Important Questions Class 12 Physics Chapter 6: Electromagnetic Induction

Electromagnetic induction is the production of induced emf or current due to changing magnetic flux. A coil gets induced current when magnetic field, area, or orientation changes with time.

Changing magnetic flux turns magnetism into electricity through a measurable induced emf. Important Questions Class 12 Physics Chapter 6 help students practise Faraday’s law, Lenz’s law, magnetic flux, motional emf, self-inductance, mutual inductance, and AC generator formulas. The CBSE 2026 chapter also connects Faraday and Henry’s experiments with generators, transformers, back emf, magnetic energy, and conservation of energy.

Key Takeaways

• Magnetic Flux: Magnetic flux through a plane surface is ΦB = BA cos θ.
• Faraday’s Law: Induced emf equals the negative rate of change of magnetic flux.
• Lenz’s Law: Induced current opposes the change in magnetic flux that produces it.
• AC Generator: A rotating coil in a magnetic field produces alternating emf.

Important Questions Class 12 Physics Chapter 6 Structure 2026

Important Questions Class 12 Physics Chapter 6 with Answers

Electromagnetic induction questions begin with flux change and end with practical power generation.
Students should identify whether the flux changes due to B, A, θ, or motion.
These electromagnetic induction class 12 important questions follow the NCERT 2026 chapter sequence.

1. What does Important Questions Class 12 Physics Chapter 6 mainly test?

Important Questions Class 12 Physics Chapter 6 mainly test magnetic flux, induced emf, Lenz’s law, motional emf, inductance, and AC generator. The chapter combines conceptual reasoning with numerical formulas.

1. Flux Skill: Use ΦB = BA cos θ.
2. Induction Skill: Apply ε = −N dΦB/dt.
3. Motion Skill: Apply ε = Blv.
4. Generator Skill: Use ε = NBAω sin ωt.
5. Final Result: The chapter tests changing magnetic flux and induced emf.

2. What is electromagnetic induction in Class 12 Physics?

Electromagnetic induction is the production of induced emf due to changing magnetic flux. A closed circuit also gets induced current.

1. Cause: Magnetic flux changes with time.
2. Effect: Induced emf appears in the circuit.
3. Closed Circuit: Induced current flows through resistance.
4. Final Result: Changing magnetic flux produces induced emf.

3. Who discovered electromagnetic induction?

Michael Faraday and Joseph Henry discovered electromagnetic induction around 1830. Their experiments showed current induction through changing magnetic fields.

1. Faraday: Worked in England.
2. Henry: Worked in the USA.
3. Main Result: Changing magnetic field induces current.
4. Final Result: Faraday and Henry established electromagnetic induction.

Faraday and Henry Experiments Class 12 Questions

Faraday and Henry used coils, magnets, batteries, and galvanometers to show induced current.
The experiments prove that relative motion is useful, but changing magnetic flux is the real cause.
This section covers Faraday and Henry experiments Class 12 through direct NCERT-style questions.

4. What happens when a bar magnet moves towards a coil connected to a galvanometer?

The galvanometer deflects when the magnet moves towards the coil. The motion changes magnetic flux through the coil.

1. Given Setup: A coil connects to a galvanometer.
2. Action: The magnet moves towards the coil.
3. Observation: The galvanometer deflects.
4. Reason: Magnetic flux through the coil changes.
5. Final Result: Current is induced only while the magnet moves.

5. What happens when the magnet is held stationary near the coil?

The galvanometer shows no deflection when the magnet stays stationary. Magnetic flux does not change with time.

1. Given Setup: Magnet remains near the coil.
2. Motion: No relative motion occurs.
3. Flux Change: dΦB/dt = 0.
4. Final Result: No induced current flows in the coil.

6. Why does faster magnet motion give larger induced current?

Faster magnet motion gives larger induced current because flux changes faster. Faraday’s law links emf with flux change rate.

1. Slow Motion: Small dΦB/dt.
2. Fast Motion: Large dΦB/dt.
3. Formula Used: ε = −dΦB/dt.
4. Final Result: Faster flux change gives larger induced emf.

7. What did Faraday’s two-coil experiment prove?

Faraday’s two-coil experiment proved that changing current in one coil induces current in another coil. The changing magnetic field causes this effect.

