PhysicsCBSE Class 12 · Electromagnetic Induction

Faraday's and Lenz's Laws Class 12: Induced EMF and Its Direction

Updated 15 October 2026 · Fact-checked

Faraday's law says an emf is induced in a circuit whenever the magnetic flux through it changes, equal to the rate of change of flux: ε = −N dΦ/dt. Lenz's law gives the direction: the induced current flows so as to oppose the change in flux that produces it, which is a consequence of conservation of energy.

Pushing the north pole in makes the coil's near face a north pole that repels it, so the induced current opposes the change; the minus sign in Faraday's law sho

Illustration drawn for QuestionBuddy.

Understand Faraday's and Lenz's Laws

Push a magnet into a coil connected to a galvanometer and the needle kicks; hold the magnet still and the needle returns to zero; pull it out and the needle kicks the other way. A current appears only while something is changing. This is electromagnetic induction.

Magnetic flux. For a plane area A in a uniform field B, the flux is $Φ_{B} = B \cdot A = B A cos θ$ , where $θ$ is the angle between $B$ and the normal to the area. Its SI unit is the weber (1 Wb = 1 T m²). For a coil of N turns, the flux linkage is $N Φ$ . Flux can change in three ways: by changing B, by changing A, or by changing $θ$ (rotating the coil).

Faraday's experiments, in short. (1) A magnet moved towards or away from a coil induces a current; the faster the motion, the larger the current. (2) Moving a current-carrying coil near a second coil does the same. (3) Even without motion, switching the current in one coil on or off induces a current in a nearby coil. The common factor in every case is a change of magnetic flux through the coil.

Faraday's law. The magnitude of the induced emf equals the rate of change of magnetic flux through the circuit. For N turns: $ε = - N \frac{d Φ _{B}}{d t}$ . If the flux changes steadily by $ΔΦ$ in time $Δ t$ , the average emf is $N \frac{ΔΦ}{Δ t}$ . The induced current is $I = \frac{ε}{R}$ , and the total charge that flows is $q = \frac{N ΔΦ}{R}$ , which depends only on the total change of flux, not on how quickly it happens.

Lenz's law and the minus sign. The induced emf (and current) has a direction such that it opposes the change in flux that produces it. If a north pole approaches a coil, the induced current makes the near face of the coil a north pole, repelling the magnet; when the north pole is pulled away, the near face becomes a south pole, attracting it back. In each case you must do work against the induced effect, and that work is what appears as electrical energy (heat in the circuit). If the induced current helped the change instead, a small push would grow on its own and create energy from nothing: Lenz's law is required by conservation of energy.

Finding the direction. (1) Decide whether the flux through the coil, in a chosen direction, is increasing or decreasing. (2) The induced current produces its own field opposite to the external field if the flux is increasing, and along it if the flux is decreasing. (3) Use the right-hand thumb rule to turn that field direction into a current direction. A closed circuit is needed for a current, but an emf is induced even in an open one.

A rotating coil. A coil of N turns and area A turning at angular speed ω in a uniform field has flux $Φ = B A cos ω t$ , so $ε = N B A ω sin ω t$ . The emf is largest when the plane of the coil is parallel to B (the flux is zero but changing fastest) and zero when the plane is perpendicular to B (flux largest, momentarily not changing). This is the principle of the AC generator, studied in the next chapter.

Faraday's and Lenz's Laws: formulas and key facts

Magnetic flux: $Φ_{B} = B \cdot A = B A cos θ$; θ between B and the normal; unit weber (Wb = T m²).
Faraday's law: $ε = - N \frac{d Φ _{B}}{d t}$; The minus sign expresses Lenz's law.
Average induced emf: $ε_{a v g} = N \frac{ΔΦ}{Δ t}$; For a steady or average change.
Induced current and charge: $I = \frac{ε}{R}, q = \frac{N ΔΦ}{R}$; Charge depends on total flux change, not on time.
Rotating coil: $ε = N B A ω sin ω t, ε_{0} = N B A ω$; Largest when the plane of the coil is parallel to B.
Flux change when a coil is flipped: $ΔΦ = 2 B A$; From +BA to −BA per turn.

