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Moving Charges and Magnetism Class 12 MCQs with Answers 2026-27
Class 12 PhysicsChapter 42026–2744 MCQsWith Answers
Moving Charges and Magnetism Class 12 MCQs with Answers — a current-syllabus practice set covering magnetic field, Lorentz force, Biot–Savart law, Ampere's law, force on conductors, parallel currents, torque, magnetic dipole and moving-coil galvanometer.
The official CBSE 2026–27 curriculum places Chapter 4 in Unit III, Magnetic Effects of Current and Magnetism. Unit III carries 17 marks collectively for Chapters 4 and 5; CBSE does not publish a fixed Chapter 4-only weightage. These are original Learn Revise Hub MCQs, not official CBSE questions or predictions.
Best way to practise: cover the answer first, solve every question yourself, then read the explanation. Mark questions you got wrong and revisit the related concept before your next attempt.
Current-syllabus coverage: magnetic field and Oersted's experiment; Lorentz force; charged-particle motion; crossed electric and magnetic fields; Biot–Savart law and circular coils; Ampere's law for an infinitely long straight wire; force on current-carrying conductors; force between parallel currents; torque and magnetic dipole; and moving-coil galvanometer current sensitivity and conversion. Straight solenoid is included only at the qualitative level specified by CBSE.
Part A — Core Concept MCQs with Answers
Q1. A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is:
A. qvB
B. qvB/2
C. zero
D. qB/v
Answer: C. F = |q|vB sin θ. For parallel motion, θ = 0°, so F = 0.
Q2. A charged particle enters a uniform magnetic field perpendicular to the field. Which path does it follow?
A. Straight line
B. Circular path
C. Parabolic path
D. Elliptical path
Answer: B. The magnetic force is perpendicular to velocity and provides centripetal force.
Q3. Which quantity remains unchanged when a charged particle moves under the action of a magnetic field alone?
A. Direction of velocity
B. Momentum direction
C. Speed
D. Radius in every magnetic field
Answer: C. Magnetic force is perpendicular to velocity, so it does no work and does not change kinetic energy or speed.
Q4. The magnetic field due to a long straight current-carrying conductor at distance r is proportional to:
A. r
B. r²
C. 1/r
D. 1/r²
Answer: C. B = μ₀I/(2πr).
Q5. The SI unit of magnetic field is:
A. Weber
B. Tesla
C. Henry
D. Coulomb
Answer: B. The SI unit of magnetic flux density B is tesla (T).
Q6. The direction of the magnetic field around a straight current-carrying conductor is given by:
A. Fleming's left-hand rule
B. Right-hand thumb rule
C. Lenz's law
D. Snell's law
Answer: B. Curl the fingers of the right hand around the conductor with the thumb pointing in the direction of conventional current.
Q7. The magnetic field at the centre of a circular coil of N turns, radius R and current I is:
A. μ₀NI/(4πR)
B. μ₀NI/(2R)
C. μ₀NI/(2πR)
D. μ₀NIR/2
Answer: B. B = μ₀NI/(2R).
Q8. For a current element Idl, the magnetic field contribution is maximum when the angle between dl and the line joining the element to the observation point is:
A. 0°
B. 30°
C. 60°
D. 90°
Answer: D. Biot–Savart law contains sin θ, which is maximum at 90°.
Q9. Ampere's circuital law for a closed path is:
A. ∮B·dl = 0 for every path
B. ∮B·dl = μ₀Ienclosed
C. ∮B·dl = q/ε₀
D. ∮B·dl = IR
Answer: B. For magnetostatics, the circulation of B around a closed loop equals μ₀ times the enclosed current.
Q10. Oersted's experiment demonstrated that:
A. a stationary charge produces a magnetic field
B. a current-carrying conductor can deflect a nearby compass needle
C. a magnetic field can create charge
D. resistance is independent of temperature
Answer: B. Oersted observed that a current-carrying conductor produces a magnetic effect capable of deflecting a nearby compass needle.
