Important Questions Class 12 Physics Chapter 4: Moving Charges and Magnetism

A magnetic field exerts force on moving charges and current-carrying conductors.
The force direction depends on charge, velocity, current direction, and the magnetic field direction.

Moving charges create magnetic fields and also experience magnetic forces in external fields. Important Questions Class 12 Physics Chapter 4 help students practise Lorentz force, circular motion in magnetic fields, Biot-Savart law, Ampere’s circuital law, solenoids, parallel current force, torque on current loops, and moving coil galvanometers. The CBSE 2026 chapter uses Oersted’s observation, right-hand rules, current loops, cyclotron motion, and galvanometer conversion from NCERT Moving Charges and Magnetism.

Key Takeaways

• Lorentz Force: A moving charge in electric and magnetic fields experiences F = q(E + v × B).
• Circular Motion: A charge moving perpendicular to B follows radius r = mv/qB.
• Long Straight Wire: The magnetic field at distance r is B = μ0I/(2πr).
• Moving Coil Galvanometer: Its equilibrium condition is kφ = NIAB.

Important Questions Class 12 Physics Chapter 4 Structure 2026

Important Questions Class 12 Physics Chapter 4 with Answers

Current-carrying wires, moving charges, and current loops form the base of this chapter.
Students should track vector direction before substituting values in formulas.
These moving charges and magnetism class 12 important questions follow the NCERT 2026 sequence.

1. What does Important Questions Class 12 Physics Chapter 4 mainly test?

Important Questions Class 12 Physics Chapter 4 mainly test magnetic force, magnetic fields due to currents, circular motion, solenoids, and galvanometers. The chapter combines direction rules with numerical formulas.

1. Force Skill: Use F = q(v × B) and F = I(l × B).
2. Field Skill: Use Biot-Savart law and Ampere’s circuital law.
3. Motion Skill: Use r = mv/qB and ω = qB/m.
4. Instrument Skill: Use kφ = NIAB.
5. Final Result: The chapter tests magnetic force and current-produced fields.

2. What did Oersted’s experiment prove?

Oersted’s experiment proved that electric current produces a magnetic field. A compass needle deflects near a current-carrying wire.

1. Setup: A wire carries current near a compass.
2. Observation: The compass needle deflects.
3. Current Reversal: The needle deflects in the opposite direction.
4. Final Result: Moving charges produce magnetic fields.

3. Why do iron filings form circles around a straight current-carrying wire?

Iron filings form circles because the magnetic field lines around a straight wire are concentric circles. The wire acts as the centre.

1. Wire Direction: Perpendicular to the paper.
2. Field Shape: Circular magnetic field lines form around the wire.
3. Direction Rule: Right-hand thumb rule gives field direction.
4. Final Result: Straight current produces circular magnetic field lines.

Lorentz Force Class 12 Physics Questions

A moving charge feels magnetic force only when velocity has a component perpendicular to magnetic field.
Electric force can change speed, but magnetic force changes only direction.
These Lorentz force Class 12 Physics questions test vector product and sign of charge.

4. What is Lorentz force in Class 12 Physics?

Lorentz force is the total force on a charge moving in electric and magnetic fields. Its formula is F = q(E + v × B).

1. Electric Part: Fe = qE.
2. Magnetic Part: Fm = q(v × B).
3. Total Force: F = q(E + v × B).
4. Final Result: Lorentz force combines electric and magnetic forces.

5. When is magnetic force on a moving charge zero?

Magnetic force is zero when the charge is stationary or velocity is parallel to magnetic field. The cross product becomes zero.

1. Formula Used: F = qvB sin θ.
2. Stationary Charge: v = 0 gives F = 0.
3. Parallel Motion: θ = 0° gives sin θ = 0.
4. Final Result: Magnetic force needs perpendicular motion.

6. Why does magnetic force do no work on a moving charge?

Magnetic force does no work because it stays perpendicular to velocity. It changes direction but not speed.

1. Force Direction: Perpendicular to velocity.
2. Work Formula: W = Fd cos 90°.
3. Result: cos 90° = 0.
4. Final Result: Magnetic force does zero work.

7. A proton moves along +x, and magnetic field acts along +y. What is force direction?

The force on the proton acts along +z direction. Use the right-hand rule for v × B.

1. Velocity: +x direction.
2. Magnetic Field: +y direction.
3. Cross Product: +x × +y = +z.
4. Charge: Proton has positive charge.
5. Final Result: Force acts along +z direction.