How to solve Faraday's and Lenz's Laws questions

Use this method for induced emf, current, charge and direction questions.

1Write the flux per turn, $Φ = B A cos θ$ , at the start and at the end (A in m², θ from the normal).
2Find $ΔΦ$ (careful with sign changes when the coil flips) and the time taken.
3Emf: $ε = N \frac{ΔΦ}{Δ t}$ for an average, or differentiate Φ(t) for an instantaneous value.
4Current: $I = \frac{ε}{R}$ ; charge: $q = \frac{N ΔΦ}{R}$ .
5Direction: decide if the flux increases or decreases; the induced field opposes an increase or supports a decrease; convert to current with the right-hand thumb rule.
6For forces on magnets or rings, remember the induced current always opposes the relative motion.
7State the units: V for emf, A for current, C for charge, Wb for flux.

Faraday's and Lenz's Laws in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Induced emf from a moving magnet

Suppose a 'shake torch' has a magnet that slides back and forth through a 500-turn coil. Each pass through the coil changes the flux in every turn by $4 \times 1 0^{- 4}$ Wb in 0.05 s.

(a) Find the average emf during one pass. (b) What happens to the emf if you shake twice as fast? (c) Why does the torch get harder to shake when its circuit is switched on?

Try it first, then show the solutionSolution

$ε = N \frac{ΔΦ}{Δ t} = 500 \times \frac{4 \times 1 0 ^{- 4}}{0.05} = 4$ V.
Shaking twice as fast halves Δt, so the emf doubles to 8 V.
With the circuit closed, the induced current produces a field that opposes the magnet's motion (Lenz's law), so you must work harder; that work becomes the electrical energy charging the torch.

Answer: (a) 4 V; (b) 8 V; (c) the induced current opposes the motion, so energy must be supplied.

Memory hook: Faster change, bigger emf; and the coil always pushes back.

Induction without motion

Imagine a phone on a wireless charging pad. The pad's coil makes a field through the phone's 10-turn receiving coil (area 10 cm²) that rises from 0 to 2 mT in 5 μs.

(a) Find the average emf induced in the phone's coil during the rise. (b) Nothing moves. Why is there still an emf? (c) What would happen if the pad's current were steady?

Try it first, then show the solutionSolution

$ε = N \frac{ΔΦ}{Δ t} = 10 \times \frac{2 \times 1 0 ^{- 3} \times 1 0 ^{- 3}}{5 \times 1 0 ^{- 6}} = 4$ V.
Faraday's law needs only a change of flux, which here comes from the changing current in the pad's coil.
A steady current gives a constant flux, so no emf would be induced.

Answer: (a) 4 V; (b) the flux changes because the field changes; (c) no emf.

Memory hook: Change is the key: no change in flux, no induced emf.

Tricks to remember Faraday's and Lenz's Laws

Short memory aids. Say each one aloud once, then use it on a question the same day.

Flux = B A cos (normal)
Angle with the normal, not with the plane.
Use it when: Flux calculations.
Lenz the lazy
The coil hates change: it fights an increase and props up a decrease.
Use it when: Finding the direction of induced current.
Approaching N makes N
Bring a north pole near: the coil face becomes north to push it back. Pull it away: the face becomes south to hold it.
Use it when: Magnet-and-coil direction questions.
Charge ignores the clock
q = NΔΦ/R: slow or fast, the same charge flows.
Use it when: Induced charge questions.

Faraday's and Lenz's Laws: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: EMF from a changing field · 2 marks · Very short answerBoard level

A coil of 200 turns and area 0.01 m² has its plane perpendicular to a magnetic field. The field falls steadily from 0.5 T to 0.1 T in 0.2 s. Find the induced emf.

