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Electromagnetic Induction

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Chapter Notes

Overview
This chapter explores the phenomenon of electromagnetic induction, where a changing magnetic field induces an electric current in a conductor. It builds on the foundational relationship between electricity and magnetism, established by Oersted and Ampere, and delves into the experiments of Faraday and Henry that conclusively demonstrated this effect. The single biggest idea a student must take away is that a time-varying magnetic flux through a circuit induces an electromotive force (emf), and the induced current always opposes the change that produced it (Lenz's law), ensuring energy conservation.
Key Concepts (10)
Electromagnetic Induction
The phenomenon in which electric current is generated by varying magnetic fields.
e.g. When a bar magnet is moved towards a coil connected to a galvanometer, a deflection is observed, indicating current generation.
Magnetic Flux
Magnetic flux through a plane of area A placed in a uniform magnetic field B is defined as Φ_B = B·A = BA cos θ, where θ is the angle between B and A.
e.g. A loop of area 0.1 m² placed perpendicular to a 0.5 T field has flux = 0.05 Wb.
Faraday's Law of Induction
The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. Mathematically, ε = -dΦ_B/dt for a single turn, and ε = -N dΦ_B/dt for a coil of N turns.
e.g. If flux through a 10-turn coil changes from 0.2 Wb to 0 in 0.1 s, induced emf = 10 × (0.2/0.1) = 20 V.
Lenz's Law
The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it.
e.g. When the north pole of a magnet approaches a coil, the induced current flows in a direction that creates a north pole facing the magnet, repelling it.
Motional Electromotive Force (emf)
The emf induced across a conductor of length l moving with velocity v perpendicular to a uniform magnetic field B is given by ε = Blv.
e.g. A 1 m rod moving at 5 m/s perpendicular to a 0.2 T field generates 1 V across its ends.
Mutual Inductance
The ratio of flux linkage in one coil to the current in another coil that produces it. For two coils, N₁Φ₁ = M I₂, where M is mutual inductance.
e.g. Two concentric coils with r₁ << r₂ have mutual inductance M = μ₀πr₁²/(2r₂).
Self-Inductance
The ratio of flux linkage to current in the same coil. NΦ_B = L I, where L is self-inductance.
e.g. A long solenoid with n turns per unit length, area A, and length l has L = μ₀n²Al.
Back emf
The self-induced emf that opposes any change in current in a circuit, given by ε = -L dI/dt.
e.g. When a switch is closed in an RL circuit, the back emf opposes the sudden rise in current.
AC Generator
A device that converts mechanical energy into electrical energy by rotating a coil in a magnetic field, producing an alternating emf.
e.g. A 100-turn coil of area 0.1 m² rotating at 0.5 Hz in a 0.01 T field generates a maximum voltage of 0.314 V.
Eddy Currents
Induced currents that flow in closed loops within a bulk conductor when it is subjected to a changing magnetic field. (Mentioned implicitly in the context of energy dissipation.)
e.g. When a metal plate swings through a magnetic field, eddy currents oppose its motion, causing damping.
Formulas & Equations (13)
Φ_B = B·A = BA cos θ
Reads as: Magnetic flux equals B dot A equals B A cosine theta
Variables: Φ_B = magnetic flux (Wb), B = magnetic field (T), A = area vector (m²), θ = angle between B and A
Use when: Calculating flux through a surface in a uniform field
⚠ Forgetting that θ is the angle between B and the area vector (normal to surface), not the surface itself.
ε = -dΦ_B/dt
Reads as: Epsilon equals negative d phi B by d t
Variables: ε = induced emf (V), Φ_B = magnetic flux (Wb), t = time (s)
Use when: Finding induced emf from changing flux for a single loop
⚠ Omitting the negative sign, which indicates Lenz's law direction.
ε = -N dΦ_B/dt
Reads as: Epsilon equals negative N d phi B by d t
Variables: ε = induced emf (V), N = number of turns, Φ_B = flux through one turn (Wb), t = time (s)
Use when: Finding induced emf in a coil of N turns
⚠ Using total flux change instead of per-turn flux change.
ε = Blv
Reads as: Epsilon equals B l v
Variables: ε = motional emf (V), B = magnetic field (T), l = length of conductor (m), v = velocity perpendicular to B (m/s)
Use when: Calculating emf induced in a rod moving perpendicular to a uniform magnetic field
⚠ Using the component of velocity parallel to the field instead of perpendicular.
N₁Φ₁ = M I₂
Reads as: N one phi one equals M I two
Variables: N₁ = number of turns in coil 1, Φ₁ = flux through coil 1 (Wb), M = mutual inductance (H), I₂ = current in coil 2 (A)
Use when: Relating flux in one coil to current in another
