NEET Physics: Magnetism, Electromagnetic Induction and AC · Chapter 3

Electromagnetic Induction

What to remember

A changing magnetic flux through a closed loop induces an emf: e = −N dΦ/dt. The negative sign is Lenz's law: the induced current opposes the change that causes it.
Motional emf of a straight conductor moving across a field is e = Blv. A rod of length l rotating about one end with angular speed ω in a perpendicular field has e = ½Bωl².
Self-inductance L = μ₀n²Al for a long solenoid; energy stored in an inductor is U = ½LI²; the induced emf in a coil is e = −L dI/dt.
An AC generator turns mechanical energy into electrical energy; its emf is e = NBAω sin ωt, with peak value e₀ = NBAω.

Experiments of Faraday and Henry

Faraday and Henry showed that a current can be produced in a closed circuit without a battery. Three simple set-ups show the effect.

A coil is connected to a galvanometer. A bar magnet is pushed towards the coil: the galvanometer deflects. The magnet is held still: no deflection. The magnet is pulled away: the deflection is in the opposite direction.
The faster the magnet moves, the larger the deflection. Moving the coil towards a fixed magnet gives the same result. Only the relative motion matters.
Two coils are placed near each other. Coil 1 is joined to a battery through a switch; coil 2 is joined to a galvanometer. The galvanometer deflects only while the current in coil 1 is being switched on or off, or while it is changing (for example, by a rheostat). A steady current in coil 1 gives no deflection in coil 2.

Conclusion: a current is induced in the circuit whenever the magnetic flux linked with it changes. The effect is present only while the flux is changing.

Magnetic flux

Magnetic flux through a plane surface of area A in a uniform field B is

Φ = B A cos θ

where θ is the angle between the field and the normal (perpendicular) to the surface. The SI unit is the weber (Wb); 1 Wb = 1 T m². Its dimensions are [M L² T⁻² A⁻¹].

Orientation of the coil	θ	Flux
Plane of coil perpendicular to B (normal along B)	0°	BA (maximum)
Plane of coil at 60° to B	30°	BA cos 30°
Plane of coil parallel to B (normal at right angles to B)	90°	zero

The flux can change in three ways: by changing B, by changing the area A, or by changing the angle θ. Any of these induces an emf.

Faraday's laws and Lenz's law

Faraday's law: the magnitude of the emf induced in a circuit is equal to the rate of change of magnetic flux through it. For a coil of N turns,

e = −N dΦ/dt

If the flux changes uniformly by ΔΦ in time Δt, the average emf is N ΔΦ/Δt. The induced current is I = e/R, and the charge that flows in time Δt is q = NΔΦ/R. The charge depends only on the total change of flux and on R, not on how fast the change occurs.

Lenz's law: the direction of the induced emf (and current) is such that it opposes the change in flux that produced it. The negative sign in Faraday's law expresses this.

When the north pole of a magnet moves towards a loop, the face of the loop towards the magnet becomes a north pole and repels the magnet.
When the north pole moves away, that face becomes a south pole and attracts the magnet.
In both cases the person moving the magnet has to do work. This work appears as electrical energy (and then heat) in the loop.

Lenz's law and conservation of energy: if the induced current aided the change instead of opposing it, the magnet would be pulled faster and faster, and the loop would get more energy without any work done. This would break the law of conservation of energy. So Lenz's law is a consequence of energy conservation.

Motional emf and energy considerations

A straight rod of length l moves with speed v perpendicular to its length and perpendicular to a uniform field B. Free charges in the rod feel the magnetic force and pile up at the ends until the electric force balances it. The emf between the ends is

e = B l v

If the rod slides on a U-shaped conducting rail of total resistance R:

Induced current I = Blv/R.
Magnetic force on the rod F = BIl = B²l²v/R. It acts opposite to the velocity (Lenz's law).
To keep the speed constant, an external force equal to B²l²v/R must be applied.
Mechanical power supplied P = Fv = B²l²v²/R. Electrical power dissipated = I²R = B²l²v²/R. The two are equal, so energy is conserved.

If the rod moves at an angle, use only the component of velocity perpendicular to both B and the rod.