Part B — Force, Motion and Applications MCQs
Q11. A charged particle moves with velocity v in a magnetic field B. The magnitude of magnetic force is:
A. |q|vB cos θ
B. |q|vB sin θ
C. qv/B
D. qB/v
Answer: B. F = |q|vB sin θ, where θ is the angle between v and B.
Q12. A particle moves in a circular path of radius r in a uniform magnetic field B. Its momentum magnitude is:
A. |q|Br
B. |q|B/r
C. r/(|q|B)
D. |q|/(rB)
Answer: A. For perpendicular motion, |q|vB = mv²/r, so the momentum magnitude is p = mv = |q|Br.
Q13. A charged particle enters a uniform magnetic field at an angle neither 0° nor 90°. Its path is:
A. Circular
B. Straight
C. Helical
D. Parabolic
Answer: C. The perpendicular velocity component causes circular motion and the parallel component causes uniform translation along the field.
Q14. A particle is undeflected while moving through crossed electric and magnetic fields. If the fields oppose each other's force, its speed is:
A. E/B
B. B/E
C. EB
D. E + B
Answer: A. For zero net force, qE = qvB, hence v = E/B.
Q15. A positive charge moves east while the magnetic field is directed upward. The magnetic force is directed:
A. east
B. west
C. north
D. south
Answer: D. For a positive charge, F = q(v × B). Taking east as +x and up as +z, x × z = −y, i.e. south.
Q16. For a long straight current-carrying conductor, the magnetic field at distance r is:
A. directly proportional to r
B. directly proportional to r²
C. inversely proportional to r
D. inversely proportional to r²
Answer: C. Ampere's law gives B = μ₀I/(2πr) for an infinitely long straight wire.
Q17. A current-carrying conductor of length L is placed in a magnetic field B. The force is maximum when the angle between the conductor and field is:
A. 0°
B. 30°
C. 60°
D. 90°
Answer: D. F = BIL sin θ, so maximum force occurs at 90°.
Q18. Two long parallel conductors carrying currents in the same direction:
A. attract each other
B. repel each other
C. exert zero force
D. always rotate about one another
Answer: A. Same-direction parallel currents attract; opposite-direction currents repel.
Part C — Torque, Magnetic Dipole and Galvanometer
Q19. The magnitude of the magnetic dipole moment of a coil of N turns, current I and area A is:
A. NIA
B. NI/A
C. IA/N
D. N/(IA)
Answer: A. m = NIA, with SI unit A m².
Q20. The torque on a magnetic dipole m in a uniform magnetic field B is maximum when:
A. m is parallel to B
B. m is antiparallel to B
C. m is perpendicular to B
D. B is zero
Answer: C. τ = mB sin θ, maximum when θ = 90°.
Q21. For a current loop in a uniform magnetic field, the net force can be zero while the torque is non-zero because:
A. all forces vanish
B. the forces form a couple
C. magnetic field does work
D. current becomes zero
Answer: B. Equal and opposite forces can have different lines of action, producing a torque without a net translational force.
Q22. Current sensitivity of a moving-coil galvanometer is:
A. k/NAB
B. NAB/k
C. N/kAB
D. AB/Nk
Answer: B. At equilibrium, NIAB = kθ, so θ/I = NAB/k.
Q23. To increase the current sensitivity of a galvanometer, which change is suitable?
A. Increase N
B. Increase k
C. Decrease A
D. Decrease B
Answer: A. Since SI = NAB/k, increasing N increases sensitivity, all else unchanged.
Q24. A galvanometer is converted into an ammeter by connecting:
A. a high resistance in series
B. a low resistance in parallel
C. a high resistance in parallel
D. a low resistance in series
Answer: B. The shunt provides a low-resistance parallel path for most of the current.
Q25. A galvanometer is converted into a voltmeter by connecting:
A. a high resistance in series
B. a low resistance in parallel
C. a high resistance in parallel
D. a zero resistance in series
Answer: A. A large series resistance limits the current through the galvanometer.