⚠ Confusing which coil's flux and current correspond to which subscript.
ε₁ = -M dI₂/dt
Reads as: Epsilon one equals minus M d I two by d t
Variables: ε₁ = induced emf in coil 1 (V), M = mutual inductance (H), I₂ = current in coil 2 (A), t = time (s)
Use when: Finding induced emf due to changing current in a neighbouring coil
⚠ Forgetting the negative sign, which indicates opposition to change.
NΦ_B = L I
Reads as: N phi B equals L I
Variables: N = number of turns, Φ_B = flux through one turn (Wb), L = self-inductance (H), I = current (A)
Use when: Relating flux linkage to current in a single coil
⚠ Using total flux instead of flux per turn.
ε = -L dI/dt
Reads as: Epsilon equals minus L d I by d t
Variables: ε = self-induced emf (V), L = self-inductance (H), I = current (A), t = time (s)
Use when: Finding back emf due to changing current in the same coil
⚠ Omitting the negative sign, which indicates opposition to current change.
L = μ₀ n² A l
Reads as: L equals mu zero n squared A l
Variables: L = self-inductance (H), μ₀ = permeability of free space (4π×10⁻⁷ T m/A), n = turns per unit length (m⁻¹), A = cross-sectional area (m²), l = length (m)
Use when: Calculating self-inductance of a long solenoid
⚠ Using total turns N instead of turns per unit length n.
U = (1/2) L I²
Reads as: U equals half L I squared
Variables: U = magnetic energy stored (J), L = self-inductance (H), I = current (A)
Use when: Finding energy stored in an inductor
⚠ Forgetting the factor of 1/2.
u_B = B²/(2μ₀)
Reads as: u B equals B squared over two mu zero
Variables: u_B = magnetic energy density (J/m³), B = magnetic field (T), μ₀ = permeability of free space
Use when: Calculating energy per unit volume in a magnetic field
⚠ Confusing with electric energy density u_E = (1/2) ε₀ E².
ε = NBAω sin ωt
Reads as: Epsilon equals N B A omega sine omega t
Variables: ε = instantaneous emf (V), N = number of turns, B = magnetic field (T), A = area (m²), ω = angular speed (rad/s), t = time (s)
Use when: Finding emf from a rotating coil in a uniform field (AC generator)
⚠ Forgetting that ω = 2πν, where ν is frequency in Hz.
ε₀ = NBAω
Reads as: Epsilon zero equals N B A omega
Variables: ε₀ = peak emf (V), N = number of turns, B = magnetic field (T), A = area (m²), ω = angular speed (rad/s)
Use when: Finding maximum emf from an AC generator
⚠ Using ω in Hz instead of rad/s.
How It Works
Faraday's Experiment 6.1: Magnet and Coil
  1. A coil C₁ is connected to a galvanometer G.
  2. The north pole of a bar magnet is pushed towards the coil.
  3. The galvanometer deflects, indicating induced current.
  4. The deflection lasts only while the magnet is in motion.
  5. When the magnet is pulled away, deflection occurs in the opposite direction.
  6. Faster motion produces larger deflection.
  7. The same effects are observed if the coil is moved instead of the magnet.
💡 It is relative motion between magnet and coil that induces current, not just motion of one.
Faraday's Experiment 6.2: Two Coils
  1. Coil C₂ (connected to a battery) is moved towards coil C₁ (connected to a galvanometer).
  2. The galvanometer shows a deflection.
  3. When C₂ is moved away, deflection occurs in the opposite direction.
  4. Deflection lasts only while C₂ is in motion.
  5. Same effects if C₁ is moved instead of C₂.
💡 Relative motion between two coils induces current; the steady current in C₂ produces a steady magnetic field.
Faraday's Experiment 6.3: Stationary Coils with Switching
  1. Two coils C₁ and C₂ are held stationary.
  2. C₁ is connected to a galvanometer, C₂ to a battery via a tapping key K.
  3. When key K is pressed, the galvanometer shows a momentary deflection.
  4. The pointer returns to zero immediately even if key is held pressed.
  5. When key is released, a momentary deflection occurs in the opposite direction.
  6. Deflection increases dramatically when an iron rod is inserted along the coil axes.
💡 Changing current in C₂ (not steady current) induces emf in C₁; the iron rod increases magnetic flux.
Lenz's Law: Determining Direction of Induced Current
  1. Identify whether magnetic flux through the loop is increasing or decreasing.
  2. The induced current will flow in a direction that opposes this change.
  3. If flux increases, induced current creates a magnetic field opposing the increase.
  4. If flux decreases, induced current creates a magnetic field opposing the decrease.
  5. Use the right-hand rule to find current direction from the induced magnetic field.
💡 Lenz's law ensures conservation of energy; if it were violated, perpetual motion would be possible.
Motional emf Generation
  1. A conductor of length l moves with velocity v perpendicular to a uniform magnetic field B.
