Class 12 Physics • Chapter 4 • 2026–27
Moving Charges and Magnetism — Important Questions
Concepts • MCQs • Short Answers • Derivations • Numericals • Competency Practice
Use this page after studying the chapter to test whether you can recall concepts, choose the correct formula, handle directions and solve board-style problems without relying on memorised answers.
Moving Charges and Magnetism is Chapter 4 of Class 12 Physics and belongs to Unit III, Magnetic Effects of Current and Magnetism. In the CBSE 2026–27 curriculum, Unit III carries 17 marks for Chapters 4 and 5 together; the curriculum does not assign a separate fixed mark allocation to Chapter 4. This page therefore focuses on broad, syllabus-aligned practice rather than claiming a fixed “chapter weightage”.
These are original exam-oriented practice questions. They are not presented as official CBSE questions or guaranteed predictions. Use the latest CBSE curriculum and Sample Question Paper as the final reference.
Self-Test First
Board-Style Practice
Best practice: Cover the answer area with your hand or scroll past it before attempting each question. The answers are provided for checking, not for first-pass memorisation.
2026–27 Focus
Original Practice
Answers Included
Numericals
Derivations
Current syllabus focus: magnetic field and Oersted experiment; Biot–Savart law and circular loop; Ampere’s law and infinitely long straight wire; straight solenoid (qualitative treatment); force on a moving charge in uniform electric and magnetic fields; force on a current-carrying conductor; force between parallel current-carrying conductors and definition of ampere; torque on a current loop; moving-coil galvanometer, current sensitivity, and conversion to ammeter and voltmeter.
How to Use These Important Questions
Do not memorise the answers alone. For every question, first identify the concept, direction rule, formula, assumptions and unit. Then compare your solution with the answer.
- Finish the chapter notes first.
- Attempt Section A without looking at the answers.
- Attempt all written sections in your notebook under a time limit.
- For numericals, show formula, substitution, unit and final answer.
- Revise the mistakes before attempting the chapter test.
Section A — 1-Mark Conceptual Questions
Rapid-recall questions covering definitions, direction rules, basic formulae and core concepts.
Q1. What experiment demonstrated the magnetic effect of electric current?
Answer: Oersted’s experiment. It showed that a current-carrying conductor produces a magnetic effect.
Q2. A charged particle moves parallel to a uniform magnetic field. What is the magnetic force on it?
Answer: Zero, because F = qvB sin θ and θ = 0°.
Q3. Under what condition is the magnetic force on a moving charge maximum?
Answer: When the velocity is perpendicular to the magnetic field, i.e. θ = 90°.
Q4. Does a magnetic field do work on a charged particle moving in it?
Answer: No. The magnetic force is perpendicular to the instantaneous velocity, so it does no work.
Q5. Write the SI unit of magnetic field.
Answer: Tesla (T).
Q6. Write the magnetic force on a charge q moving with velocity v in electric and magnetic fields E and B.
Answer: F = q(E + v × B).
Q7. What is the shape of the path of a charged particle entering a uniform magnetic field perpendicular to the field?
Answer: Circular path.
Q8. What happens to the speed of a charged particle when only a magnetic field acts on it?
Answer: Its speed remains constant because the magnetic field does no work.
Q9. State the direction of the magnetic field around a straight current-carrying conductor.
Answer: It forms concentric circles around the conductor; its direction is given by the right-hand thumb rule.
Q10. What is the SI unit of magnetic dipole moment of a current loop?
Answer: A m².
Q11. Write the expression for the magnetic dipole moment of a plane current loop.
Answer: m = NIA, where N is the number of turns, I the current and A the area.
Q12. What is the condition for maximum torque on a current-carrying loop in a uniform magnetic field?
Answer: The angle between the magnetic moment and magnetic field must be 90°.
Section B — Multiple Choice Questions
Choose the option first; then use the explanation to identify the exact concept or relationship being tested.
Q13. A charged particle enters a uniform magnetic field with velocity perpendicular to the field. Which quantity remains unchanged?
A. Direction of velocity
B. Speed
C. Momentum direction
D. Radius if the magnetic field changes
Answer: B. The magnetic force changes the direction of velocity but not its magnitude.
Q14. The magnetic field at the centre of a circular loop of radius R carrying current I is:
A. μ₀I/(2R)
B. μ₀I/(2πR)
C. μ₀IR/2
D. μ₀I/R²
Answer: A.