1. Coil C2: Connected to a battery.
2. Coil C1: Connected to a galvanometer.
3. Key Pressed: Current in C2 changes.
4. Observation: Galvanometer in C1 deflects momentarily.
5. Final Result: Changing current induces emf in a neighbouring coil.

Magnetic Flux Class 12 Questions

Magnetic flux measures how much magnetic field passes through a surface.
The value depends on magnetic field, surface area, and angle between B and area vector.
These magnetic flux Class 12 questions build the base for Faraday law Class 12 Physics questions.

8. What is magnetic flux in Class 12 Physics?

Magnetic flux is the scalar product of magnetic field and area vector. Its formula is ΦB = BA cos θ.

1. Magnetic Field: B.
2. Area Vector: A.
3. Angle: θ between B and A.
4. Formula Used: ΦB = BA cos θ.
5. Final Result: Magnetic flux = BA cos θ.

9. What is the SI unit of magnetic flux?

The SI unit of magnetic flux is weber, written as Wb. It also equals tesla metre squared.

1. Quantity: Magnetic flux.
2. SI Unit: Weber.
3. Equivalent Unit: T m².
4. Final Result: 1 Wb = 1 T m².

10. When is magnetic flux maximum through a plane surface?

Magnetic flux is maximum when B is parallel to the area vector. The angle θ equals 0°.

1. Formula Used: ΦB = BA cos θ.
2. Maximum Value: cos 0° = 1.
3. Flux: ΦB = BA.
4. Final Result: Maximum flux = BA.

11. When is magnetic flux zero through a plane surface?

Magnetic flux is zero when B is perpendicular to the area vector. The angle θ equals 90°.

1. Formula Used: ΦB = BA cos θ.
2. At θ = 90°: cos 90° = 0.
3. Flux: ΦB = 0.
4. Final Result: Magnetic flux becomes zero.

12. Find magnetic flux through area 0.02 m² in field 0.5 T at 60°.

The magnetic flux is 0.005 Wb. Use ΦB = BA cos θ.

1. Given Data:
   B = 0.5 T
   A = 0.02 m²
   θ = 60°
2. Formula Used: ΦB = BA cos θ.
3. Calculation:
   ΦB = 0.5 × 0.02 × cos 60°
   ΦB = 0.5 × 0.02 × 0.5
   ΦB = 0.005 Wb
4. Final Result: ΦB = 5 × 10^-3 Wb.

Faraday Law Class 12 Physics Questions

Faraday’s law gives the magnitude of induced emf through the rate of flux change.
The negative sign comes from Lenz’s law and gives the direction of induced current.
This section uses Faraday law Class 12 Physics questions with solved numerical steps.

13. State Faraday’s law of electromagnetic induction.

Faraday’s law states that induced emf equals the negative rate of change of magnetic flux. For N turns, ε = −N dΦB/dt.

1. Single Turn: ε = −dΦB/dt.
2. N Turns: ε = −N dΦB/dt.
3. Meaning: Faster flux change gives larger emf.
4. Final Result: Induced emf depends on dΦB/dt.

14. Why does a coil with more turns produce larger induced emf?

A coil with more turns produces larger induced emf because each turn contributes emf. Total emf is N times the single-turn value.

1. Single Turn: ε = −dΦB/dt.
2. N Turns: ε = −N dΦB/dt.
3. Effect: Increasing N increases induced emf.
4. Final Result: More turns give larger induced emf.

15. A square loop of side 10 cm has B = 0.10 T at 45°. Field becomes zero in 0.70 s. Find emf.

The induced emf is 1.0 mV. Use change in magnetic flux over time.

1. Given Data:
   Side = 10 cm = 0.10 m
   A = 0.01 m²
   B = 0.10 T
   θ = 45°
   Δt = 0.70 s
2. Initial Flux:
   Φi = BA cos θ
   Φi = 0.10 × 0.01 × cos 45°
   Φi = 7.07 × 10^-4 Wb
3. Induced EMF:
   ε = ΔΦB/Δt
   ε = 7.07 × 10^-4/0.70
   ε = 1.01 × 10^-3 V
4. Final Result: ε ≈ 1.0 mV.

16. A loop has resistance 0.5 Ω and induced emf 1.0 mV. Find induced current.

The induced current is 2 mA. Use Ohm’s law.

1. Given Data:
   ε = 1.0 mV = 1.0 × 10^-3 V
   R = 0.5 Ω
2. Formula Used: I = ε/R.
3. Calculation:
   I = (1.0 × 10^-3)/0.5
   I = 2.0 × 10^-3 A
4. Final Result: I = 2 mA.