8. What is force direction for an electron moving along +x in magnetic field +y?

The force on the electron acts along −z direction. Negative charge reverses the v × B direction.

1. Velocity: +x direction.
2. Magnetic Field: +y direction.
3. v × B Direction: +z direction.
4. Electron Charge: Negative.
5. Final Result: Force acts along −z direction.

Magnetic Force on Current Carrying Conductor Class 12 Questions

A current-carrying conductor contains many drifting charge carriers.
The magnetic force on each carrier adds to give force on the whole wire.
These magnetic force on current carrying conductor Class 12 questions use F = I(l × B).

9. What is magnetic force on a current-carrying conductor?

The magnetic force on a straight conductor is F = I(l × B). Its magnitude is F = IlB sin θ.

1. Current: I.
2. Length Vector: l points along current.
3. Magnetic Field: B.
4. Formula Used: F = I(l × B).
5. Final Result: Force magnitude = IlB sin θ.

10. When is force on a current-carrying wire maximum?

The force is maximum when the wire is perpendicular to the magnetic field. The angle θ equals 90°.

1. Formula Used: F = IlB sin θ.
2. Maximum Case: sin 90° = 1.
3. Maximum Force: Fmax = IlB.
4. Final Result: Maximum force equals IlB.

11. When is force on a current-carrying wire zero?

The force is zero when the wire is parallel to the magnetic field. The angle θ equals 0°.

1. Formula Used: F = IlB sin θ.
2. Parallel Case: sin 0° = 0.
3. Result: F = 0.
4. Final Result: Parallel current and field give zero force.

12. A 3.0 cm wire carries 10 A perpendicular to 0.27 T field. Find force.

The magnetic force is 0.081 N. Use F = IlB.

1. Given Data:
   I = 10 A
   l = 3.0 cm = 0.03 m
   B = 0.27 T
   θ = 90°
2. Formula Used: F = IlB.
3. Calculation:
   F = 10 × 0.03 × 0.27
   F = 0.081 N
4. Final Result: F = 0.081 N.

13. What field suspends a 200 g, 1.5 m wire carrying 2 A?

The magnetic field is 0.65 T. The magnetic force balances the weight.

1. Given Data:
   m = 0.2 kg
   g = 9.8 m/s²
   I = 2 A
   l = 1.5 m
2. Formula Used: mg = IlB.
3. Calculation:
   B = mg/(Il)
   B = (0.2 × 9.8)/(2 × 1.5)
   B = 0.65 T
4. Final Result: B = 0.65 T.

Motion in Magnetic Field Class 12 Questions

A magnetic field bends a moving charge when velocity has a perpendicular component.
Pure perpendicular motion gives a circle, while mixed motion gives a helix.
These motion in magnetic field Class 12 questions cover radius, frequency, and pitch.

14. Why does a charged particle move in a circle in a uniform magnetic field?

A charged particle moves in a circle when velocity is perpendicular to magnetic field. Magnetic force acts as centripetal force.

1. Magnetic Force: F = qvB.
2. Centripetal Force: F = mv²/r.
3. Equating Forces: qvB = mv²/r.
4. Final Result: Magnetic force provides centripetal force.

15. What is radius of circular path in a magnetic field?

The radius is r = mv/qB for perpendicular motion. Larger momentum gives a larger circle.

1. Formula Used: qvB = mv²/r.
2. Rearrange: r = mv/qB.
3. Dependence: r increases with mass and speed.
4. Final Result: Radius = mv/qB.

16. What is cyclotron frequency in Class 12 Physics?

Cyclotron frequency is the frequency of circular motion of a charge in a magnetic field. It equals ν = qB/(2πm).

1. Angular Frequency: ω = qB/m.
2. Frequency Relation: ν = ω/(2π).
3. Formula Used: ν = qB/(2πm).
4. Final Result: Cyclotron frequency does not depend on speed.

17. What is pitch in helical motion of a charged particle?

Pitch is the distance moved along magnetic field in one complete rotation. Its formula is p = 2πmv||/qB.

1. Parallel Velocity: v||.
2. Time Period: T = 2πm/qB.
3. Formula Used: p = v||T.
4. Final Result: Pitch = 2πmv||/qB.

18. Find radius of an electron moving at 3 × 10⁷ m/s in 6 × 10⁻⁴ T field.

The radius is 0.28 m. Use r = mv/qB.