Try it first, then show the solutionSolution

Flux per turn changes by $ΔΦ = Δ B \times A = 0.4 \times 0.01 = 4 \times 1 0^{- 3}$ Wb. (1)
$ε = N \frac{ΔΦ}{Δ t} = 200 \times \frac{4 \times 1 0 ^{- 3}}{0.2} = 4$ V. (1)

Answer: 4 V.

Example 2: Lenz's law and energy · 3 marks · Short answerBoard level

(a) State Lenz's law. (b) A bar magnet is pushed with its north pole towards a closed coil. Find the polarity of the near face of the coil and the direction of the current as seen from the magnet. (c) Show that Lenz's law follows from conservation of energy.

Try it first, then show the solutionSolution

(a) The induced current flows in a direction that opposes the change in magnetic flux that produces it. (1)
(b) The flux towards the coil increases, so the induced current makes the near face a north pole, repelling the magnet: the current is anticlockwise as seen from the magnet. (1)
(c) Because the coil repels the approaching magnet, work must be done to push it in; this work appears as electrical energy in the coil. If the current helped the motion, energy would be created from nothing, violating conservation of energy. (1)

Answer: (a) Induced current opposes the change in flux; (b) near face north, current anticlockwise seen from the magnet; (c) the work done against the opposition supplies the electrical energy.

Example 3: Charge through a flipped coil · 3 marks · Short answerHarder board level

A coil of 100 turns, area 20 cm² and resistance 5 Ω lies with its plane perpendicular to a field of 0.5 T. It is turned over through 180°. Find the charge that flows through it. Does it depend on how quickly it is turned?

Try it first, then show the solutionSolution

Flux per turn goes from $+ B A$ to $- B A$ : $ΔΦ = 2 B A = 2 \times 0.5 \times 2 \times 1 0^{- 3} = 2 \times 1 0^{- 3}$ Wb. (1)
$q = \frac{N ΔΦ}{R} = \frac{100 \times 2 \times 1 0 ^{- 3}}{5} = 0.04$ C. (1)
No: $q = \int I d t = \frac{N}{R} \int \frac{d Φ}{d t} d t = \frac{N ΔΦ}{R}$ , whatever the time taken; turning faster gives a larger current for a shorter time. (1)

Answer: 0.04 C, independent of the time taken.

Example 4: Time-varying field · 2 marks · Very short answerHarder board level

The field through a 50-turn coil of area 0.05 m² (plane perpendicular to the field) varies as B = 0.2 + 0.05t² tesla, with t in seconds. The coil has resistance 10 Ω. Find the emf and current at t = 2 s.

Try it first, then show the solutionSolution

$\frac{d B}{d t} = 0.1 t = 0.2$ T/s at t = 2 s, so $ε = N A \frac{d B}{d t} = 50 \times 0.05 \times 0.2 = 0.5$ V. (1)
$I = \frac{0.5}{10} = 0.05$ A. (1)

Answer: 0.5 V and 0.05 A.

Example 5: Rotating coil: peak and average emf · 5 marks · Long answerHOTS / competency

A coil of 100 turns and area 0.01 m² rotates at 50 revolutions per second about an axis perpendicular to a uniform field of 0.2 T.
(a) Write the flux through it as a function of time and derive the emf. (2)
(b) Find the peak emf and the positions where it is largest and zero. (2)
(c) Find the average emf over the quarter turn from the position of largest flux to zero flux, and compare it with the peak value. (1)