17. Why does a steady magnetic field not induce emf in a stationary loop?

A steady magnetic field does not induce emf because magnetic flux does not change. Faraday’s law needs dΦB/dt.

1. Condition: Magnetic field remains constant.
2. Loop: Stationary loop has fixed area and angle.
3. Flux Rate: dΦB/dt = 0.
4. Final Result: No induced emf appears.

Lenz Law Class 12 Important Questions

Lenz’s law gives the direction of induced emf and induced current.
The induced current always opposes the change in magnetic flux that produces it.
These Lenz law Class 12 important questions also connect induction with conservation of energy.

18. State Lenz’s law in electromagnetic induction.

Lenz’s law states that induced current opposes the change in magnetic flux that produces it. It explains the negative sign in Faraday’s law.

1. Cause: Magnetic flux changes through a circuit.
2. Effect: Induced current appears.
3. Direction: Induced current opposes the flux change.
4. Final Result: Lenz’s law follows conservation of energy.

19. Why is Lenz’s law consistent with conservation of energy?

Lenz’s law is consistent with energy conservation because external work is needed against induced effects. That work becomes electrical energy or heat.

1. Approaching Magnet: Coil opposes the approach.
2. External Work: A person pushes the magnet.
3. Energy Form: Induced current produces Joule heating.
4. Final Result: Lenz’s law prevents free energy generation.

20. What is the induced current direction when a north pole approaches a coil?

The coil face near the magnet becomes a north pole. The induced current opposes the increasing flux.

1. Situation: North pole approaches the coil.
2. Flux Change: Magnetic flux increases.
3. Opposition: Coil repels the approaching north pole.
4. Current Direction: Anticlockwise from the magnet side.
5. Final Result: The coil face becomes north.

21. What is the induced current direction when a north pole moves away from a coil?

The coil face near the magnet becomes a south pole. The induced current opposes the decreasing flux.

1. Situation: North pole moves away.
2. Flux Change: Magnetic flux decreases.
3. Opposition: Coil attracts the receding north pole.
4. Current Direction: Clockwise from the magnet side.
5. Final Result: The coil face becomes south.

22. Why is no current induced when a loop stays fully inside a uniform magnetic field?

No current is induced because magnetic flux through the loop remains constant. A constant flux gives zero emf.

1. Loop Position: Fully inside the field.
2. Field: Uniform and steady.
3. Flux Change: dΦB/dt = 0.
4. Final Result: Induced current is zero.

Motional EMF Class 12 Questions

Motional emf appears when a conductor moves through a magnetic field.
It can be explained through Faraday’s law or the Lorentz force on charges.
These motional emf Class 12 questions cover rods, loops, wheels, and rotating conductors.

23. What is motional emf in Class 12 Physics?

Motional emf is emf induced across a conductor moving in a magnetic field. Its formula is ε = Blv.

1. Magnetic Field: B.
2. Rod Length: l.
3. Speed: v.
4. Formula Used: ε = Blv.
5. Final Result: Motional emf = Blv.

24. How does Lorentz force explain motional emf?

Lorentz force pushes charges along a moving conductor. Charge separation creates potential difference between its ends.

1. Charge Force: F = qvB.
2. Work Done: W = qvBl.
3. EMF: ε = W/q.
4. Final Result: ε = Blv.

25. A rod of length 0.8 m moves at 5 m/s in 0.2 T field. Find motional emf.

The motional emf is 0.8 V. Use ε = Blv.

1. Given Data:
   B = 0.2 T
   l = 0.8 m
   v = 5 m/s
2. Formula Used: ε = Blv.
3. Calculation:
   ε = 0.2 × 0.8 × 5
   ε = 0.8 V
4. Final Result: ε = 0.8 V.

26. A 1 m rod rotates at 400 rad/s in 0.5 T field. Find emf between centre and ring.

The emf is 100 V. Use ε = 1/2 BωR².

1. Given Data:
   B = 0.5 T
   ω = 400 rad/s
   R = 1 m
2. Formula Used: ε = 1/2 BωR².
3. Calculation:
   ε = 1/2 × 0.5 × 400 × 1²
   ε = 100 V
4. Final Result: ε = 100 V.

27. Why does the number of spokes not affect emf in a rotating wheel?

The number of spokes does not affect emf because spoke emfs connect in parallel. Each spoke gives the same axle-rim emf.