1. Given Data:
   m = 9 × 10^-31 kg
   v = 3 × 10^7 m/s
   q = 1.6 × 10^-19 C
   B = 6 × 10^-4 T
2. Formula Used: r = mv/qB.
3. Calculation:
   r = (9 × 10^-31 × 3 × 10^7)/(1.6 × 10^-19 × 6 × 10^-4)
   r = 0.28 m
4. Final Result: r = 28 cm.

Biot Savart Law Class 12 Questions

Biot-Savart law gives the magnetic field due to a small current element.
It works like the magnetic analogue of Coulomb’s law for steady currents.
These Biot Savart law Class 12 questions cover magnitude, direction, and angle dependence.

19. State Biot-Savart law.

Biot-Savart law gives magnetic field due to a current element. Its vector form is dB = μ0I(dl × r)/(4πr³).

1. Current Element: I dl.
2. Position Vector: r from element to point.
3. Vector Formula: dB = μ0I(dl × r)/(4πr³).
4. Final Result: Biot-Savart law gives field due to a current element.

20. What is magnitude form of Biot-Savart law?

The magnitude form is dB = μ0I dl sin θ/(4πr²). Here θ is the angle between dl and r.

1. Formula Used: dB = μ0I dl sin θ/(4πr²).
2. Maximum Field: θ = 90°.
3. Zero Field: θ = 0° or 180°.
4. Final Result: The field depends on sin θ.

21. How is Biot-Savart law similar to Coulomb’s law?

Both laws show inverse-square dependence and obey superposition. Their sources differ.

1. Coulomb Source: Electric charge.
2. Biot-Savart Source: Current element.
3. Distance Dependence: Both vary as 1/r².
4. Final Result: Both laws describe fields from sources.

22. Why is magnetic field zero along the direction of current element?

Magnetic field is zero along the current element direction because θ = 0°. The cross product becomes zero.

1. Formula Used: dB = μ0I dl sin θ/(4πr²).
2. Along dl: θ = 0°.
3. Result: sin 0° = 0.
4. Final Result: Field along current element direction is zero.

23. Find magnetic field due to a 1 cm element carrying 10 A at 0.5 m on y-axis.

The magnetic field is 4 × 10^-8 T. The direction is along +z.

1. Given Data:
   I = 10 A
   dl = 1 cm = 10^-2 m
   r = 0.5 m
   θ = 90°
2. Formula Used: dB = μ0I dl sin θ/(4πr²).
3. Calculation:
   dB = 10^-7 × 10 × 10^-2/(0.5)²
   dB = 4 × 10^-8 T
4. Final Result: dB = 4 × 10^-8 T along +z.

Magnetic Field Due to Circular Loop Class 12 Questions

A circular current loop produces a magnetic field along its axis.
At large distances, the loop behaves like a magnetic dipole.
These magnetic field due to circular loop Class 12 questions cover centre field and axial field.

24. What is magnetic field at the centre of a circular current loop?

The magnetic field at the centre is B = μ0I/(2R) for one turn. For N turns, B = μ0NI/(2R).

1. One Turn: B = μ0I/(2R).
2. N Turns: B = μ0NI/(2R).
3. Direction: Right-hand thumb rule gives direction.
4. Final Result: Centre field = μ0NI/(2R).

25. What is magnetic field on the axis of a circular loop?

The axial magnetic field is B = μ0IR²/[2(x² + R²)^(3/2)]. It acts along the loop axis.

1. Loop Radius: R.
2. Axial Distance: x.
3. Formula Used: B = μ0IR²/[2(x² + R²)^(3/2)].
4. Final Result: Axial field depends on x and R.

26. A 100-turn coil of radius 10 cm carries 1 A. Find field at centre.

The magnetic field is 6.28 × 10^-4 T. Use B = μ0NI/(2R).

1. Given Data:
   N = 100
   R = 0.10 m
   I = 1 A
2. Formula Used: B = μ0NI/(2R).
3. Calculation:
   B = (4π × 10^-7 × 100 × 1)/(2 × 0.10)
   B = 6.28 × 10^-4 T
4. Final Result: B = 6.28 × 10^-4 T.

27. Why do perpendicular components cancel on the axis of a circular loop?

Perpendicular components cancel because opposite current elements produce equal and opposite transverse fields. Axial components add.

1. Opposite Elements: Equal current elements lie opposite each other.
2. Transverse Fields: Same magnitude and opposite direction.
3. Axial Fields: Same direction along axis.
4. Final Result: Only axial field survives.