Try it first, then show the solutionSolution

(a) $ω = 2 π \times 50 = 100 π$ rad/s. Taking the normal along B at t = 0: $Φ = B A cos ω t$ per turn. (1)
$ε = - N \frac{d Φ}{d t} = N B A ω sin ω t$ . (1)
(b) $ε_{0} = N B A ω = 100 \times 0.2 \times 0.01 \times 100 π \approx 62.8$ V. (1)
Largest when the plane of the coil is parallel to B (flux zero but changing fastest); zero when the plane is perpendicular to B (flux largest, momentarily constant). (1)
(c) In a quarter turn ( $\frac{1}{200}$ s) the flux linkage falls by $N B A = 0.2$ Wb, so $ε_{a v g} = \frac{0.2}{0.005} = 40$ V, which is $\frac{2}{π} \approx 0.64$ of the peak. (1)

Answer: (a) ε = NBAω sin ωt; (b) about 62.8 V, largest with the plane parallel to B, zero with it perpendicular; (c) 40 V, about 0.64 of the peak.

Example 6: Magnet dropped through a ring · 4 marks · Case-basedHarder board level

Read the passage and answer.

A bar magnet is dropped, north pole first, along the axis of a horizontal metal ring. In a second trial the ring has a narrow cut in it.
(a) Describe the direction of the induced current (seen from above) as the magnet approaches the closed ring. (1)
(b) How does the magnet's acceleration compare with g as it approaches and as it leaves the closed ring? (1)
(c) What changes when the ring is cut? (1)
(d) Where does the 'lost' kinetic energy go in the first trial? (1)

Try it first, then show the solutionSolution

(a) Downward flux through the ring increases; the induced current makes the top face a north pole, i.e. anticlockwise as seen from above. (1)
(b) Both while approaching (repelled) and leaving (attracted back), the force on the magnet opposes its motion, so its acceleration is less than g. (1)
(c) An emf is still induced, but no current flows in the cut ring, so there is no opposing force: the magnet falls with acceleration g. (1)
(d) It becomes electrical energy and then heat in the ring. (1)

Answer: (a) anticlockwise from above; (b) less than g on both sides; (c) emf but no current, so acceleration g; (d) heat in the ring.

Where students lose marks in Faraday's and Lenz's Laws

The slips examiners see most often, and what to write instead.

Mistake: Using the angle with the plane of the coil in Φ = BA cos θ.
Questions often state the angle with the plane.
Wrong working
Plane at 30° to B: Φ = BA cos 30°
Right working
Normal at 60° to B: Φ = BA cos 60°
Do this: Always measure θ from the normal.
Usually costs: 1 mark.
Mistake: Taking ΔΦ = 0 when a coil is flipped.
The size of the flux is the same before and after.
Wrong working
Flux was BA and is still BA, so ΔΦ = 0.
Right working
Flux goes from +BA to −BA: ΔΦ = 2BA.
Do this: Keep the sign of the flux relative to a fixed normal.
Usually costs: Most of the numerical.
Mistake: Forgetting the number of turns N.
Using flux per turn as the total.
Do this: Write ε = N dΦ/dt with N shown every time.
Usually costs: The answer mark.
Mistake: Explaining Lenz's law without mentioning energy.
Only the direction rule is stated.
Do this: Add: work done against the opposing force becomes electrical energy, so Lenz's law follows from energy conservation.
Usually costs: 1 mark in a 3-mark answer.
Mistake: Saying no emf is induced in an open coil.
Confusing emf with current.
Do this: An emf is induced whenever flux changes; a current needs a closed circuit.
Usually costs: 1 mark in assertion-reason items.

Exam tips

Define flux with units before using it; in derivations write Φ = BA cos θ explicitly.
For direction questions, write three short lines: flux increasing or decreasing; induced field direction; current direction by the right-hand rule.
Recent board papers set questions on Faraday's and Lenz's laws in 2023 (55/5/1 Q4), 2025 (55/1/1 Q4 and Q13) and 2026 (55/5/1 Q13 and Q18).
For charge questions, use q = NΔΦ/R directly; you do not need the time.
Eddy currents are not in the 2026-27 syllabus; explain ring and magnet effects with Lenz's law only.