Q15. For a long straight current-carrying conductor, magnetic field at distance r is proportional to:
A. r
B. r²
C. 1/r
D. 1/r²
Answer: C.
Q16. The force on a current-carrying conductor of length L in a magnetic field B is maximum when the angle between L and B is:
A. 0°
B. 30°
C. 60°
D. 90°
Answer: D.
Q17. Two long parallel wires carrying currents in the same direction:
A. Repel each other
B. Attract each other
C. Exert no force
D. Always rotate
Answer: B.
Q18. The current sensitivity of a moving-coil galvanometer is increased by:
A. Decreasing the number of turns
B. Decreasing the area of the coil
C. Increasing the number of turns
D. Increasing the torsional constant
Answer: C. Since current sensitivity is proportional to N A B/k.
Q19. To convert a galvanometer into an ammeter, the required resistance is connected:
A. In series and is large
B. In parallel and is small
C. In series and is small
D. In parallel and is large
Answer: B.
Q20. To convert a galvanometer into a voltmeter, the required resistance is connected:
A. In series and is large
B. In parallel and is small
C. In series and is small
D. In parallel and is large
Answer: A.
Section C — 2-Mark Questions
Practise concise explanations, laws, conditions and short derivations with the essential equation included.
Q21. State Biot–Savart law and explain the meaning of each physical quantity in its magnitude form.
Answer: dB = (μ₀/4π) (I dl sin θ/r²). Here dB is the magnetic field due to current element I dl, r is the distance from the element to the observation point, θ is the angle between dl and the line joining the element to the point, and μ₀ is the permeability of free space. Direction is given by the cross product dl × r̂.
Q22. Write the magnetic field due to a long straight current-carrying conductor and state its direction.
Answer: B = μ₀I/(2πr). The field lines are concentric circles around the conductor, with direction given by the right-hand thumb rule.
Q23. Why does a charged particle move in a circular path in a uniform magnetic field when its velocity is perpendicular to the field?
Answer: The magnetic force is perpendicular to the velocity and acts as the centripetal force: qvB = mv²/r. Hence r = mv/(|q|B).
Q24. State Ampere’s circuital law and give its integral form.
Answer: The line integral of magnetic field around a closed path equals μ₀ times the net current enclosed: ∮ B·dl = μ₀ Ienclosed.
Q25. Why is a radial magnetic field used in a moving-coil galvanometer?
Answer: It keeps the plane of the coil parallel to the magnetic field, so the coil's normal (and hence magnetic moment) remains perpendicular to the field for all deflections. Thus magnetic torque remains proportional to current and the current–deflection relation is linear.
Q26. Distinguish between current sensitivity and voltage sensitivity of a galvanometer.
Answer: Current sensitivity is deflection per unit current, SI = θ/I = NBA/k. Voltage sensitivity is deflection per unit potential difference, SV = θ/V. Since V = IG for the galvanometer, SV = SI/G; therefore voltage sensitivity also depends on the galvanometer resistance G.
Q27. A charged particle enters a magnetic field at an angle other than 0° or 90°. What type of path does it follow?
Answer: A helical path. The velocity component perpendicular to B produces circular motion, while the component parallel to B remains unchanged, producing forward motion along the field.
Q28. Why do two parallel current-carrying conductors carrying currents in the same direction attract each other?
Answer: Each conductor produces a magnetic field at the location of the other. The magnetic force on each conductor due to the other is directed towards the other conductor when the currents are parallel and in the same direction.
Section D — 3-Mark Questions
Write these in logical steps: principle or law → equation → substitution/derivation → conclusion.
Q29. Derive the radius and time period of a charged particle moving perpendicular to a uniform magnetic field.
Answer:
Magnetic force provides centripetal force:
qvB = mv²/r
Therefore,
r = mv/(|q|B)
The time period is:
T = 2πr/v = 2πm/(|q|B)
Thus, for a non-relativistic particle, T is independent of speed.
Q30. Obtain the expression for the magnetic field at the centre of a circular coil of N turns and radius R carrying current I.
Answer: Applying Biot–Savart law to every element of the circular loop and using symmetry, the field due to one complete turn at the centre is
B = μ₀I/(2R). For N turns, fields add:
B = μ₀NI/(2R)
Direction is perpendicular to the plane of the loop according to the right-hand rule.
Q31. A current-carrying rectangular loop is placed in a uniform magnetic field. Explain why its net force can be zero while a torque acts on it.