Part D — Numerical and Reasoning MCQs with Answers
Q26. A long straight wire carries a current of 5 A. The magnetic field at a point 10 cm from the wire is:
A. 1 × 10−5 T
B. 1 × 10−6 T
C. 5 × 10−5 T
D. 5 × 10−6 T
Answer: A. B = μ₀I/(2πr) = (4π×10−7×5)/(2π×0.10) = 1×10−5 T.
Q27. If the current in a circular coil is doubled while its radius remains unchanged, the magnetic field at its centre becomes:
A. half
B. unchanged
C. double
D. four times
Answer: C. B ∝ I.
Q28. If the radius of a circular coil is doubled while current and number of turns remain unchanged, the field at its centre becomes:
A. double
B. half
C. four times
D. unchanged
Answer: B. B ∝ 1/R.
Q29. A proton and an alpha particle enter the same magnetic field with equal speeds, both perpendicular to B. The ratio of their path radii rα/rp is:
A. 1/2
B. 1
C. 2
D. 4
Answer: C. For the alpha particle, m = 4mp and q = 2e, so r ∝ m/q gives rα/rp = 2.
Q30. A 0.20 T magnetic field acts on a 2 A current-carrying wire of length 0.50 m at 90°. The force is:
A. 0.02 N
B. 0.10 N
C. 0.20 N
D. 2.0 N
Answer: C. F = BIL = 0.20 × 2 × 0.50 = 0.20 N.
Q31. A 100-turn coil of area 0.01 m² carries 2 A in a 0.5 T field. Its maximum torque is:
A. 0.05 N m
B. 0.5 N m
C. 1.0 N m
D. 5.0 N m
Answer: C. τmax = NIAB = 100 × 2 × 0.01 × 0.5 = 1.0 N m.
Q32. A galvanometer has G = 50 Ω and full-scale current Ig = 2 mA. The shunt needed for a 2 A ammeter is approximately:
A. 0.005 Ω
B. 0.05 Ω
C. 5 Ω
D. 50 Ω
Answer: B. S = IgG/(I − Ig) = (0.002×50)/(2−0.002) ≈ 0.050 Ω.
Part E — Higher-Order MCQs
Q33. The force per unit length between two long parallel conductors carrying currents I1 and I2, separated by distance d, is:
A. μ₀I1I2/(2πd)
B. μ₀I1I2d/(2π)
C. μ₀d/(2πI1I2)
D. 2πd/(μ₀I1I2)
Answer: A. F/L = μ₀I1I2/(2πd) for long parallel conductors.
Q34. A particle has its velocity component parallel to B doubled while the perpendicular component remains unchanged. In an ideal uniform magnetic field, its circular radius:
A. doubles
B. halves
C. remains unchanged
D. becomes zero
Answer: C. The radius depends on v⊥, not on the velocity component parallel to B.
Q35. Two long parallel wires carry equal currents in opposite directions. The force between them is:
A. attractive
B. repulsive
C. zero in every case
D. perpendicular to the plane containing the wires
Answer: B. Opposite-direction parallel currents repel.
Q36. A moving-coil galvanometer has its torsional constant reduced to half, with N, A and B unchanged. Its current sensitivity becomes:
A. half
B. unchanged
C. double
D. four times
Answer: C. SI = NAB/k; halving k doubles sensitivity.
Q37. In a moving-coil galvanometer, a radial magnetic field is useful mainly because:
A. it makes the magnetic field zero at all positions
B. it keeps the torque relation linear with current
C. it prevents current from flowing
D. it removes the need for a restoring torque
Answer: B. A radial field keeps the relevant angle such that magnetic torque remains proportional to current over the deflection range.
Q38. Two long parallel wires carry equal currents in the same direction. If their separation is doubled, the force per unit length between them becomes:
A. double
B. half
C. four times
D. unchanged
Answer: B. Since F/L ∝ 1/d, doubling the separation halves the force per unit length.
Q39. Which statement about a magnetic field acting alone on a charged particle is correct?
