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Physics · Class 12 · Chapter 6

Electromagnetic Induction

A changing magnetic flux produces an emf. That one sentence, together with Lenz's rule for the direction, explains generators, transformers and induction cooktops. The chapter rewards careful thinking about signs and directions.

In this chapter: experiments of Faraday and Henry, magnetic flux, Faraday's law, Lenz's law and energy conservation, motional emf (sliding and rotating rods), eddy currents, mutual inductance, self-inductance, energy stored in an inductor, the AC generator.

The experiments of Faraday and Henry

Faraday and Henry showed, around 1830, that a current is induced in a coil whenever the magnetic field through it changes. Moving a magnet towards or away from a coil connected to a galvanometer deflects the needle; the deflection lasts only while the magnet moves, and reverses when the direction of motion or the pole is reversed. The same happens with a current-carrying coil in place of the magnet, and even with both coils at rest if the current in one of them is switched on or off. A soft iron rod placed through both coils makes the effect much larger.

Magnetic flux

ΦB = B · A = BA cos θθ is the angle between B and the normal to the area; unit weber (Wb) = T m2; flux is a scalar

For a non-uniform field or curved surface, add B · dA over small elements. Flux can be changed by changing B, the area, or the angle θ.

Faraday's law

ε = −N dΦBdtN = number of turns; the magnitude of the emf equals the rate of change of flux linkage NΦ

The induced emf drives a current I = ε/R if the circuit is closed. The total charge that flows, Q = NΔΦ/R, depends only on the change in flux and not on how quickly it happens, which is a common numerical question.

Lenz's law

The polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. That is the meaning of the minus sign in Faraday's law.

Bring the north pole of a magnet towards a coil: the flux through the coil increases, so the induced current makes the near face of the coil a north pole to repel the magnet. Pull the magnet away and the near face becomes a south pole to attract it back. Either way you have to do work against the induced effect, and that work is what appears as electrical energy. If the induced current helped the change instead, a small push would give endless energy, which is impossible. So Lenz's law is a consequence of the conservation of energy.

Lenz's law: north pole approaching a coilwww.iitmedicoguide.comGNSmagnet moves inNinducedB from induced currentCoil (front of each turn solid)current in front runs downwardThe near face of the coil becomes N and repels the approaching N pole.Seen from the magnet, the induced current is anticlockwise.www.iitmedicoguide.com
As the north pole approaches, the induced current makes the near face of the coil a north pole. The induced field inside the coil points towards the magnet and opposes the increase of flux.

Motional emf

A straight conductor of length l moving with velocity v perpendicular to a uniform field B has an emf across its ends:

ε = Blv

You can derive it two ways. With the rod sliding on a U-shaped frame, the enclosed area changes at the rate lv, so dΦ/dt = Blv. Or look at the free charges in the rod: each feels the Lorentz force qvB along the rod, and the work done per unit charge over the length l is vBl. In the circuit below, the rod acts like a cell of emf Blv with no internal resistance (if the rod's resistance is neglected).

Motional emf in a rod sliding on railswww.iitmedicoguide.comB ⊗(into page)RPQvIIIF = IlBlemf ε = Blv, P at higher potential; current I = Blv/R flows up the rodF on the rod opposes v (Lenz), so an external force must keep it movingwww.iitmedicoguide.com
A rod sliding to the right in a field directed into the page. The flux through the circuit increases, so the induced current flows anticlockwise, and the force on the rod opposes its motion.

With a resistance R in the circuit, I = Blv/R, and the magnetic force on the rod is F = IlB = B2l2v/R, directed against v. To keep the rod moving at constant speed, an external agent must supply power P = Fv = B2l2v2/R, which is exactly the Joule heat I2R in the resistor.

Worked example: A rod of length 0.2 m slides at 5 m s−1 on rails connected by a 2 Ω resistor, in a field of 0.5 T normal to the plane of the rails. Find the emf, the current, the retarding force and the power needed.
Solution: ε = Blv = 0.5 × 0.2 × 5 = 0.5 V. I = 0.5/2 = 0.25 A. F = IlB = 0.25 × 0.2 × 0.5 = 0.025 N. Power = Fv = 0.025 × 5 = 0.125 W, the same as I2R = 0.0625 × 2 = 0.125 W.

Rod rotating about one end

A rod of length l rotating with angular velocity ω about one end, in a field B parallel to the axis of rotation, has different speeds at different points (v = ωr). Adding Bv dr from 0 to l gives

ε = ½Bωl2
Worked example: A 1.0 m metal rod rotates at 400 rad s−1 about an axis through one end, perpendicular to the rod, in a field of 0.5 T parallel to the axis. Find the emf between the ends.
Solution: ε = ½ × 0.5 × 400 × (1.0)2 = 100 V.