  2. Free charges in the conductor experience Lorentz force F = q(v × B), directed along the conductor.
  3. This force moves charges, creating a potential difference across the ends.
  4. The induced emf is ε = Blv.
  5. Alternatively, consider the change in area of a closed loop: flux changes as the conductor moves.
💡 Motional emf can be derived from both Lorentz force and Faraday's law; both give the same result.
AC Generator Operation
  1. A coil of N turns and area A is rotated with constant angular speed ω in a uniform magnetic field B.
  2. The angle between the area vector A and B changes as θ = ωt.
  3. Magnetic flux through the coil is Φ_B = BA cos ωt.
  4. From Faraday's law, induced emf is ε = -N dΦ_B/dt = NBAω sin ωt.
  5. The emf varies sinusoidally with time, producing alternating current.
  6. Slip rings and brushes connect the rotating coil to an external circuit.
💡 The emf is maximum when the plane of the coil is parallel to the field (θ = 90° or 270°), where flux change is greatest.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Flux Change with Flux
✗ Induced emf is proportional to the magnetic flux through the coil.
→ Faraday's law states that induced emf is proportional to the rate of change of flux, not the flux itself. A large constant flux induces no emf.
✓ Induced emf is proportional to the time rate of change of magnetic flux through the coil.
⚠ Misapplying Lenz's Law Direction
✗ The induced current opposes the magnetic field that produces it.
→ Lenz's law says the induced current opposes the change in magnetic flux, not the magnetic field itself. For example, if flux increases, the induced field opposes the increase, not the original field.
✓ The induced current opposes the change in magnetic flux that produces it.
⚠ Forgetting the Negative Sign in Faraday's Law
✗ ε = dΦ_B/dt
→ The negative sign is crucial; it represents Lenz's law and indicates the direction of induced emf. Omitting it loses conceptual marks.
✓ ε = -dΦ_B/dt
⚠ Motional emf with Wrong Velocity Component
✗ ε = Blv, where v is the speed of the rod in any direction.
→ The formula ε = Blv applies only when the velocity is perpendicular to both the magnetic field and the conductor. If v has a component parallel to B, that component does not contribute to emf.
✓ ε = Blv, where v is the component of velocity perpendicular to both B and the conductor.
⚠ Confusing Self and Mutual Inductance
✗ Self-inductance is the same as mutual inductance for a single coil.
→ Self-inductance relates flux and current in the same coil, while mutual inductance relates flux in one coil to current in another. They are different physical quantities.
✓ Self-inductance L = NΦ_B/I for a single coil; mutual inductance M = N₁Φ₁/I₂ between two coils.
Compare & Contrast
Cause of induced emf
Motional emf
Transformer emf
Motional emf is due to motion of a conductor in a static magnetic field; transformer emf is due to a time-varying magnetic field acting on a stationary conductor.
Coils involved
Self-Inductance
Mutual Inductance
Self-inductance involves a single coil where changing current induces emf in the same coil; mutual inductance involves two coils where changing current in one induces emf in the other.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 12 marks
Most asked: Faraday's law and Lenz's law (direction of induced current) · Motional emf (ε = Blv) problems · AC generator principle and emf equation · Self-inductance of a solenoid · Energy stored in an inductor
Don't skip: Numerical problems on motional emf with rotating rods or moving conductors; Derivation of induced emf in a rotating coil (AC generator); Application of Lenz's law to determine current direction in various scenarios
One-Line Revision (read aloud before the exam)
  1. Electromagnetic induction is the generation of electric current by a changing magnetic field.
  2. Magnetic flux Φ_B = BA cos θ, measured in weber (Wb).
  3. Faraday's law: Induced emf ε = -N dΦ_B/dt.
  4. Lenz's law: Induced current opposes the change in flux that produces it.
  5. Motional emf across a conductor moving in a magnetic field: ε = Blv.
  6. Mutual inductance M relates flux in one coil to current in another: N₁Φ₁ = M I₂.
  7. Self-inductance L relates flux and current in the same coil: NΦ_B = L I.
  8. Self-induced emf (back emf) is ε = -L dI/dt.
  9. Self-inductance of a long solenoid: L = μ₀ n² A l.
  10. Energy stored in an inductor: U = (1/2) L I².
  11. Magnetic energy density: u_B = B²/(2μ₀).
  12. An AC generator produces sinusoidal emf: ε = NBAω sin ωt.
  13. The peak emf of an AC generator is ε₀ = NBAω.
  14. Lenz's law ensures conservation of energy in electromagnetic induction.
  15. Eddy currents are induced currents in bulk conductors that oppose motion.