Rod rotating in a field: a rod of length l rotates with angular speed ω about one end, in a plane perpendicular to a uniform field B. The speed of a point at distance r is ωr, which changes along the rod. The average speed is ωl/2, so

e = ½ B ω l²

The centre of rotation and the free end are the two terminals. For a wheel with many spokes, each spoke has this emf, and the spokes are in parallel, so the total emf is the same as for one spoke.

Eddy currents

When a solid piece of metal is in a changing magnetic flux (or moves in a magnetic field), circulating currents are induced inside it. They are called eddy currents (Foucault currents). Their direction follows Lenz's law, so they oppose the motion or the change.

Aspect	Details
Cause	Changing flux through a bulk conductor
Effect	Heating of the conductor; opposing force on a moving conductor
Harmful effect	Energy loss as heat in transformer cores and motor armatures
Reduction	Laminated cores: thin sheets insulated from each other, with the planes of the sheets parallel to the field
Useful applications	Magnetic braking in trains; electromagnetic damping in galvanometers and balances; induction furnace for melting metals; electric power meters

Lamination breaks up the large loops of eddy current into small ones with much higher resistance, so the heat loss is small. Slots cut in a solid plate also reduce the eddy current. In an induction furnace, a rapidly changing current in a coil heats the metal inside by eddy currents alone.

Inductance

Mutual inductance. When the current in coil 1 changes, the flux linked with coil 2 changes. For coil 2, the flux linkage is N₂Φ₂ = M₂₁ I₁, and the emf induced is

e₂ = −M dI₁/dt

M is the mutual inductance. Its SI unit is the henry (H); 1 H = 1 V s A⁻¹ = 1 Wb A⁻¹. For two long coaxial solenoids, the inner one of radius r₁ with n₁ turns per metre and the outer one with n₂ turns per metre and length l, both of the same length l:

M = μ₀ n₁ n₂ π r₁² l

M depends on the geometry, the number of turns and the medium between the coils (it increases with a soft iron core). Also M₁₂ = M₂₁.

Self-inductance. A changing current in a coil changes the flux through the coil itself. The flux linkage is NΦ = LI, and the self-induced emf is

e = −L dI/dt

L is the self-inductance. Its unit is also the henry. The self-induced emf opposes the change of current, so inductance is the electrical equivalent of inertia: it opposes a change in current just as mass opposes a change in velocity.

For a long solenoid of length l, area of cross-section A and n turns per metre (N = nl):

L = μ₀ n² A l

With a core of relative permeability μr, L = μ₀ μr n² A l.

Energy stored in an inductor. To build up a current I in an inductor, the source does work against the self-induced emf. This work is stored in the magnetic field:

U = ½ L I²

Energy density of the magnetic field: u = B²/(2μ₀).

AC generator

Principle: electromagnetic induction. A coil rotating in a magnetic field has a changing flux, so an emf is induced.

Construction (in words): a rectangular coil of N turns and area A is rotated at a constant angular speed ω about an axis perpendicular to a uniform field B. The ends of the coil are connected to slip rings. Carbon brushes touching the slip rings carry the current to the outside circuit. Mechanical energy (from water, steam, wind, or a turbine) turns the coil.

At time t, with θ = ωt measured from the position where the coil is perpendicular to the field, the flux is Φ = NBA cos ωt. So

e = −dΦ/dt = NBAω sin ωt = e₀ sin ωt

Peak value e₀ = NBAω. The frequency is ν = ω/2π, and the emf reverses direction twice in every cycle.

Position of coil	Flux	Induced emf
Plane perpendicular to B (θ = 0°)	maximum	zero
Plane parallel to B (θ = 90°)	zero	maximum
Plane perpendicular again (θ = 180°)	maximum (reversed)	zero

Worked examples

Example 1. A coil of 100 turns and area 0.02 m² has its plane perpendicular to a field. The field rises uniformly from 0.1 T to 0.5 T in 0.2 s. Find the induced emf.

ΔΦ per turn = 0.02 × (0.5 − 0.1) = 0.008 Wb. e = NΔΦ/Δt = 100 × 0.008/0.2 = 4 V.