1. Wheel Setup: Spokes rotate in the same magnetic field.
2. Each Spoke: Same length and angular speed.
3. Connection: Spokes connect between axle and rim.
4. Final Result: Axle-rim emf remains the same for all spokes.

Electromagnetic Induction Numericals Class 12

Numericals in this chapter often need direct substitution after unit conversion.
Students must check whether the question uses flux change, motional emf, or inductance.
This section gives electromagnetic induction numericals Class 12 with step-by-step calculation.

28. A circular coil has radius 10 cm, 500 turns, and B = 3.0 × 10^-5 T. It rotates 180° in 0.25 s. Find emf.

The induced emf is 3.8 × 10^-3 V. The flux changes from +BA to −BA.

1. Given Data:
   r = 0.10 m
   N = 500
   B = 3.0 × 10^-5 T
   Δt = 0.25 s
2. Flux Change:
   A = πr² = π × 10^-2 m²
   ΔΦB = 2BA = 2 × 3.0 × 10^-5 × π × 10^-2
3. Induced EMF:
   ε = NΔΦB/Δt
   ε = 500 × 6π × 10^-7/0.25
   ε = 3.8 × 10^-3 V
4. Final Result: ε = 3.8 × 10^-3 V.

29. If the coil resistance is 2 Ω, find induced current in the same coil.

The induced current is 1.9 × 10^-3 A. Use I = ε/R.

1. Given Data:
   ε = 3.8 × 10^-3 V
   R = 2 Ω
2. Formula Used: I = ε/R.
3. Calculation:
   I = 3.8 × 10^-3/2
   I = 1.9 × 10^-3 A
4. Final Result: I = 1.9 mA.

30. A loop moves out of a 0.3 T field with speed 1 cm/s. Longer side is 8 cm. Find emf.

The induced emf is 2.4 × 10^-4 V when the longer side cuts the field boundary. Use ε = Blv.

1. Given Data:
   B = 0.3 T
   l = 8 cm = 0.08 m
   v = 1 cm/s = 0.01 m/s
2. Formula Used: ε = Blv.
3. Calculation:
   ε = 0.3 × 0.08 × 0.01
   ε = 2.4 × 10^-4 V
4. Final Result: ε = 2.4 × 10^-4 V.

31. A current falls from 5 A to 0 A in 0.1 s and average emf is 200 V. Find self-inductance.

The self-inductance is 4 H. Use |ε| = L |dI/dt|.

1. Given Data:
   ΔI = 5 A
   Δt = 0.1 s
   |ε| = 200 V
2. Formula Used: |ε| = L ΔI/Δt.
3. Calculation:
   L = |ε|Δt/ΔI
   L = 200 × 0.1/5
   L = 4 H
4. Final Result: L = 4 H.

Self Inductance Class 12 Questions

Self-inductance measures how strongly a coil opposes current change in itself.
A changing current changes flux through the same coil and produces back emf.
These self inductance Class 12 questions cover formulas, solenoids, and magnetic energy.

32. What is self-inductance in Class 12 Physics?

Self-inductance is the property by which a coil induces emf in itself. It opposes change in current.

1. Flux Linkage: NΦB = LI.
2. Induced EMF: ε = −L dI/dt.
3. Unit: Henry.
4. Final Result: Self-inductance measures electrical inertia.

33. Why is self-induced emf called back emf?

Self-induced emf is called back emf because it opposes current change. It resists both rise and fall of current.

1. Current Increasing: Back emf opposes increase.
2. Current Decreasing: Back emf opposes decrease.
3. Law Used: Lenz’s law.
4. Final Result: Back emf opposes dI/dt.

34. What is self-inductance of a long solenoid?

The self-inductance of a long solenoid is L = μ0n²Al. With magnetic core, L = μrμ0n²Al.

1. Air Core Formula: L = μ0n²Al.
2. Magnetic Core Formula: L = μrμ0n²Al.
3. Variables: n is turns per unit length.
4. Final Result: Solenoid inductance depends on geometry and permeability.

35. What is energy stored in an inductor?

The energy stored in an inductor is U = 1/2 LI². It gets stored as magnetic energy.

1. Given Current: I.
2. Self-Inductance: L.
3. Formula Used: U = 1/2 LI².
4. Final Result: Magnetic energy = 1/2 LI².

36. What is magnetic energy density in a solenoid?

Magnetic energy density in a solenoid is uB = B²/(2μ0). It gives energy stored per unit volume.