MCQs — 10 Questions with Answers

Q1. A coil of area 100 cm² is placed in a uniform magnetic field of 0.5 T perpendicular to the coil plane. What is the magnetic flux through the coil?

  • A. 5 mWb ✓
  • B. 50 mWb
  • C. 500 mWb
  • D. 5000 mWb

Answer: A — ΦB = BA cosθ = 0.5 × 0.01 × cos(0°) = 0.005 Wb = 5 mWb (area must be converted to m²).

Q2. In Faraday's Experiment 6.1, when the magnet is held stationary inside the coil, the galvanometer shows no deflection. This is because:

  • A. The magnetic field inside the coil is zero
  • B. The rate of change of magnetic flux is zero (dΦB/dt = 0) ✓
  • C. The magnetic flux through the coil is zero
  • D. The coil is not made of a conducting material

Answer: B — Faraday's law states that induced emf depends on dΦB/dt; when the magnet is stationary, flux is constant, so its rate of change is zero, producing no induced emf or current.

Q3. A magnetic flux through a coil increases from 0 to 10 Wb in 2 seconds. What is the magnitude of the induced emf?

  • A. 20 V
  • B. 5 V ✓
  • C. 10 V
  • D. 0.2 V

Answer: B — |ε| = |dΦB/dt| = |ΔΦB/Δt| = |10 − 0|/2 = 5 V.

Q4. In Experiment 6.3, when the key K is pressed, the galvanometer shows a momentary deflection. When the key is held continuously pressed, there is no deflection. Which statement correctly explains this observation?

  • A. The current in coil C2 increases and then becomes steady; induced emf appears only during the increasing phase (dΦB/dt ≠ 0) ✓
  • B. The galvanometer is faulty and stops responding
  • C. The magnetic field produced by C2 is too weak to induce current
  • D. The key prevents the flow of current in coil C1

Answer: A — Initially, current in C2 rises from zero to maximum, changing the flux through C1 and inducing emf; once steady, flux becomes constant and dΦB/dt = 0, so no induced emf.

Q5. In Experiment 6.2, the deflection in the galvanometer is larger when coil C2 is moved towards C1 faster. Which of the following correctly explains this?

  • A. Faster motion increases the magnetic field produced by C2
  • B. Faster motion increases the rate of change of magnetic flux (|dΦB/dt|), thereby increasing the magnitude of induced emf ✓
  • C. Faster motion decreases the resistance of the coil
  • D. Faster motion increases the number of turns in the coil

Answer: B — By Faraday's law, |ε| = |dΦB/dt|; faster motion causes flux to change more rapidly, increasing |dΦB/dt| and thus the induced emf and current.

Q6. Which of the following statements is NOT correct about magnetic flux?

  • A. Magnetic flux is a scalar quantity
  • B. Magnetic flux through a closed surface is always zero
  • C. The SI unit of magnetic flux is weber (Wb)
  • D. Magnetic flux is directly proportional to the sine of the angle between B and the area vector ✓

Answer: D — Magnetic flux ΦB = BA cosθ is proportional to cosine, not sine, of the angle θ between B and the area vector.