Example 2. A rod of length 0.5 m slides at 4 m s⁻¹ on rails of total resistance 2 Ω in a field of 0.2 T perpendicular to the plane. Find the emf, the current, the force needed and the power.

e = Blv = 0.2 × 0.5 × 4 = 0.4 V. I = 0.4/2 = 0.2 A. F = BIl = 0.2 × 0.2 × 0.5 = 0.02 N. P = Fv = 0.02 × 4 = 0.08 W. Check: I²R = 0.04 × 2 = 0.08 W.

Example 3. A rod of length 0.5 m rotates at 20 rad s⁻¹ about one end in a field of 0.1 T perpendicular to its plane. e = ½Bωl² = ½ × 0.1 × 20 × 0.25 = 0.25 V.

Example 4. A coil has L = 0.2 H. The current in it rises from 0 to 3 A in 0.05 s. Average emf = L ΔI/Δt = 0.2 × 3/0.05 = 12 V. Energy stored = ½ × 0.2 × 9 = 0.9 J.

Example 5. A coil of 50 turns and area 0.04 m² rotates at 50 rad s⁻¹ in a field of 0.5 T. Peak emf = NBAω = 50 × 0.5 × 0.04 × 50 = 50 V.

Formula summary

Quantity	Formula	Units
Flux	Φ = BA cos θ	Wb
Induced emf	e = −N dΦ/dt	V
Charge flowed	q = NΔΦ/R	C
Motional emf	e = Blv	V
Rotating rod	e = ½Bωl²	V
Mutual emf	e = −M dI/dt	V
Self emf	e = −L dI/dt	V
Solenoid inductance	L = μ₀n²Al	H
Energy in inductor	U = ½LI²	J
Generator emf	e = NBAω sin ωt	V

Constant: μ₀ = 4π × 10⁻⁷ T m A⁻¹.

NEET traps

Flux depends on the angle between B and the normal to the coil, not between B and the plane. A coil with its plane parallel to B has zero flux.
Zero flux does not mean zero emf. The emf depends on the rate of change of flux. In a rotating coil, the emf is maximum when the flux is zero.
A steady current in a primary coil induces no emf in a nearby coil. Only a changing current does.
Lenz's law says "opposes the change", not "opposes the flux". If the flux is increasing, the induced field is opposite to it; if the flux is decreasing, the induced field is in the same direction.
The induced charge q = NΔΦ/R does not depend on the time taken.
A rod moving along the field lines (parallel to B) has no motional emf.
Rotating rod: e = ½Bωl², not Bωl². The factor ½ comes from the average speed.
Self-inductance depends on geometry and core, not on the current flowing.
Laminating a core reduces eddy-current loss; it does not remove it fully, and the sheets must be parallel to the field.
Energy stored in an inductor is ½LI², not LI². Doubling the current increases the energy four times.

One-liners

1. SI unit of magnetic flux: weber (Wb) = T m².
2. Induced emf depends on the rate of change of flux.
3. Lenz's law follows from conservation of energy.
4. Motional emf e = Blv.
5. Rod rotating about one end: e = ½Bωl².
6. Eddy currents are induced in bulk metal by changing flux.
7. Laminated cores reduce eddy-current loss.
8. Magnetic braking and induction furnaces use eddy currents.
9. SI unit of inductance: henry (H) = V s A⁻¹.
10. Solenoid: L = μ₀n²Al.
11. Inductance is the electrical analogue of mass (inertia).
12. Energy stored in an inductor: ½LI².
13. Magnetic energy density: B²/(2μ₀).
14. AC generator emf: e = NBAω sin ωt.
15. In a generator, emf is maximum when the coil plane is parallel to the field.