Answer: Opposite sides experience equal and opposite magnetic forces, so the vector sum of forces can be zero. Because the forces act along different lines, they form a couple and produce torque. The magnitude is τ = NIAB sin θ.
Q32. Explain how a moving-coil galvanometer is converted into an ammeter.
Answer: A low resistance shunt S is connected in parallel with the galvanometer of resistance G. If I is the desired full-scale current and I
g is the galvanometer full-scale current, then the shunt carries I − I
g. Since the potential difference across the galvanometer and shunt is the same:
IgG = (I − Ig)S
Hence,
S = IgG/(I − Ig)
Section E — 3-Mark Numericals
For exam practice, write the formula, substitute SI values, retain units and state the final answer clearly.
Q33. A long straight wire carries a current of 10 A. Calculate the magnetic field at a point 5 cm from the wire. Take μ₀ = 4π × 10
−7 T m A
−1.
Solution:
B = μ₀I/(2πr)
Here I = 10 A and r = 0.05 m.
B = (4π × 10⁻⁷ × 10)/(2π × 0.05) = 4 × 10⁻⁵ T
Answer: 4 × 10
−5 T.
Q34. A circular coil has 50 turns, radius 0.10 m and carries a current of 2 A. Find the magnetic field at its centre.
Solution:
B = μ₀NI/(2R)
B = (4π × 10⁻⁷ × 50 × 2)/(2 × 0.10)
B = 2π × 10⁻⁴ T ≈ 6.28 × 10⁻⁴ T
Answer: 6.28 × 10
−4 T.
Q35. A proton moves perpendicular to a uniform magnetic field of 0.20 T with speed 4.0 × 10
6 m/s. Calculate the radius of its path. Take m
p = 1.7 × 10
−27 kg and e = 1.6 × 10
−19 C.
Solution:
r = mv/(qB)
r = (1.7 × 10⁻²⁷ × 4.0 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.20)
r = 0.2125 m
Answer: 0.2125 m.
Q36. A galvanometer has resistance 50 Ω and gives full-scale deflection at 2 mA. Calculate the shunt resistance needed to convert it into a 2 A ammeter.
Solution:
S = IgG/(I − Ig)
S = (0.002 × 50)/(2 − 0.002) ≈ 0.05005 Ω
Answer: Approximately 0.050 Ω.
Q37. A rectangular coil of 100 turns has area 0.02 m² and carries 0.50 A in a uniform magnetic field of 0.40 T. Find the maximum torque on the coil.
Solution:
Maximum torque occurs for sin θ = 1:
τmax = NIAB
τmax = 100 × 0.50 × 0.02 × 0.40 = 0.40 N m
Answer: 0.40 N m.
Section F — 5-Mark / Long-Answer Practice
Practise complete board-style responses with equations and the reasoning connecting each step.
Q38. Using Ampere’s circuital law, obtain the magnetic field due to an infinitely long straight current-carrying conductor.
Answer: Choose a circular Amperian path of radius r centred on the conductor. By symmetry, B has the same magnitude at every point of the path and is tangential to it. Therefore:
∮ B·dl = B(2πr) = μ₀I
Hence:
B = μ₀I/(2πr)
The direction is tangential to the circular field lines and is determined by the right-hand thumb rule.
Q39. Derive the expression for torque on a rectangular current loop in a uniform magnetic field and relate it to magnetic dipole moment.
Answer: For a rectangular loop carrying current I in a uniform field B, the forces on two opposite sides form a couple. The net force is zero, but the torque is:
τ = NIAB sin θ
Define magnetic dipole moment:
m = NIA
Therefore:
τ = mB sin θ
The vector form is:
⃗τ = ⃗m × ⃗B
The torque tends to align the magnetic dipole moment with the magnetic field.
Q40. Derive the working equation of a moving-coil galvanometer and obtain its current sensitivity. Explain the principle of conversion into a voltmeter.
Answer: In a radial magnetic field, magnetic torque is:
τm = NIAB
Restoring torque of the suspension is:
τr = kθ
At equilibrium:
NIAB = kθ
Thus:
θ/I = NAB/k
This is the current sensitivity. To convert the galvanometer into a voltmeter, a large resistance R is connected in series. If V is the desired range and I
g is the full-scale current:
V = Ig(G + R)
Therefore:
R = V/Ig − G
Q41. Explain the force between two long parallel current-carrying conductors and state the basis of the traditional definition of ampere.