Q7. A coil is placed in a magnetic field such that the magnetic flux through it is increasing with time. According to Lenz's law, the induced current in the coil will produce a magnetic field that:

  • A. Is in the same direction as the external magnetic field
  • B. Opposes the increase in the external magnetic flux ✓
  • C. Is perpendicular to the external magnetic field
  • D. Is always equal in magnitude to the external magnetic field

Answer: B — Lenz's law states that the induced current creates a magnetic field opposing the change in flux, so if flux is increasing, the induced field opposes this increase.

Q8. In Faraday's experiments, an iron rod is inserted along the axis of two coils. The induced emf in the nearby coil increases dramatically. This is primarily because:

  • A. Iron is a good conductor and reduces resistance
  • B. Iron has very high permeability, greatly increasing the magnetic field strength and flux change ✓
  • C. Iron generates its own magnetic field independent of the coil
  • D. Iron increases the number of free electrons in the coil

Answer: B — Iron has high permeability (μ >> μ₀), which significantly amplifies the magnetic field produced by the coil carrying current, leading to a much larger change in flux and induced emf.

Q9. Two identical coils are placed near each other. In coil 1, the current increases from 0 to 2 A in 0.1 s. In coil 2, which is stationary, the induced emf is observed. If in another scenario the same current change occurs in 0.2 s instead, how will the induced emf in coil 2 change? (Assertion: The induced emf will remain the same. Reason: The induced emf depends only on the change in current, not on the rate of change.)

  • A. Both assertion and reason are correct, and reason explains the assertion
  • B. Both assertion and reason are correct, but reason does not explain the assertion
  • C. Assertion is correct, but reason is incorrect
  • D. Assertion is incorrect; the induced emf will be halved because it depends on dΦB/dt, which decreases when the time interval increases ✓

Answer: D — Induced emf depends on dΦB/dt (rate of change of flux); if the same flux change occurs over a longer time (0.2 s instead of 0.1 s), the rate decreases to half, so induced emf is halved.

Q10. A rectangular coil of dimensions 20 cm × 10 cm is rotated in a uniform magnetic field of 2 T such that the angle θ between the magnetic field and the normal to the coil changes from 0° to 90° in 0.5 seconds. Calculate the average induced emf. (Initial flux: ΦB,i = BA cos(0°); Final flux: ΦB,f = BA cos(90°))

  • A. 0.8 V ✓
  • B. 0.4 V
  • C. 1.6 V
  • D. 0.08 V

Answer: A — A = 0.2 × 0.1 = 0.02 m². ΦB,i = 2 × 0.02 × 1 = 0.04 Wb. ΦB,f = 2 × 0.02 × 0 = 0 Wb. |ε_avg| = |ΔΦB/Δt| = |0 − 0.04|/0.5 = 0.8 V.

Flashcards

What is magnetic flux and what is its SI unit?

Magnetic flux ΦB = B·A = BA cosθ is the scalar product of magnetic field and area vector, with SI unit weber (Wb) or T·m².

State Faraday's law of electromagnetic induction.

The induced emf in a coil is equal to the negative rate of change of magnetic flux through it: ε = −dΦB/dt.

What is the key observation from Experiment 6.1?

Moving a bar magnet towards or away from a coil induces current only while the magnet is in motion; the direction of current reverses when the magnet's direction of motion reverses.

Why does Experiment 6.3 show deflection even without motion?

When the key is pressed, the current in coil C2 changes from zero to maximum, causing the magnetic flux through C1 to change rapidly, inducing emf momentarily.

What is the role of an iron rod in Faraday's experiments?

Inserting an iron rod along the axis of the coils dramatically increases the induced emf because iron has high permeability and greatly enhances the magnetic field.

What does Lenz's law state about induced currents?

Lenz's law states that the direction of the induced current is always such that it opposes the change in magnetic flux that produces it (indicated by the negative sign in Faraday's law).

Is absolute motion of a magnet necessary to induce current, or is relative motion sufficient?

Only relative motion between the magnet and coil is necessary; if the magnet is held fixed and the coil moves, the same induced current is observed.

What is the difference between magnetic flux and induced emf?

Magnetic flux ΦB measures the total magnetic field passing through a surface, while induced emf is generated by the rate of change of this flux with time.

In Experiment 6.3, why does the galvanometer show no deflection when the key is held continuously pressed?

Once the current in coil C2 becomes steady (constant), the magnetic flux through C1 becomes constant, so dΦB/dt = 0, producing zero induced emf and no deflection.