Practice questions

A uniform magnetic field of 0.4 T makes an angle of 60° with the normal to a plane surface of area 0.05 m². What is the magnetic flux through the surface?
1. 1.0 × 10⁻² Wb
2. 1.7 × 10⁻² Wb
3. 2.0 × 10⁻² Wb
4. 3.5 × 10⁻² Wb
Answer
A. 1.0 × 10⁻² Wb
Φ = BA cos θ = 0.4 × 0.05 × 0.5 = 1.0 × 10⁻² Wb.
The flux through each turn of a 200-turn coil falls uniformly by 4 × 10⁻³ Wb in 0.1 s. What is the magnitude of the average induced emf?
1. 16 V
2. 0.04 V
3. 8 V
4. 80 V
Answer
C. 8 V
e = NΔΦ/Δt = 200 × 0.004/0.1 = 8 V.
A coil of 100 turns and resistance 20 Ω is joined to a galvanometer of negligible resistance. The flux through each turn changes by 0.02 Wb. What charge flows through the circuit?
1. 0.2 C
2. 0.001 C
3. 2 C
4. 0.1 C
Answer
D. 0.1 C
q = NΔΦ/R = 100 × 0.02/20 = 0.1 C.
A straight conductor of length 0.4 m moves at 5 m s⁻¹ perpendicular to its length and perpendicular to a uniform field of 0.25 T. What emf is induced between its ends?
1. 2 V
2. 0.5 V
3. 0.25 V
4. 1.25 V
Answer
B. 0.5 V
e = Blv = 0.25 × 0.4 × 5 = 0.5 V.
A metal rod of length 0.6 m rotates at 50 rad s⁻¹ about one end in a plane perpendicular to a uniform field of 0.2 T. What is the emf between its ends?
1. 6 V
2. 3 V
3. 1.8 V
4. 3.6 V
Answer
C. 1.8 V
e = ½Bωl² = 0.5 × 0.2 × 50 × 0.36 = 1.8 V.
A long air-cored solenoid has 1000 turns per metre, cross-sectional area 5 × 10⁻⁴ m² and length 0.4 m. Take μ₀ = 4π × 10⁻⁷ T m A⁻¹. Its self-inductance is about
1. 2.5 × 10⁻⁷ H
2. 2.5 × 10⁻⁴ H
3. 6.3 × 10⁻⁴ H
4. 4.0 × 10⁻⁵ H
Answer
B. 2.5 × 10⁻⁴ H
L = μ₀n²Al = 4π × 10⁻⁷ × 10⁶ × 5 × 10⁻⁴ × 0.4 ≈ 2.5 × 10⁻⁴ H.
An inductor of 0.4 H carries a steady current of 5 A. How much energy is stored in it?
1. 2 J
2. 1 J
3. 10 J
4. 5 J
Answer
D. 5 J
U = ½LI² = 0.5 × 0.4 × 25 = 5 J.
The current in a 2 H inductor falls uniformly from 4 A to zero in 0.08 s. What is the magnitude of the induced emf?
1. 100 V
2. 200 V
3. 8 V
4. 50 V
Answer
A. 100 V
e = LΔI/Δt = 2 × 4/0.08 = 100 V.
A rectangular coil of 100 turns and area 0.05 m² rotates at 10 revolutions per second about an axis perpendicular to a uniform field of 0.2 T. Taking π = 3.14, the peak emf is about
1. 10 V
2. 20π V (about 63 V)
3. 40π V (about 126 V)
4. 10π V (about 31 V)
Answer
B. 20π V (about 63 V)
ω = 2π × 10 = 20π rad s⁻¹. e₀ = NBAω = 100 × 0.2 × 0.05 × 20π = 20π V ≈ 63 V.
Which of the following is equal to one henry?
1. 1 V s A⁻¹
2. 1 V A s⁻¹
3. 1 V s² A⁻¹
4. 1 A s V⁻¹
Answer
A. 1 V s A⁻¹
From e = L dI/dt, L = e Δt/ΔI, so 1 H = 1 V s A⁻¹.
Lenz's law is a direct consequence of the law of conservation of
1. charge
2. linear momentum
3. energy
4. mass
Answer
C. energy
The induced current opposes the change; otherwise energy would be created without work.
Which method is used to reduce eddy-current loss in the core of a transformer?
1. Increasing the number of turns in the coil
2. Making the core of a single solid block
3. Using a core of non-magnetic material
4. Making the core of thin insulated sheets (laminations)
Answer
D. Making the core of thin insulated sheets (laminations)
Laminations raise the resistance of eddy-current paths and so cut the heating.
The mutual inductance of two coils is 0.5 H. The current in one coil changes uniformly by 2 A in 0.01 s. What is the magnitude of the emf induced in the other coil?
1. 200 V
2. 100 V
3. 1 V
4. 50 V
Answer
B. 100 V
e = MΔI/Δt = 0.5 × 2/0.01 = 100 V.
The current in a long solenoid is doubled while its dimensions and core stay the same. The energy stored in it becomes
1. four times
2. unchanged
3. two times
4. half
Answer
A. four times
U = ½LI² and L is constant, so U ∝ I².