Answer: Each conductor produces a magnetic field at the position of the other. For two parallel conductors separated by distance d and carrying currents I₁ and I₂:
B₁ = μ₀I₁/(2πd)
Force on length L of the second conductor:
F = I₂LB₁ = μ₀I₁I₂L/(2πd)
Therefore:
F/L = μ₀I₁I₂/(2πd)
Parallel currents in the same direction attract; opposite directions repel. The historical definition of the ampere was based on the force between ideal parallel conductors. The present SI definition is based on the elementary charge e, so students should distinguish the historical definition from the modern SI definition.
Competency & Application Practice
Apply the chapter ideas to comparisons, unfamiliar situations and linked concepts rather than recalling a formula in isolation.
Q42. A proton and an alpha particle enter the same uniform magnetic field with equal speeds perpendicular to the field. Compare their radii of circular paths.
Answer: r = mv/(qB). For the alpha particle, m = 4mp and q = 2e. Hence rα/rp = (4mp/2e)/(mp/e) = 2. Therefore the alpha particle has twice the radius.
Q43. A charged particle passes undeflected through crossed electric and magnetic fields. What relation between E, v and B follows when the electric and magnetic forces oppose each other?
Answer: For zero net force, qE = qvB. Therefore v = E/B, provided the fields and velocity are oriented so that the two forces oppose each other.
Q44. A student says that increasing the magnetic field always increases the speed of a charged particle moving in that field. Is the statement correct? Explain.
Answer: No. A magnetic field alone does no work on the charge. It changes the direction of velocity and, for a given speed, changes the radius of the path, but it does not change the speed.
Case-Based / Competency Practice
These original case-based sets are designed to practise the type of linked, application-focused reasoning used in the current CBSE assessment pattern.
Case Study 1. A charged particle enters a region containing mutually perpendicular electric and magnetic fields. The particle travels undeflected when its speed is adjusted to a particular value. After leaving the selector region, it enters a uniform magnetic field perpendicular to its velocity and follows a circular path.
(a) What condition makes the particle pass undeflected through the crossed fields?
Answer: The electric and magnetic forces must be equal and opposite: qE = qvB.
(b) What is the selected speed?
Answer: v = E/B.
(c) In the second region, what provides the centripetal force?
Answer: The magnetic force provides the centripetal force: |q|vB = mv²/r.
(d) Write the radius of the circular path.
Answer: r = mv/(|q|B).
Case Study 2. A moving-coil galvanometer is used in a laboratory. Its coil has N turns, area A and is placed in a radial magnetic field B. The suspension has torsional constant k. The instrument is first used as a galvanometer and is then converted into an ammeter and a voltmeter.
(a) Write the equilibrium relation for the galvanometer.
Answer: NIAB = kθ.
(b) Write its current sensitivity.
Answer: SI = θ/I = NBA/k.
(c) How is it converted into an ammeter?
Answer: A low resistance shunt is connected in parallel with the galvanometer.
(d) How is it converted into a voltmeter?
Answer: A high resistance is connected in series with the galvanometer.
Important: “Important Questions” does not mean guaranteed questions. The purpose of this page is structured practice across the current syllabus and common board-style question forms. Official CBSE documents remain the final authority.
2026–27 syllabus boundary: This page excludes older Chapter 4 topics that are not part of the current CBSE scope. Straight solenoid is retained only at qualitative level.
Quick Revision Checklist
| Topic | Can you do it without notes? |
|---|
| Lorentz force and direction | □ |
| Charged particle in uniform B: radius, time period and path | □ |
| Velocity selector / crossed E and B fields | □ |
| Biot–Savart law and circular loop | □ |
| Ampere’s law and long straight wire | □ |
| Straight solenoid — qualitative treatment | □ |
| Force on current-carrying conductor | □ |
| Force between parallel conductors | □ |
| Torque and magnetic dipole moment | □ |
| Moving-coil galvanometer | □ |
| Current sensitivity and instrument conversion | □ |
Continue Your Chapter 4 Preparation
Final Exam-Readiness Check
✓ I can explain the direction of magnetic force using the appropriate rule.
✓ I can derive and use r = mv/(|q|B) and T = 2πm/(|q|B).
✓ I can calculate magnetic fields for a long straight wire and circular loop.
✓ I can apply Ampere’s law correctly.
✓ I can solve force, torque and galvanometer-conversion numericals.
✓ I can solve case-based questions by linking two or more chapter concepts.
✓ I can distinguish original practice questions from official CBSE/PYQ questions.
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