1. Magnetic Energy: U = B²Al/(2μ0).
2. Volume: V = Al.
3. Energy Density: uB = U/V.
4. Final Result: uB = B²/(2μ0).

Mutual Inductance Class 12 Questions

Mutual inductance describes induction between two nearby coils.
A changing current in one coil changes flux through the other coil.
These mutual inductance Class 12 questions focus on coil pairs and solenoids.

37. What is mutual inductance in Class 12 Physics?

Mutual inductance is the property by which changing current in one coil induces emf in another coil. Its unit is henry.

1. Flux Linkage: N1Φ1 = MI2.
2. Induced EMF: ε1 = −M dI2/dt.
3. Unit: Henry.
4. Final Result: Mutual inductance links two coils magnetically.

38. What is the relation between M12 and M21?

The relation is M12 = M21 = M. Mutual inductance is the same in both directions.

1. Current in Coil 2: Produces flux through Coil 1.
2. Current in Coil 1: Produces flux through Coil 2.
3. General Equality: M12 = M21.
4. Final Result: Mutual inductance is reciprocal.

39. What is mutual inductance of two long coaxial solenoids?

The mutual inductance is M = μ0n1n2πr1²l for air core solenoids. Here r1 is inner solenoid radius.

1. Turns Densities: n1 and n2.
2. Inner Area: A1 = πr1².
3. Length: l.
4. Formula Used: M = μ0n1n2πr1²l.
5. Final Result: M depends on turns density, area, and length.

40. A pair of coils has M = 1.5 H. Current changes from 0 to 20 A. Find flux linkage change.

The flux linkage change is 30 Wb-turn. Use Δ(NΦ) = MΔI.

1. Given Data:
   M = 1.5 H
   ΔI = 20 A
2. Formula Used: Δ(NΦ) = MΔI.
3. Calculation:
   Δ(NΦ) = 1.5 × 20
   Δ(NΦ) = 30 Wb-turn
4. Final Result: Change in flux linkage = 30 Wb-turn.

AC Generator Class 12 Physics Questions

An AC generator converts mechanical energy into electrical energy using electromagnetic induction.
A rotating coil changes magnetic flux periodically and produces alternating emf.
These AC generator Class 12 Physics questions cover working, formula, and NCERT numericals.

41. What is the principle of an AC generator?

An AC generator works on electromagnetic induction. A rotating coil in a magnetic field produces changing flux and induced emf.

1. Input Energy: Mechanical energy.
2. Conversion: Mechanical energy becomes electrical energy.
3. Law Used: Faraday’s law.
4. Final Result: An AC generator works on electromagnetic induction.

42. What is the induced emf in an AC generator?

The induced emf is ε = NBAω sin ωt. The maximum emf is ε0 = NBAω.

1. Flux: ΦB = BA cos ωt.
2. Faraday’s Law: ε = −N dΦB/dt.
3. Result: ε = NBAω sin ωt.
4. Maximum EMF: ε0 = NBAω.
5. Final Result: ε = ε0 sin ωt.

43. Why does an AC generator produce alternating current?

An AC generator produces alternating current because induced emf changes polarity periodically. The sine function changes sign with time.

1. Equation: ε = ε0 sin ωt.
2. Positive Half-Cycle: Current flows one way.
3. Negative Half-Cycle: Current reverses direction.
4. Final Result: AC changes direction periodically.

44. A 100-turn coil of area 0.10 m² rotates at 0.5 Hz in B = 0.01 T. Find maximum voltage.

The maximum voltage is 0.314 V. Use ε0 = NBA(2πν).

1. Given Data:
   N = 100
   A = 0.10 m²
   B = 0.01 T
   ν = 0.5 Hz
2. Formula Used: ε0 = NBA(2πν).
3. Calculation:
   ε0 = 100 × 0.01 × 0.10 × 2π × 0.5
   ε0 = 0.314 V
4. Final Result: Maximum voltage = 0.314 V.

45. What is the frequency of AC generators used in India?

The frequency of AC generators used in India is 50 Hz. Some countries use 60 Hz.

1. India: 50 Hz.
2. USA: 60 Hz.
3. Meaning: Coil rotation or field alternation sets AC frequency.
4. Final Result: India uses 50 Hz AC supply.

