Moving Charges and Magnetism Class 12 Chapter Test 2026-27
Moving Charges and Magnetism — Chapter Test
A complete chapter-level test covering concepts, directions, numericals, magnetic fields, force, torque, dipole moment and moving-coil galvanometer.
Test instructions
| Time | 60 minutes |
|---|---|
| Maximum marks | 40 |
| Calculator | Not allowed, following the current CBSE Class XII 2026–27 sample-paper instruction. |
| Suggested attempt | Attempt without notes first. Show formula, substitution, unit and direction wherever required. |
Important: This is an original Learn Revise Hub chapter test, not an official CBSE paper or prediction. The official 2026–27 sample paper retains the 33-question, five-section, 70-mark assessment design for the full Physics paper. This chapter test adapts the same broad skill mix to a single chapter rather than pretending that Chapter 4 has a fixed official mark allocation.
A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is:
The magnetic field at the centre of a circular coil of N turns, radius R and current I is:
Two long parallel conductors carry currents in opposite directions. The force between them is:
A current loop has magnetic dipole moment m. Its maximum torque in a magnetic field B is:
A galvanometer is converted into an ammeter by connecting:
A particle enters a uniform magnetic field with velocity perpendicular to B. If its speed is doubled, its circular-path radius:
Assertion (A): A magnetic field alone cannot change the speed of a charged particle.
Reason (R): Magnetic force is perpendicular to the instantaneous velocity.
Assertion (A): Increasing the torsional constant k of a galvanometer increases its current sensitivity.
Reason (R): Current sensitivity is θ/I = NBA/k.
Assertion (A): A current-carrying conductor experiences maximum magnetic force when it is perpendicular to the magnetic field.
Reason (R): F = BIL sinθ.
Assertion (A): A galvanometer is converted into a voltmeter by connecting a large resistance in parallel.
Reason (R): A voltmeter should have high resistance.
Why does a magnetic field do no work on a moving charged particle? State the consequence for its speed when the magnetic field is the only force acting.
A long straight wire carries 10 A current. Calculate the magnetic field at a point 20 cm away from it. Take μ₀ = 4π × 10−7 T m A−1.
A 50-turn coil of area 0.020 m² carries 0.40 A current. Calculate its magnetic dipole moment.
State the two changes required to convert a galvanometer into (i) an ammeter and (ii) a voltmeter. Mention how the external resistance is connected in each case.
A proton enters a uniform magnetic field of 0.20 T perpendicular to its velocity with speed 4.0 × 105 m/s. Calculate the radius of its circular path.
A circular coil has 100 turns, radius 10 cm and carries a current of 1.5 A. Find the magnetic field at its centre. Show the substitution clearly.
A rectangular current-carrying loop has magnetic dipole moment 0.80 A m². It is placed in a uniform magnetic field of 0.50 T with its magnetic moment at 30° to the field. Find the torque.
A galvanometer has resistance 40 Ω and full-scale deflection current 2 mA. Calculate the series resistance required to convert it into a 10 V voltmeter.
(a) Write the magnitudes of the electric and magnetic forces for v ⟂ B. (1)
(b) Derive the condition for undeflected motion. (1)
(c) If E = 2.4 × 104 N/C and B = 0.20 T, calculate the selected speed. (1)
(d) For a particle entering only the magnetic field perpendicular to B, state the expression for its path radius. (1)
A moving-coil galvanometer has resistance G = 50 Ω and gives full-scale deflection at Ig = 2 mA.
(a) Explain the principle of operation of a moving-coil galvanometer and write the equilibrium relation between magnetic and restoring torque. (2)
(b) Write the expression for current sensitivity and state two ways of increasing it, with other relevant quantities considered independently. (2)
(c) Calculate the resistance of the series resistor required to convert this galvanometer into a voltmeter of range 10 V. (2)
Answer Key & Solutions
Section A — Answers
| Q | Answer | Reason |
|---|---|---|
| 1 | C | F = |q|vB sin0° = 0. |
| 2 | B | B = μ₀NI/(2R). |
| 3 | B | Opposite parallel currents repel. |
| 4 | B | τmax = mB. |
| 5 | B | A low-resistance shunt is connected in parallel. |
| 6 | C | r = mv/(|q|B), so r ∝ v. |
| 7 | A | F ⟂ v, so magnetic force does no work. |
| 8 | D | Reason is true, but increasing k decreases θ/I. |
| 9 | A | sinθ is maximum at 90°. |
| 10 | D | A is false; the large resistance is connected in series. R is true. |
Section B — Answers
Q11: Magnetic force is perpendicular to velocity, so F · v = 0 and work done is zero. Therefore kinetic energy and speed remain unchanged when magnetic force is the only force.
Q12: B = μ₀I/(2πr) = [(4π × 10−7)×10]/[2π×0.20] = 1.0 × 10−5 T.
Q13: m = NIA = 50×0.40×0.020 = 0.40 A m².
Q14: Ammeter: connect a low-resistance shunt in parallel. Voltmeter: connect a high resistance in series.
Section C — Solutions
Q15: r = mv/(eB) = [(1.7×10−27)(4.0×105)]/[(1.6×10−19)(0.20)] = 2.125 × 10−2 m ≈ 2.13 cm.
Q16: B = μ₀NI/(2R) = [(4π×10−7)×100×1.5]/(2×0.10) = 3π×10−4 T ≈ 9.42×10−4 T.
Q17: τ = mB sinθ = 0.80×0.50×sin30° = 0.20 N m.
Q18: R = V/Ig − G = 10/0.002 − 40 = 4960 Ω = 4.96 kΩ.
Section D — Solution
(a) FE = |q|E and FB = |q|vB.
(b) For no deflection, |q|E = |q|vB, hence v = E/B.
(c) v = (2.4×104)/0.20 = 1.2×105 m/s.
(d) r = mv/(|q|B).
Section E — Solution
(a) A current-carrying coil in a magnetic field experiences torque. At equilibrium, magnetic torque balances restoring torque: NIAB = kθ.
(b) Current sensitivity: θ/I = NBA/k. It can be increased by increasing N, A or B, or by decreasing k, with the other quantities held unchanged.
(c) R = V/Ig − G = 10/0.002 − 50 = 4950 Ω = 4.95 kΩ.
Score Guide
| Score | Revision action |
|---|---|
| 34–40 | Chapter-ready. Recheck only errors and formula slips. |
| 28–33 | Good preparation. Revise the topics behind every lost mark. |
| 20–27 | Repeat Formula Sheet + Numericals before attempting the test again. |
| Below 20 | Return to Complete Notes and rebuild the chapter topic-by-topic before retesting. |
Continue the Chapter 4 Workflow
Research note: The current CBSE 2026–27 Physics curriculum and official 2026–27 SQP/MS were checked while preparing this test. The full CBSE sample paper has 33 questions, 70 marks and five sections; this page adapts that assessment style into a focused 40-mark Chapter 4 test.
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