28. What is magnetic moment of a circular current loop?

Magnetic moment of a current loop is m = IA. For N turns, m = NIA.

1. Current: I.
2. Area Vector: A.
3. N-Turn Coil: m = NIA.
4. Final Result: Magnetic moment equals current times area.

Ampere Circuital Law Class 12 Questions

Ampere’s circuital law relates magnetic circulation to current enclosed by a closed loop.
It works best for highly symmetric current distributions.
These Ampere circuital law Class 12 questions cover straight wires and thick conductors.

29. State Ampere’s circuital law.

Ampere’s circuital law states that ∮ B · dl = μ0I. Here I is the current enclosed by the loop.

1. Closed Path: Amperian loop.
2. Magnetic Circulation: ∮ B · dl.
3. Enclosed Current: I.
4. Final Result: ∮ B · dl = μ0I.

30. What is magnetic field due to a long straight wire?

The magnetic field at distance r is B = μ0I/(2πr). Field lines are concentric circles.

1. Amperian Loop: Circle of radius r.
2. Ampere’s Law: B(2πr) = μ0I.
3. Formula: B = μ0I/(2πr).
4. Final Result: Field varies inversely with distance.

31. A long wire carries 35 A. Find field at 20 cm.

The magnetic field is 3.5 × 10^-5 T. Use B = μ0I/(2πr).

1. Given Data:
   I = 35 A
   r = 20 cm = 0.20 m
2. Formula Used: B = μ0I/(2πr).
3. Calculation:
   B = (4π × 10^-7 × 35)/(2π × 0.20)
   B = 3.5 × 10^-5 T
4. Final Result: B = 3.5 × 10^-5 T.

32. How does magnetic field vary inside a thick current-carrying wire?

Inside a thick wire with uniform current, magnetic field increases linearly with distance from centre. It follows B ∝ r.

1. Region: r < a.
2. Enclosed Current: Ie = Ir²/a².
3. Field Formula: B = μ0Ir/(2πa²).
4. Final Result: Inside field increases linearly with r.

33. How does magnetic field vary outside a thick current-carrying wire?

Outside the wire, magnetic field varies inversely with distance from centre. It follows B ∝ 1/r.

1. Region: r > a.
2. Enclosed Current: Ie = I.
3. Field Formula: B = μ0I/(2πr).
4. Final Result: Outside field decreases as 1/r.

Solenoid Class 12 Physics Questions

A long solenoid produces a strong and nearly uniform magnetic field inside it.
Outside field becomes very weak for an ideal long solenoid.
These solenoid Class 12 Physics questions cover field formula and NCERT numericals.

34. What is a solenoid in Class 12 Physics?

A solenoid is a long wire wound closely in helical turns. Each turn behaves like a circular current loop.

1. Structure: Closely wound helical coil.
2. Current: Same current passes through all turns.
3. Field: Strong and uniform inside a long solenoid.
4. Final Result: A solenoid produces an axial magnetic field.

35. What is magnetic field inside a long solenoid?

The magnetic field inside a long solenoid is B = μ0nI. Here n is turns per unit length.

1. Turns Density: n = N/l.
2. Current: I.
3. Formula Used: B = μ0nI.
4. Final Result: Long solenoid field equals μ0nI.

36. A solenoid has 500 turns in 0.5 m and current 5 A. Find B.

The magnetic field is 6.28 × 10^-3 T. Use B = μ0nI.

1. Given Data:
   N = 500
   l = 0.5 m
   I = 5 A
2. Turns Density:
   n = N/l = 500/0.5 = 1000 m^-1
3. Formula Used: B = μ0nI.
4. Calculation:
   B = 4π × 10^-7 × 1000 × 5
   B = 6.28 × 10^-3 T
5. Final Result: B = 6.28 mT.

37. Why is field inside a long solenoid uniform?

The field inside is uniform because contributions from many turns add along the axis. The exterior field nearly cancels.

1. Inside Region: Axial components add.
2. Outside Region: Field lines spread and weaken.
3. Long Solenoid Limit: End effects become small.
4. Final Result: A long solenoid gives uniform internal field.

Force Between Parallel Currents Class 12 Questions

Two current-carrying wires exert magnetic forces on each other.
Parallel currents attract, while anti-parallel currents repel.
These force between parallel currents Class 12 questions also explain the old ampere definition.

38. What is force per unit length between two parallel currents?

The force per unit length is f = μ0IaIb/(2πd). It acts attractively for parallel currents.