The flux through a single-turn loop varies as Φ = (3t² + 2t) Wb, where t is in seconds. What is the magnitude of the induced emf at t = 2 s?
1. 16 V
2. 10 V
3. 12 V
4. 14 V
Answer
D. 14 V
e = dΦ/dt = 6t + 2 = 14 V.
A straight conductor moves with a uniform velocity along the direction of a uniform magnetic field, with its length perpendicular to the field. The emf induced between its ends is
1. Bl/v
2. ½Blv
3. zero
4. Blv
Answer
C. zero
The velocity is parallel to B, so v × B = 0 and there is no magnetic force on the charges.
Given below are two statements: Statement I: In an AC generator the induced emf is maximum when the plane of the coil is parallel to the magnetic field. Statement II: At that instant the flux through the coil is zero. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
e = NBAω sin ωt is maximum when the flux NBA cos ωt is zero.
Given below are two statements: Statement I: A steady current in the primary coil induces a steady emf in a nearby secondary coil. Statement II: The induced emf depends on the rate of change of magnetic flux. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
A steady current gives no change of flux, so no emf is induced.
Given below are two statements: Statement I: The self-inductance of a long solenoid depends on the number of turns per unit length. Statement II: The self-inductance of a solenoid depends on the current flowing through it. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
L = μ₀n²Al is fixed by geometry and core, not by the current.
Given below are two statements: Statement I: Eddy currents can be produced only in liquid conductors. Statement II: Laminating a core increases the eddy currents in it. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Eddy currents arise in any bulk conductor, and lamination reduces them.
Given below are two statements: Statement I: The energy stored in an inductor carrying current I is ½LI². Statement II: This energy is stored in the magnetic field of the inductor. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are standard results.
Given below are two statements: Statement I: The motional emf of a rod is independent of the speed of the rod. Statement II: The motional emf of a rod is proportional to its length. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
e = Blv, so it is proportional to both v and l.
Given below are two statements: Statement I: A bar magnet at rest inside a coil produces a steady deflection in a galvanometer joined to the coil. Statement II: The induced current in a loop does not depend on the relative motion between the magnet and the loop. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Only a changing flux induces a current, and a change needs relative motion.
Given below are two statements: Statement I: The mutual inductance M₁₂ of coil 1 with coil 2 equals M₂₁. Statement II: The mutual inductance of two coils depends on the currents flowing in them. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
M depends on geometry, turns and medium, not on current.
Which of the following statements are correct? 1. The SI unit of magnetic flux is the weber. 2. One henry is equal to one weber per ampere. 3. Magnetic flux is a vector quantity.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
Flux is a scalar (BA cos θ). Statements 1 and 2 are correct.
Which of the following statements are correct? 1. Eddy currents heat the core of a transformer. 2. Lamination of the core reduces eddy currents. 3. Eddy currents can never be put to any use.
1. 1 and 3 only
2. 2 and 3 only
3. 1 and 2 only