Class 12 Physics Chapter 6 Questions and Answers

NCERT questions often combine Lenz’s law, Faraday’s law, motional emf, and inductance.
Students should identify whether the answer needs direction, magnitude, or physical reasoning.
These class 12 physics chapter 6 questions and answers match the 2026 exercise style.

46. Can a stationary closed loop generate current in a steady magnetic field?

No, a stationary closed loop cannot generate current in a steady magnetic field. Magnetic flux stays constant.

1. Loop: Stationary.
2. Magnetic Field: Steady.
3. Flux Change: dΦB/dt = 0.
4. Final Result: No induced current appears.

47. Can changing electric flux induce current in a loop?

No, changing electric flux alone does not induce current in a loop in this chapter. Changing magnetic flux induces emf.

1. Faraday’s Law: Depends on magnetic flux.
2. Electric Flux: Does not induce current here.
3. Required Change: dΦB/dt.
4. Final Result: Magnetic flux change induces emf.

48. Why is induced emf constant for a rectangular loop leaving a uniform magnetic field?

Induced emf is constant because area leaves the field at a constant rate. The side length crossing the boundary remains constant.

1. Loop Shape: Rectangle.
2. Velocity: Constant.
3. Area Change Rate: Constant.
4. Final Result: Induced emf remains constant during exit.

49. Why is induced emf not constant for a circular loop leaving a uniform magnetic field?

Induced emf is not constant because area leaving the field changes at a variable rate. The chord length changes continuously.

1. Loop Shape: Circle.
2. Velocity: Constant.
3. Effective Length: Changes with position.
4. Final Result: Induced emf varies during exit.

50. Why does inserting an iron rod increase galvanometer deflection in Faraday’s experiment?

An iron rod increases galvanometer deflection because it increases magnetic field strength. Larger flux change gives larger induced emf.

1. Iron Rod: Increases magnetic flux through the coil.
2. Flux Rate: dΦB/dt becomes larger.
3. Faraday’s Law: Larger flux rate gives larger emf.
4. Final Result: Iron core increases induced current.

NCERT Class 12 Physics Chapter 6 Questions

NCERT Class 12 Physics Chapter 6 questions require both formula accuracy and direction logic.
The exercises use real cases like moving loops, rotating rods, falling wires, and changing current.
These NCERT Class 12 Physics Chapter 6 questions stay aligned with the 2026 textbook.

51. A solenoid has 15 turns/cm, loop area 2.0 cm², and current changes from 2 A to 4 A in 0.1 s. Find induced emf.

The induced emf is 7.5 × 10^-6 V. Use B = μ0nI and ε = A ΔB/Δt.

1. Given Data:
   n = 15 turns/cm = 1500 turns/m
   A = 2.0 cm² = 2.0 × 10^-4 m²
   ΔI = 2 A
   Δt = 0.1 s
2. Formula Used:
   ΔB = μ0nΔI
   ε = AΔB/Δt
3. Calculation:
   ε = Aμ0nΔI/Δt
   ε = 2.0 × 10^-4 × 4π × 10^-7 × 1500 × 2/0.1
   ε = 7.5 × 10^-6 V
4. Final Result: ε = 7.5 μV.

52. A 10 m wire falls at 5 m/s through Earth’s horizontal field 0.30 × 10^-4 T. Find emf.

The induced emf is 1.5 × 10^-3 V. Use ε = Blv.

1. Given Data:
   B = 0.30 × 10^-4 T
   l = 10 m
   v = 5 m/s
2. Formula Used: ε = Blv.
3. Calculation:
   ε = 0.30 × 10^-4 × 10 × 5
   ε = 1.5 × 10^-3 V
4. Final Result: ε = 1.5 mV.

53. What is the direction of emf in a wire falling east-west through Earth’s field?

The west end is at higher potential when the wire falls downward through the northward horizontal field. Use v × B direction for positive charges.

1. Velocity: Downward.
2. Magnetic Field: Northward.
3. Force Direction: q(v × B) points westward.
4. Final Result: West end becomes positive.

54. What is induced emf across an open circuit loop when flux changes?

An induced emf appears across open ends even when current cannot flow. The emf still follows Faraday’s law.

1. Circuit: Open loop.
2. Flux Change: Magnetic flux changes with time.
3. Current: No continuous current flows.
4. Final Result: Open circuits can have induced emf.

Class 12 Physics Chapter-Wise Important Questions