How do the results differ when the North-pole versus South-pole of the magnet is moved towards the coil?

Both poles produce the same magnitude of induced emf, but the direction of the induced current reverses because the direction of the magnetic field (and hence the direction of flux change) is opposite.

Important Board Questions

Define magnetic flux. A coil of 50 turns has an area of 200 cm² and is placed perpendicular to a magnetic field of 0.4 T. Calculate the total magnetic flux through the coil. [2 marks]

Magnetic flux is ΦB = B·A = BA cosθ (scalar); total flux for N-turn coil is N × BA. Convert area to m² and use θ = 0° since perpendicular.

Explain Faraday's law of electromagnetic induction with reference to Experiment 6.3. Why is there a momentary deflection in the galvanometer when the key is pressed and released, but no deflection when the key is held continuously pressed? [5 marks]

State Faraday's law (ε = −dΦB/dt). Explain: key press → current in C2 changes → flux through C1 changes → dΦB/dt ≠ 0 → induced emf. Held pressed → current becomes steady → dΦB/dt = 0 → no emf. Key release → current drops → flux decreases → opposite deflection.

Derive the expression for magnetic flux through a non-uniform magnetic field over a curved surface. How does Faraday's law (in its differential form) relate this concept to the induced emf? Explain with the help of Experiment 6.1 why the induced current is larger when the magnet is moved faster towards the coil. [6 marks]

For non-uniform field: ΦB = ∫ B·dA (integral form). Faraday's law: ε = −dΦB/dt. Faster motion → magnetic field and flux change more rapidly → larger |dΦB/dt| → larger |ε| → larger induced current. Justify each step with Experiment 6.1 observations.

Assertion-Reason Questions for Electromagnetic Induction

Q1. Assertion (A):

In Experiment 6.1, the galvanometer shows a deflection when the magnet is moved towards the coil, but no deflection when the magnet is held stationary.

Reason (R):

A stationary magnet produces a constant magnetic field and constant flux through the coil, resulting in zero induced emf according to Faraday's law.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true, and the reason correctly explains the assertion. Faraday's law shows that induced emf depends on dΦ/dt; a constant flux (dΦ/dt = 0) produces zero emf and hence no galvanometer deflection. This is the direct experimental observation.

Q2. Assertion (A):

The magnitude of induced emf in a coil is greater when a magnet is pushed towards it faster.

Reason (R):

Faster motion of the magnet increases the magnetic field strength at the coil.

Answer: C — A is true but R is false — The assertion is true—faster motion increases dΦ/dt, increasing induced emf. However, the reason is false. Faster motion does not increase B itself; rather, it increases the rate at which B changes at the coil's location, thus increasing dΦ/dt. The reason confuses 'field strength' with 'rate of change of field'.

Q3. Assertion (A):

In Experiment 6.3, when the tapping key K is pressed and held, the galvanometer shows deflection only at the instant the key is pressed and again when it is released.

Reason (R):

The induced emf depends on the rate of change of current in the primary coil, not the steady-state current itself.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both are true. When the key is pressed, current in C2 rises from 0 (dI/dt ≠ 0), inducing emf in C1. During steady conduction (dI/dt = 0), no emf is induced. Upon release, current drops (dI/dt ≠ 0 again), inducing emf in opposite direction. The reason correctly explains the assertion using the concept of rate of change.

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Case-Based Questions (CBQ) for Electromagnetic Induction

Case Study:

A hydroelectric power station uses the principle of electromagnetic induction. Water flowing through a dam rotates a turbine connected to a coil of 500 turns. The coil rotates in a uniform magnetic field of 2 T. The coil has an area of 0.4 m² and completes 50 rotations per second. At a particular instant during rotation, the plane of the coil is perpendicular to the magnetic field. Engineers need to calculate the induced emf to design the power transmission system.

Q1. At the instant when the coil plane is perpendicular to the magnetic field, what is the magnetic flux through the coil?

  • A. 0.4 Wb
  • B. 400 Wb
  • C. 0 Wb ✓
  • D. 0.8 Wb

Answer: C

Q2. What is the maximum magnetic flux through the coil during one complete rotation?

  • A. 0.4 Wb
  • B. 0.8 Wb ✓
  • C. 1.6 Wb
  • D. 2 Wb

Answer: B

Q3. If the magnetic field is doubled to 4 T while keeping the rotation rate constant, how does the maximum induced emf change?