4. 3 only
Answer
C. 1 and 2 only
Eddy currents are used in induction furnaces and magnetic braking, so statement 3 is wrong.
Assertion (A): Laminated cores are used in transformers. Reason (R): Lamination increases the resistance of the paths of eddy currents.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Higher resistance gives smaller eddy currents and less heat.
Assertion (A): Lenz's law is a consequence of conservation of energy. Reason (R): The SI unit of magnetic flux is the weber.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both true, but the unit has nothing to do with the reason.
Assertion (A): The self-inductance of a coil depends on the current flowing in it. Reason (R): The self-inductance depends on the geometry of the coil and the core material.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
A is false: L does not depend on current.
Assertion (A): The emf of a rod rotating about one end in a perpendicular field is ½Bωl². Reason (R): The average speed of the points on the rod is ωl/2.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
e = B l × (average speed) = B l (ωl/2).
Assertion (A): The energy stored in an inductor is ½LI². Reason (R): An inductor opposes a change in the current through it.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
The energy is the work done against the self-induced emf; R is true but is a different statement of inductance, not the source of the ½ factor.
Match List I with List II. List I: a. Magnetic flux b. Inductance c. Magnetic flux density d. Induced emf List II: i. Wb m⁻² ii. Wb s⁻¹ iii. V s A⁻¹ iv. T m²
1. a-iv, b-i, c-iii, d-ii
2. a-ii, b-iii, c-i, d-iv
3. a-iii, b-iv, c-ii, d-i
4. a-iv, b-iii, c-i, d-ii
Answer
D. a-iv, b-iii, c-i, d-ii
Flux = T m²; L = V s A⁻¹; B = Wb m⁻²; emf = Wb s⁻¹.
Match List I with List II. List I: a. Sliding rod b. Rotating rod c. Long solenoid inductance d. Energy in inductor List II: i. ½LI² ii. μ₀n²Al iii. Blv iv. ½Bωl²
1. a-i, b-iv, c-ii, d-iii
2. a-iii, b-iv, c-ii, d-i
3. a-iii, b-iv, c-i, d-ii
4. a-iv, b-iii, c-ii, d-i
Answer
B. a-iii, b-iv, c-ii, d-i
Standard formulae.
Match List I with List II. List I: a. Induction furnace b. Laminated core c. AC generator d. Lenz's law List II: i. Rate of change of flux of a rotating coil ii. Heating by eddy currents iii. Reduces eddy currents iv. Conservation of energy
1. a-i, b-iii, c-ii, d-iv
2. a-iii, b-ii, c-i, d-iv
3. a-ii, b-iii, c-iv, d-i
4. a-ii, b-iii, c-i, d-iv
Answer
D. a-ii, b-iii, c-i, d-iv
Furnace: eddy-current heating; lamination: reduces eddy currents; generator: changing flux; Lenz: energy conservation.
Match List I with List II. List I: a. Rod of length 1 m moving at 2 m s⁻¹ across 0.5 T b. Rod of length 1 m rotating at 10 rad s⁻¹ about one end in 1 T c. 10-turn coil, flux change 0.1 Wb in 0.5 s d. 100-turn coil, area 0.01 m², 0.5 T, 20 rad s⁻¹ (peak) List II: i. 1 V ii. 2 V iii. 5 V iv. 10 V
1. a-i, b-iii, c-iv, d-ii
2. a-ii, b-iii, c-i, d-iv
3. a-i, b-iii, c-ii, d-iv
4. a-i, b-ii, c-iii, d-iv
Answer
C. a-i, b-iii, c-ii, d-iv
a: 0.5 × 1 × 2 = 1 V. b: ½ × 1 × 10 × 1 = 5 V. c: 10 × 0.1/0.5 = 2 V. d: 100 × 0.5 × 0.01 × 20 = 10 V.
Match List I with List II. List I: a. Normal along the field (θ = 0°) b. Normal at 60° to the field c. Normal perpendicular to the field d. Normal opposite to the field (θ = 180°) List II: i. −BA ii. zero iii. BA iv. BA/2
1. a-iii, b-ii, c-iv, d-i
2. a-iii, b-iv, c-i, d-ii
3. a-iv, b-iii, c-ii, d-i
4. a-iii, b-iv, c-ii, d-i
Answer
D. a-iii, b-iv, c-ii, d-i
Φ = BA cos θ: BA, BA/2, 0, −BA.