A. It always increases kinetic energy.
B. It always decreases kinetic energy.
C. It can change direction of motion without changing speed.
D. It always brings the particle to rest.
Answer: C. The magnetic force is perpendicular to velocity and therefore does no work.
Q40. A galvanometer is converted into a voltmeter by adding resistance R in series. If the desired voltage range increases while the galvanometer remains unchanged, R should:
A. decrease
B. increase
C. become zero
D. remain necessarily unchanged
Answer: B. R = V/Ig − G; a larger voltage range requires a larger series resistance.
Part F — Assertion–Reason MCQs
Directions: Choose the correct option: (A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is not the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.
Q41. Assertion (A): A magnetic field does no work on a moving charged particle.
Reason (R): Magnetic force is perpendicular to the instantaneous velocity of the particle.
Reason (R): Magnetic force is perpendicular to the instantaneous velocity of the particle.
Answer: A. Both statements are true, and the perpendicularity explains why magnetic force does no work.
Q42. Assertion (A): Two parallel conductors carrying currents in the same direction attract each other.
Reason (R): Each conductor experiences a magnetic force due to the field produced by the other.
Reason (R): Each conductor experiences a magnetic force due to the field produced by the other.
Answer: A. Both are true and the interaction of each current with the other's magnetic field gives the attraction.
Q43. Assertion (A): Increasing the number of turns of a galvanometer coil can increase current sensitivity.
Reason (R): Current sensitivity is proportional to NAB/k.
Reason (R): Current sensitivity is proportional to NAB/k.
Answer: A. Both are true and the expression directly explains the assertion.
Q44. Assertion (A): A charged particle entering a uniform magnetic field parallel to the field follows a circular path.
Reason (R): Magnetic force is qvB sin θ.
Reason (R): Magnetic force is qvB sin θ.
Answer: D. The assertion is false because θ = 0°, so the magnetic force is zero. The reason is true.
Score Yourself
| Score out of 44 | Suggested next step |
|---|---|
| 40–44 | Review only the questions you missed and the formulas behind them. |
| 32–39 | Revisit weak topics, then retry the incorrect questions. |
| 22–31 | Do another concept-revision cycle before taking the Chapter 4 test. |
| Below 22 | Review the chapter notes topic by topic, then attempt this set again. |
Quick Topic Checklist — CBSE 2026–27 Scope
| Topic | Practised? |
|---|---|
| Magnetic field and Oersted's experiment | □ |
| Lorentz force and direction | □ |
| Charged-particle motion in a magnetic field | □ |
| Crossed electric and magnetic fields | □ |
| Biot–Savart law and circular coil | □ |
| Ampere's law and long straight wire | □ |
| Straight solenoid — qualitative treatment | □ |
| Force on current-carrying conductor | □ |
| Parallel current-carrying conductors | □ |
| Torque and magnetic dipole | □ |
| Moving-coil galvanometer | □ |
| Current sensitivity and conversion | □ |
Syllabus boundary: Cyclotron is not part of the CBSE 2026–27 Chapter 4 curriculum, so it is intentionally excluded from the questions. Straight solenoid is retained only at qualitative level.
Continue Chapter 4 Preparation
Complete Notes: Moving Charges and Magnetism Class 12 Physics Notes
Important Questions: Moving Charges and Magnetism Important Questions
Case-Based Questions: Moving Charges and Magnetism Case-Based Questions
Assertion–Reason: Moving Charges and Magnetism Assertion–Reason Questions
PYQs: Moving Charges and Magnetism Previous Year Questions
Numericals: Moving Charges and Magnetism Numericals
Formula Sheet: Moving Charges and Magnetism Formula Sheet
Chapter Test: Moving Charges and Magnetism Class 12 Chapter Test
Official CBSE: Physics Curriculum 2026–27 · Class XII 2026–27 SQP & Marking Scheme
Important: These are original practice MCQs. They are not official CBSE questions, and no question should be treated as a guaranteed board-exam prediction. Use the latest CBSE curriculum and official SQP/MS as the final reference.
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