  • A. It doubles ✓
  • B. It remains the same
  • C. It halves
  • D. It quadruples

Answer: A

Q4. Which factor among the following would NOT affect the magnitude of induced emf in this generator?

  • A. The number of turns in the coil
  • B. The area of the coil
  • C. The colour of the magnetic field source ✓
  • D. The rotation rate (frequency)

Answer: C

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Previous Year Questions (PYQ) for Electromagnetic Induction

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 55/1/1 · 2 marks

A wire of length L is bent round into (i) a square coil having N turns and (ii) a circular coil having N turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.

CBSE 2026 · Set 55/1/1 · 5 marks

State Faraday's law of electromagnetic induction. Derive an expression for the self-inductance of an air-filled long solenoid of length l and cross-sectional area A having N turns. A conducting rod of length 50 cm, with one end pivoted, is rotated with angular speed of 60 rpm in a uniform magnetic field of 4.0 mT directed perpendicular to the plane of rotation of rod. Find the emf induced in the rod.

CBSE 2026 · Set 55/1/2 · 5 marks

State Faraday's law of electromagnetic induction.

CBSE 2026 · Set 55/1/3 · 5 marks

(a) State Faraday's law of electromagnetic induction. (b) Derive an expression for the self-inductance of an air-filled long solenoid of length l and cross-sectional area A having N turns. (c) A conducting rod of length 50 cm, with one end pivoted, is rotated with angular speed of 60 rpm in a uniform magnetic field of 4.0 mT directed perpendicular to the plane of rotation of rod. Find the emf induced in the rod.

CBSE 2026 · Set 55/2/1 · 1 mark

The magnetic flux Φ (in Wb) linked with a coil is related to time t (in s) as Φ = 5 At² + Bt – 2C. The SI units of A and B are respectively

  • Wb s⁻², Wb s⁻¹
  • Wb s⁻¹, Wb
  • Wb s⁻², Wb s⁻¹
  • Wb s⁻¹, Wb s⁻²

CBSE 2026 · Set 55/2/1 · 1 mark

A magnet held vertically, with its north pole down is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

  • the induced current will flow in the anticlockwise direction.
  • the induced current will flow in the clockwise direction.
  • no induced current will flow in the solenoid.
  • the magnet will fall with a constant velocity.

CBSE 2026 · Set 55/2/1 · 3 marks

(i) Define mutual inductance of a pair of coils. Write its SI unit. (ii) A long solenoid of radius R and length L has n turns per unit length. A circular loop of radius r(<R) is placed inside at the centre of the solenoid such that its axis coincides with the axis of the solenoid. Obtain the mutual inductance of the solenoid and the loop.

CBSE 2026 · Set 55/2/2 · 1 mark

Figure shows a magnet dropped through a small loop with a small cut. Which of the following statements is correct?

  • The speed of the falling magnet increases as it approaches the loop and starts decreasing as it crosses the loop.
  • Acceleration of magnet increases as it approaches the loop and starts decreasing as it crosses the loop.
  • Speed of magnet remains uniform as it moves through the loop.
  • Acceleration of the magnet remains uniform as it moves through the loop.

CBSE 2026 · Set 55/2/3 · 1 mark

The magnetic flux ϕ (in Wb) linked with a coil is related to time t (in s) as ϕ = 5 At² + Bt − 2C. The SI units of A and B are respectively

  • Wb s⁻², Wb s⁻¹
  • Wb s⁻¹, Wb
  • Wb s², Wb s
  • Wb s⁻¹, Wb s⁻²

CBSE 2026 · Set 55/2/3 · 1 mark

Two identical circular loops A and B of metal wire are arranged on a table close to each other. Loop A is connected to a battery and the current in it increases with time. In response, the loop B

  • remains stationary
  • is attracted by loop A
  • is repelled by loop A
  • rotates about its centre of mass which is fixed

CBSE 2026 · Set 55/2/3 · 3 marks

(i) Define mutual inductance of a pair of coils. Write its SI unit.

CBSE 2026 · Set 55(B)_Physics (Theory) · 1 mark

An electric current can be induced in a wire loop if –

  • the wire loop is wound around a bar magnet
  • a bar magnet is placed in contact with the wire loop
  • a bar magnet is placed near the wire loop
  • a bar magnet is moved rapidly towards the wire loop

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Class 12 Physics Formula Sheet

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Class 12 Physics Derivations

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