current circuits.

Magnetic flux

For a flat area A in a uniform field B, with θ the angle between B and the normal to the area:

Φ = BA cos θ, measured in webers (1 Wb = 1 T m²).

Flux can change in three ways: change B, change the area inside the field, or change the angle. Every induction problem uses one or more of these.

Faraday's and Lenz's laws

For a coil of N turns:

e = −N dΦ/dt

The minus sign is Lenz's law: the induced current flows so as to oppose the change in flux that produces it. Push the north pole of a magnet towards a coil, and the near face of the coil becomes a north pole and repels it. Pull the magnet away, and the near face becomes south and attracts it.

Lenz's law is conservation of energy. If the induced current helped the change, you would get electrical energy for nothing.

Induced charge. With total resistance R, the current is e/R, and the charge that flows is

q = NΔΦ/R

This depends only on the total change in flux, not on how quickly it happens.

Motional emf

A rod of length l moving at speed v perpendicular to a field B has an emf e = Blv across it. Two ways to see it:

• Force on charges: each charge feels qvB along the rod. Charges pile up at the ends until the electric force qE balances it, so E = vB and e = El = Blv.
• Flux: if the rod slides on rails, the enclosed area grows at the rate lv, so dΦ/dt = Blv.

With the circuit closed through a resistance R, the current is I = Blv/R. The field then pushes back on the rod with F = BIl = B²l²v/R, opposite to its motion. Keeping the rod moving at a steady speed takes power Fv, which appears as heat I²R.

A rod sliding down vertical rails reaches terminal velocity when this force equals its weight: v = mgR/(B²l²).

Rotating rod. A rod of length l spins about one end with angular speed ω in a field B parallel to the axis. An element at distance r moves at ωr, so

e = ∫Bωr dr, from 0 to l, which gives e = ½Bωl²

A metal disc of radius R spinning the same way gives ½BωR² between the centre and the rim.

Rotating coil. A coil of N turns and area A spinning at ω in a uniform field has Φ = NBA cos ωt, so e = NBAω sin ωt. This is the AC generator.

Eddy currents

A changing flux through a solid piece of metal drives circulating currents inside it. They heat the metal and oppose the motion that causes them.

• Useful: magnetic braking in trains, induction furnaces, electromagnetic damping of galvanometer coils, and some speedometers.
• Unwanted: heating of transformer and motor cores. Making the core from thin insulated sheets (lamination) cuts the eddy current paths and reduces the loss.

Self and mutual inductance

Self-inductance. The flux a coil produces through itself is proportional to its current: NΦ = LI. When the current changes,

e = −L dI/dt

The inductor opposes changes in current, much as mass opposes changes in velocity. For a long solenoid of n turns per metre, cross-section A and length l:

L = μ₀n²Al = μ₀N²A/l

Energy stored is U = ½LI², with energy density B²/(2μ₀) in the field.

Mutual inductance. A current I₁ in one coil puts flux through a second: N₂Φ₂ = MI₁, and e₂ = −M dI₁/dt. M is the same whichever coil carries the current. For two long coaxial solenoids, the inner of radius r₁, M = μ₀n₁n₂πr₁²l. In general, M ≤ √(L₁L₂).

Worked problems

Problem 1: Faraday's law and charge. The flux through each turn of a 100-turn coil falls from 0.02 Wb to 0.01 Wb in 0.1 s. The coil's resistance is 5 Ω. Find the average emf and the charge that flows.

e = 100 × 0.01/0.1 = 10 V.
q = NΔΦ/R = 100 × 0.01/5 = 0.2 C (check: current 2 A for 0.1 s).

Problem 2: rotating rod. A 1 m rod spins about one end at 10 rad/s in a 0.2 T field along the axis. Find the emf between its ends.

e = ½ × 0.2 × 10 × 1² = 1 V.

Problem 3: sliding rod. A 0.5 m rod slides at 5 m/s on rails in a 0.4 T field, and the circuit resistance is 2 Ω. Find the current, the force needed to keep the speed steady, and the power.

e = 0.4 × 0.5 × 5 = 1 V, so I = 0.5 A.
F = BIl = 0.4 × 0.5 × 0.5 = 0.1 N.
P = Fv = 0.5 W, which equals I²R = 0.25 × 2.

Problem 4 (numerical answer): self-induced emf. The current in a 50 mH coil falls steadily from 4 A to zero in 0.01 s. Find the induced emf in volts.

e = L ΔI/Δt = 0.05 × 4/0.01 = 20 V. The energy released is ½ × 0.05 × 4² = 0.4 J.

Practice set

1. Find the energy stored in a 2 H inductor carrying 3 A.
2. A loop of area 0.02 m² lies perpendicular to a field that changes at 0.5 T/s. Find the emf.
3. The north pole of a magnet is pushed towards a coil. Seen from the magnet's side, which way does the induced current flow?
4. A 50-turn coil of area 0.01 m² and resistance 10 Ω lies perpendicular to a 0.1 T field. It is flipped through 180°. Find the charge that flows.
5. A rod of mass 20 g and length 0.5 m slides down smooth vertical rails in a horizontal 1 T field; the circuit resistance is 1 Ω. Find its terminal speed (g = 10 m/s²).
6. An aircraft with a 20 m wingspan flies horizontally at 250 m/s where the vertical component of the Earth's field is 4 × 10⁻⁵ T. Find the emf between its wingtips.
7. Why are transformer cores laminated?
8. The number of turns of a solenoid is doubled, keeping its length and area fixed. What happens to its self-inductance?

Answers

1. ½ × 2 × 3² = 9 J.
2. e = A dB/dt = 0.02 × 0.5 = 0.01 V (10 mV).
3. Anticlockwise, making the near face a north pole that repels the magnet.
4. ΔΦ = 2BA per turn, so q = 2NBA/R = 2 × 50 × 0.1 × 0.01/10 = 0.01 C.
5. v = mgR/(B²l²) = (0.02 × 10 × 1)/(1 × 0.25) = 0.8 m/s.
6. e = Blv = 4 × 10⁻⁵ × 20 × 250 = 0.2 V.
7. To break up the paths of eddy currents and reduce heating.
8. L ∝ N², so it becomes 4 times.

What to do next

• For each of the three ways flux can change, solve five problems and name the mechanism before you start.
• Derive e = Blv and e = ½Bωl² from the force on charges.
• Practise Lenz's law directions with magnets, loops entering fields and pairs of coils.
• Move on to AC circuits, and revise magnetic effects of current for the field of a solenoid.