1. Currents: Ia and Ib.
2. Separation: d.
3. Formula Used: f = μ0IaIb/(2πd).
4. Final Result: Parallel current force varies as 1/d.

39. Why do parallel currents attract each other?

Parallel currents attract because each wire lies in the magnetic field of the other. The Lorentz force points towards the other wire.

1. Wire A: Produces magnetic field at wire B.
2. Wire B: Experiences force due to that field.
3. Same Direction Currents: Force points inward.
4. Final Result: Parallel currents attract.

40. Why do anti-parallel currents repel?

Anti-parallel currents repel because the magnetic force reverses when one current direction reverses. The wires push away.

1. Current Direction: Opposite in the two wires.
2. Magnetic Field: Same source-field idea applies.
3. Force Direction: Reverses compared with parallel case.
4. Final Result: Anti-parallel currents repel.

41. Two wires carry 8 A and 5 A, separated by 4 cm. Find force on 10 cm length.

The force is 2 × 10^-5 N. Use F = μ0IaIbL/(2πd).

1. Given Data:
   Ia = 8 A
   Ib = 5 A
   d = 4 cm = 0.04 m
   L = 10 cm = 0.10 m
2. Formula Used: F = μ0IaIbL/(2πd).
3. Calculation:
   F = (4π × 10^-7 × 8 × 5 × 0.10)/(2π × 0.04)
   F = 2 × 10^-5 N
4. Final Result: Force = 2 × 10^-5 N.

42. How was ampere defined using parallel current force?

Ampere was defined using force between two long parallel wires. A 1 A current in each wire gives 2 × 10^-7 N/m at 1 m separation.

1. Wire Condition: Very long and parallel wires.
2. Separation: 1 m.
3. Force Per Length: 2 × 10^-7 N/m.
4. Final Result: This gives the old force-based ampere definition.

Torque on Current Loop Class 12 Questions

A current loop in a uniform magnetic field experiences zero net force but a torque.
The torque tends to align the magnetic moment with the field.
These torque on current loop Class 12 questions cover coil torque and magnetic dipole moment.

43. What is torque on a current loop in a magnetic field?

The torque on a current loop is τ = m × B. Its magnitude is τ = NIAB sin θ.

1. Magnetic Moment: m = NIA.
2. Vector Formula: τ = m × B.
3. Magnitude: τ = NIAB sin θ.
4. Final Result: Torque aligns the loop with magnetic field.

44. When is torque on a current loop maximum?

Torque is maximum when the magnetic moment is perpendicular to the magnetic field. The angle θ equals 90°.

1. Formula Used: τ = NIAB sin θ.
2. Maximum Case: sin 90° = 1.
3. Maximum Torque: τmax = NIAB.
4. Final Result: Maximum torque equals NIAB.

45. When is torque on a current loop zero?

Torque is zero when magnetic moment is parallel or anti-parallel to the magnetic field. The angle is 0° or 180°.

1. Formula Used: τ = mB sin θ.
2. Parallel Case: sin 0° = 0.
3. Anti-parallel Case: sin 180° = 0.
4. Final Result: Torque becomes zero in both alignments.

46. A square coil has side 10 cm, 20 turns, 12 A current, and B = 0.80 T at 30°. Find torque.

The torque is 0.96 N m. Use τ = NIAB sin θ.

1. Given Data:
   N = 20
   side = 10 cm = 0.10 m
   A = 0.01 m²
   I = 12 A
   B = 0.80 T
   θ = 30°
2. Formula Used: τ = NIAB sin θ.
3. Calculation:
   τ = 20 × 12 × 0.01 × 0.80 × 0.5
   τ = 0.96 N m
4. Final Result: τ = 0.96 N m.

47. What is magnetic dipole moment of a current loop?

Magnetic dipole moment of a current loop is m = IA. For N turns, m = NIA.

1. Current: I.
2. Area Vector: A.
3. N-Turn Coil: m = NIA.
4. Final Result: Magnetic moment unit is A m².

Moving Coil Galvanometer Class 12 Questions

A moving coil galvanometer measures small currents through magnetic torque on a coil.
Its radial magnetic field keeps torque proportional to current.
These moving coil galvanometer Class 12 questions cover principle, sensitivity, and conversion.

48. What is the principle of moving coil galvanometer?

A moving coil galvanometer works on torque on a current-carrying coil in a magnetic field. The deflection is proportional to current.

1. Magnetic Torque: τ = NIAB.
2. Restoring Torque: τ = kφ.
3. Equilibrium: kφ = NIAB.
4. Final Result: Galvanometer deflection is proportional to current.