Match List I with List II. List I: a. Flux linkage of a coil b. Self-induced emf c. Mutual-induced emf d. Energy density of a magnetic field List II: i. −M dI/dt ii. B²/2μ₀ iii. LI iv. −L dI/dt
1. a-iv, b-iii, c-i, d-ii
2. a-iii, b-iv, c-ii, d-i
3. a-iii, b-iv, c-i, d-ii
4. a-iii, b-i, c-iv, d-ii
Answer
C. a-iii, b-iv, c-i, d-ii
NΦ = LI; e = −L dI/dt; e₂ = −M dI₁/dt; u = B²/2μ₀.
A rod of length 0.5 m slides at a constant 4 m s⁻¹ on a U-shaped rail in a perpendicular field of 0.2 T. The total resistance of the circuit is 2 Ω. What external force keeps the speed constant?
1. 0.2 N
2. 0.02 N
3. 0.01 N
4. 0.04 N
Answer
B. 0.02 N
I = Blv/R = 0.2 A; F = BIl = 0.2 × 0.2 × 0.5 = 0.02 N.
A rod of length 0.4 m moves at a constant 5 m s⁻¹ across a perpendicular field of 0.5 T on rails. The total resistance of the circuit is 0.2 Ω. At what rate is electrical energy dissipated as heat?
1. 5 W
2. 1 W
3. 25 W
4. 10 W
Answer
A. 5 W
e = 0.5 × 0.4 × 5 = 1 V; P = e²/R = 1/0.2 = 5 W. The same as the mechanical power Fv.
An aircraft with a wing span of 40 m flies horizontally at 250 m s⁻¹. The vertical component of the Earth's magnetic field is 5 × 10⁻⁵ T. What is the emf between the wing tips?
1. 0.05 V
2. 5 V
3. 0.5 V
4. 2.5 V
Answer
C. 0.5 V
e = Blv = 5 × 10⁻⁵ × 40 × 250 = 0.5 V.
The north pole of a bar magnet is moved vertically downwards towards a horizontal circular loop kept below it. Looking from above, the induced current in the loop and the force on the magnet are
1. clockwise, and the loop repels the magnet
2. clockwise, and the loop attracts the magnet
3. anticlockwise, and the loop attracts the magnet
4. anticlockwise, and the loop repels the magnet
Answer
D. anticlockwise, and the loop repels the magnet
Downward flux increases, so the loop makes an upward field (anticlockwise current seen from above). Its upper face acts as a north pole and repels the magnet.
The current in a coil of self-inductance 0.5 H increases uniformly as I = 4t amperes, where t is in seconds. What is the energy stored in the coil at t = 2 s?
1. 32 J
2. 16 J
3. 4 J
4. 2 J
Answer
B. 16 J
I = 8 A at t = 2 s. U = ½LI² = 0.5 × 0.5 × 64 = 16 J.
A long solenoid has a self-inductance of 10 mH. Keeping the length, area and core the same, the number of turns is doubled. The new self-inductance is
1. 40 mH
2. 80 mH
3. 5 mH
4. 20 mH
Answer
A. 40 mH
L ∝ N² for fixed length and area, so L becomes 4 × 10 = 40 mH.
A generator coil rotates at 50 Hz and gives a peak emf of 200 V, with e = e₀ sin ωt. What is the instantaneous emf at t = 1/600 s?
1. 200 V
2. 100 V
3. 173 V
4. 141 V
Answer
B. 100 V
ωt = 2π × 50/600 = π/6, so e = 200 sin 30° = 100 V.

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1.The phenomenon of electromagnetic induction occurs due to:
2.Induced EMF in a coil during the phenomenon of electromagnetic induction is directly proportional to:
3.Which of the following devices is based on the phenomenon of electromagnetic induction?
4.Which of the following works on the basis of electromagnetic induction?
5.Electromagnetic Induction was discovered by:
6.__________ is defined as the characteristic of wavelength of electromagnetic radiation emitted or absorbed by an object or substance, atom or molecule.
7._______________ is a discrete packet of energy related to electromagnetic radiation (light), in which energy is E which is proportional to frequency of radiation ν.
8.Which of the following are electromagnetic waves?
9.What is self-induction?
10.Which equation demonstrates that electricity, magnetism and light are all manifestations of the electromagnetic field?

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