Eddy currents

When a bulk piece of metal moves in a magnetic field, or the flux through it changes, circulating currents are induced in the body of the metal. These eddy currents oppose the motion (a copper plate swinging between magnet poles stops quickly) and produce heat. They are reduced by using laminated cores or cutting slots in the metal, which lengthens the current paths. Useful applications include magnetic braking in trains, electromagnetic damping in galvanometers, induction furnaces and electric power meters.

Inductance

Flux through a coil is proportional to the current producing it, so we can write flux linkage as a constant times current. That constant is the inductance, measured in henry (1 H = 1 Wb A−1 = 1 V s A−1). It depends only on the geometry of the coils and the medium.

Mutual inductance

For two coils, the flux linkage in coil 1 due to current I2 in coil 2 is N1Φ1 = M12I2, and a changing I2 induces ε1 = −M dI2/dt. It can be shown that M12 = M21 = M. For a long solenoid of radius r1 placed coaxially inside another of radius r2, both of length l:

M = μ0n1n2πr12lwith a medium of relative permeability μr, multiply by μr

Self-inductance

A changing current in a coil induces an emf in the same coil, ε = −L dI/dt. This back emf opposes any change in current, so inductance plays the role of inertia in circuits. For a long solenoid of n turns per unit length, cross-section A and length l:

L = μ0n2Al = μ0N2AlEnergy stored: U = ½LI2

Compare ½LI2 with ½mv2: L is the analogue of mass and I of velocity. The energy is stored in the magnetic field of the coil.

AC generator

A coil of N turns and area A is rotated with angular speed ω in a uniform field B. The angle between the normal and the field is θ = ωt, so Φ = NBA cos ωt and

ε = NBAω sin ωt = ε0 sin ωt,   ε0 = NBAω

The emf is maximum when the plane of the coil is parallel to B (flux zero but changing fastest) and zero when the plane is perpendicular to B (flux maximum). In India the mains frequency is 50 Hz. Slip rings and brushes connect the rotating coil to the external circuit; in hydroelectric generators falling water turns the coil, and in thermal generators steam does.

Common mistakes: (1) Thinking a large flux means a large emf; the emf depends on the rate of change of flux. (2) Getting the Lenz direction by guesswork; decide whether flux is increasing or decreasing, and make the induced field oppose that change. (3) Using ε = Blv for a rotating rod; the speed varies along the rod, so use ½Bωl2. (4) Taking the emf of a generator as maximum when the flux is maximum; it is the other way round. (5) Forgetting that doubling the turns of a solenoid of the same length multiplies L by four.

JEE and NEET focus

  • Faraday's law with a changing field, area or angle, including emf from a given Φ(t).
  • Direction of induced current by Lenz's law in loops, rings and falling magnets.
  • Motional emf of sliding and rotating rods, with the force and power needed.
  • Self and mutual inductance of solenoids and energy stored in an inductor.
  • Peak emf of an AC generator and its variation with time.

Practice questions

The flux through a coil varies as Φ = (5t2 + 3t) Wb. The magnitude of the induced emf at t = 2 s is:

  1. 13 V
  2. 20 V
  3. 23 V
  4. 26 V
Show answer
C. dΦ/dt = 10t + 3 = 23 V at t = 2 s.

Lenz's law is a consequence of the conservation of:

  1. charge
  2. momentum
  3. energy
  4. mass
Show answer
C. Work done against the induced effect becomes electrical energy.

A 100-turn coil of area 0.01 m2 is normal to a field that falls uniformly from 0.5 T to zero in 0.1 s. The average induced emf is:

  1. 0.5 V
  2. 5 V
  3. 50 V
  4. 0.05 V
Show answer
B. ε = NAΔB/Δt = 100 × 0.01 × 0.5/0.1 = 5 V.

The number of turns of a solenoid is doubled, keeping its length and area the same. Its self-inductance becomes:

  1. L/2
  2. 2L
  3. 4L
  4. L
Show answer
C. L ∝ N2 for fixed l and A.

The energy stored in a 2 H inductor carrying 3 A is:

  1. 3 J
  2. 6 J
  3. 9 J
  4. 18 J
Show answer
C. U = ½ × 2 × 32 = 9 J.

A 50-turn coil of area 0.02 m2 rotates at 100 rad s−1 in a field of 0.4 T. The peak emf is:

  1. 4 V
  2. 40 V
  3. 400 V
  4. 0.4 V
Show answer
B. ε0 = NBAω = 50 × 0.4 × 0.02 × 100 = 40 V.

The henry is equivalent to:

  1. V A s−1
  2. V s A−1
  3. Wb A
  4. T m2 s
Show answer
B. From ε = L dI/dt, L = ε/(dI/dt).

A bar magnet is dropped along the axis of a horizontal copper ring. While approaching the ring, the magnet's acceleration is:

  1. equal to g
  2. greater than g
  3. less than g
  4. zero
Show answer
C. The induced current in the ring opposes the approach, giving an upward force.
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