49. Why does moving coil galvanometer use radial magnetic field?

It uses radial magnetic field to keep sin θ equal to 1. The torque remains proportional to current.

1. Torque Formula: τ = NIAB sin θ.
2. Radial Field: Plane of coil stays parallel to field.
3. Effect: sin θ = 1.
4. Final Result: Radial field gives linear scale.

50. What is current sensitivity of a galvanometer?

Current sensitivity is deflection per unit current. Its formula is φ/I = NAB/k.

1. Equilibrium: kφ = NIAB.
2. Rearrange: φ/I = NAB/k.
3. Meaning: Larger value gives more deflection.
4. Final Result: Current sensitivity = NAB/k.

51. How is a galvanometer converted into an ammeter?

A galvanometer becomes an ammeter by connecting a small shunt resistance in parallel. Most current passes through the shunt.

1. Problem: Galvanometer carries only small current.
2. Shunt: Low resistance in parallel.
3. Effect: Total instrument resistance becomes small.
4. Final Result: Ammeter needs low resistance.

52. How is a galvanometer converted into a voltmeter?

A galvanometer becomes a voltmeter by connecting a large resistance in series. It then draws very small current.

1. Connection: Voltmeter connects in parallel.
2. Series Resistance: Large resistance limits current.
3. Effect: Circuit disturbance remains small.
4. Final Result: Voltmeter needs high resistance.

53. Why does increasing turns increase current sensitivity?

Increasing turns increases current sensitivity because magnetic torque becomes larger for the same current. Torque is proportional to N.

1. Torque Formula: τ = NIAB.
2. More Turns: Larger N.
3. Deflection: φ = NIAB/k.
4. Final Result: Current sensitivity increases with N.

NCERT Class 12 Physics Chapter 4 Questions

NCERT questions combine direct formulas with vector directions and unit conversions.
Students should check whether the problem uses force, field, torque, or galvanometer sensitivity.
These NCERT Class 12 Physics Chapter 4 questions follow the 2026 exercise pattern.

54. A circular coil has 100 turns, radius 8 cm, and current 0.40 A. Find centre field.

The centre field is 3.14 × 10^-4 T. Use B = μ0NI/(2R).

1. Given Data:
   N = 100
   R = 8 cm = 0.08 m
   I = 0.40 A
2. Formula Used: B = μ0NI/(2R).
3. Calculation:
   B = (4π × 10^-7 × 100 × 0.40)/(2 × 0.08)
   B = 3.14 × 10^-4 T
4. Final Result: B = 3.14 × 10^-4 T.

55. A wire carries 8 A at 30° to a 0.15 T field. Find force per unit length.

The force per unit length is 0.60 N/m. Use f = IB sin θ.

1. Given Data:
   I = 8 A
   B = 0.15 T
   θ = 30°
2. Formula Used: f = IB sin θ.
3. Calculation:
   f = 8 × 0.15 × 0.5
   f = 0.60 N/m
4. Final Result: f = 0.60 N/m.

56. A solenoid is 80 cm long with 2000 turns and current 8 A. Find field near centre.

The field is 2.51 × 10^-2 T. Use B = μ0nI.

1. Given Data:
   N = 2000
   l = 0.80 m
   I = 8 A
2. Turns Density:
   n = N/l = 2000/0.80 = 2500 m^-1
3. Formula Used: B = μ0nI.
4. Calculation:
   B = 4π × 10^-7 × 2500 × 8
   B = 2.51 × 10^-2 T
5. Final Result: B = 2.51 × 10^-2 T.

57. The horizontal Earth field is 3 × 10^-5 T. A 1 A wire runs east-west. Find force per length.

The force per unit length is 3 × 10^-5 N/m. The force acts downward.

1. Given Data:
   I = 1 A
   B = 3 × 10^-5 T
   θ = 90°
2. Formula Used: f = IB sin θ.
3. Calculation:
   f = 1 × 3 × 10^-5 × 1
   f = 3 × 10^-5 N/m
4. Final Result: f = 3 × 10^-5 N/m downward.

58. What force per length acts when the same wire runs south-north?

The force per unit length is zero. Current and magnetic field are parallel.

1. Given Data: θ = 0°.
2. Formula Used: f = IB sin θ.
3. Calculation: f = IB sin 0° = 0.
4. Final Result: Force per unit length = 0.

Class 12 Physics Chapter-Wise Important Questions