Practice questions on Faraday's and Lenz's Laws

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The SI unit of magnetic flux is
(a) tesla (b) weber (c) henry (d) T/m²
Hint
Φ = BA.
Answer
(b) weber (1 Wb = 1 T m²).
Q2 · 2 marks · Very short answerBoard level
State Faraday's law of electromagnetic induction. What does the negative sign in ε = −dΦ/dt indicate?
Hint
Think of Lenz.
Answer
The induced emf in a circuit equals the rate of change of magnetic flux linked with it. The negative sign shows that the induced emf opposes the change in flux (Lenz's law).
Q3 · 2 marks · Very short answerHarder board level
The field perpendicular to a 50-turn coil of area 40 cm² falls uniformly from 0.8 T to zero in 0.4 s. Find the induced emf.
Hint
Convert the area to m².
Answer
$ε = 50 \times \frac{4 \times 1 0 ^{- 3} \times 0.8}{0.4} = 0.4$ V.
Q4 · 1 mark · Very short answerHarder board level
The flux through a single loop varies as Φ = (3t² + 2t) mWb. Find the emf at t = 2 s.
Hint
Differentiate.
Answer
$\frac{d Φ}{d t} = 6 t + 2 = 14$ mWb/s at t = 2 s, so the emf is 14 mV.
Q5 · 2 marks · Very short answerHarder board level
A north pole is pulled away from the face of a closed coil. What polarity does that face of the coil acquire, and what is the direction of the current as seen from the magnet?
Hint
The coil tries to keep the magnet.
Answer
The flux towards the coil decreases, so the coil tries to keep it: the near face becomes a south pole, attracting the magnet. The current is clockwise as seen from the magnet.
Q6 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): When a magnet is moved towards a closed coil, the coil exerts a repulsive force on the magnet.
Reason (R): The induced current in the coil opposes the change in flux that produces it.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Lenz's law.
Answer
(a).
Q7 · 3 marks · Short answerHOTS / competency
Two identical coils, one of copper and one of a high-resistance alloy, are moved in exactly the same way through the same magnetic field. Compare (a) the induced emfs, (b) the induced currents and (c) the effort needed to move them.
Hint
The emf depends on the flux change only.
Answer
(a) Same emf, since the flux changes identically. (b) The copper coil has the larger current (I = ε/R). (c) Moving the copper coil needs more effort: the larger current gives a larger opposing force, and more electrical energy (I²R heat) must be supplied.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define magnetic flux.Show answer
Φ = B · A = BA cos θ, the number of field lines through an area; unit weber.
2.State Faraday's law.Show answer
Induced emf equals the rate of change of magnetic flux linkage: ε = −N dΦ/dt.
3.State Lenz's law.Show answer
The induced current opposes the change in flux that produces it.
4.Which conservation law underlies Lenz's law?Show answer
Conservation of energy.
5.Charge through a coil when flux changes by ΔΦ?Show answer
q = NΔΦ/R.
6.Peak emf of a rotating coil?Show answer
NBAω.

Faraday's and Lenz's Laws: frequently asked questions

What is the difference between Faraday's law and Lenz's law?

Faraday's law gives the size of the induced emf (the rate of change of flux linkage). Lenz's law gives its direction (opposing the change). Together they are written ε = −N dΦ/dt.

Why does Lenz's law follow from conservation of energy?

The induced current always opposes the motion or change causing it, so work must be done against it. That work is the source of the electrical energy. If the current aided the change, energy would appear from nothing.

Is an emf induced if the circuit is open?

Yes. An emf is induced whenever the flux changes. Without a closed circuit, however, no current flows and there is no opposing force.

Why does the total charge not depend on the speed of the change?

A faster change gives a bigger emf and current, but for a shorter time. The product (charge) equals NΔΦ/R either way.

Are eddy currents in the CBSE 2026-27 syllabus?

No. Eddy currents are listed as not to be assessed. Faraday's laws, Lenz's law and self and mutual induction are in the syllabus.