Moving Charges and Magnetism Class 12 Numericals 2026-27

CBSE Class 12 Physics Chapter 4 2026–27

Moving Charges and Magnetism — Numericals

Step-by-step numerical practice covering the current CBSE syllabus, with formulas, substitutions, units and final answers.

Why these numericals matter: The official CBSE 2026–27 Physics curriculum places Chapter 4 inside Unit III, “Magnetic Effects of Current and Magnetism”, with 17 marks assigned to the unit containing Chapters 4 and 5. The curriculum also explicitly emphasises problem-solving and applications of Physics concepts. This page therefore focuses on calculation skills that directly match the listed Chapter 4 content rather than adding older, out-of-scope topics.

Skill area	Practice included here
Magnetic force & charged-particle motion	Force, circular motion, crossed E–B fields
Magnetic field of current	Long straight wire and circular coil
Force between currents	Conductor force and parallel-current force
Magnetic dipole & torque	Dipole moment and torque on a current loop
Moving-coil galvanometer	Current sensitivity, ammeter and voltmeter conversion

2026–27 syllabus alert: Chapter 4 is part of Unit III, which carries 17 marks collectively with Chapter 5; CBSE does not assign a fixed standalone mark allocation to Chapter 4. The current syllabus includes magnetic force, Biot–Savart law for a circular loop, Ampere’s law for an infinitely long straight wire, force on conductors, parallel currents, torque on a current loop, magnetic dipole moment and moving-coil galvanometer. Straight solenoid is listed for qualitative treatment only. These practice numericals therefore do not use solenoid calculations or cyclotron questions.

How to use this page: First try each question without opening the solution. Then compare your working line by line. In board-style Physics numericals, marks are often earned through the correct formula, substitution, unit conversion and final unit—not only the final number.

Complete Notes Important Questions MCQs Case-Based Questions Assertion–Reason PYQs CBSE 2026–27 Physics Curriculum CBSE Class XII 2026–27 SQP & Marking Scheme

Essential Formula Bank

Concept	Formula
Magnetic force on charge	F = |q|vB sinθ
Lorentz force	F⃗ = q(E⃗ + v⃗ × B⃗)
Field due to long straight wire	B = μ0I / (2πr)
Field at centre of N-turn circular coil	B = μ0NI / (2R)
Force on current-carrying conductor	F = BIL sinθ
Force per unit length between parallel currents	F/L = μ0I1I2 / (2πd)
Magnetic dipole moment of current loop	m = NIA
Torque on current loop	τ = NIAB sinθ = mB sinθ
Galvanometer → ammeter	S = IgG / (I − Ig)
Galvanometer → voltmeter	R = V/Ig − G

Use: μ₀ = 4π × 10⁻⁷ T m A⁻¹, e = 1.6 × 10⁻¹⁹ C, m_e = 9.1 × 10⁻³¹ kg and m_p = 1.7 × 10⁻²⁷ kg when required.

Section A — Foundation Numericals

1. Magnetic force on a moving electron

2–3 marks

An electron moves with speed 4.0 × 10⁶ m/s perpendicular to a uniform magnetic field of 0.25 T. Calculate the magnitude of the magnetic force on it. Also find the radius of its circular path.

Given: |q| = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg, v = 4.0 × 10⁶ m/s, B = 0.25 T, θ = 90°.

Force: F = qvB sin90° = (1.6 × 10⁻¹⁹)(4.0 × 10⁶)(0.25).

F = 1.6 × 10⁻¹³ N

Radius: qvB = mv²/r, so r = mv/(qB).

r = (9.1 × 10⁻³¹ × 4.0 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.25).

r = 9.1 × 10⁻⁵ m

2. Field due to a long straight conductor

2 marks

A long straight wire carries a current of 12 A. Find the magnetic field at a point 15 cm from the wire.

r = 15 cm = 0.15 m.

B = μ₀I/(2πr) = [(4π × 10⁻⁷) × 12]/[2π × 0.15].

B = 1.6 × 10⁻⁵ T

Direction: Around the wire, given by the right-hand thumb rule.

3. Field at the centre of a circular coil

2 marks

A circular coil has 50 turns, radius 10 cm and carries 2.0 A current. Calculate the magnetic field at its centre.

R = 10 cm = 0.10 m.

B = μ₀NI/(2R).

B = [(4π × 10⁻⁷) × 50 × 2]/(2 × 0.10).

B = 6.28 × 10⁻⁴ T

4. Force per unit length between parallel currents

2 marks

Two long parallel conductors carry currents 8 A and 5 A in the same direction. Their separation is 20 cm. Find the force per metre length between them.

d = 0.20 m.

F/L = μ₀I₁I₂/(2πd).

F/L = [(4π × 10⁻⁷) × 8 × 5]/(2π × 0.20).

F/L = 4.0 × 10⁻⁵ N m⁻¹

Because the currents are in the same direction, the force is attractive.

5. Force on a current-carrying conductor

2 marks

A straight conductor of length 30 cm carries 4 A current in a uniform magnetic field of 0.50 T. The conductor makes an angle of 30° with the field. Find the magnetic force.

L = 0.30 m.

F = BIL sinθ = (0.50)(4)(0.30)(sin30°).

F = 0.30 N

Section B — Board-Style Application

6. Proton accelerated through a potential difference

3 marks

A proton is accelerated from rest through a potential difference of 2.0 kV and then enters a uniform magnetic field of 0.20 T perpendicular to its velocity. Calculate its speed and the radius of its circular path.

Given: m_p = 1.7 × 10⁻²⁷ kg, e = 1.6 × 10⁻¹⁹ C.

Kinetic energy gained = qV.

½mv² = qV, so v = √(2qV/m).

V = 2.0 × 10³ V.

v = √[2(1.6 × 10⁻¹⁹)(2.0 × 10³)/(1.7 × 10⁻²⁷)]

v ≈ 6.14 × 10⁵ m/s

For perpendicular entry, r = mv/(qB).

r = [(1.7 × 10⁻²⁷)(6.14 × 10⁵)]/[(1.6 × 10⁻¹⁹)(0.20)].

r ≈ 3.26 × 10⁻² m = 3.26 cm

7. Crossed electric and magnetic fields

2 marks

A charged particle passes undeflected through mutually perpendicular electric and magnetic fields. If E = 3.0 × 10⁴ N/C and B = 0.20 T, find the speed of the particle. Assume the electric and magnetic forces oppose each other.

For no deflection, qE = qvB.

Therefore, v = E/B = (3.0 × 10⁴)/(0.20).

v = 1.5 × 10⁵ m/s

8. Torque on a current loop

3 marks

A coil of 100 turns and area 2.0 × 10⁻² m² carries a current of 0.50 A. It is placed in a uniform magnetic field of 0.30 T such that the normal to the coil makes 60° with the field. Calculate the torque.

τ = NIAB sinθ.

τ = (100)(0.50)(2.0 × 10⁻²)(0.30)(sin60°).

τ ≈ 0.260 N m

9. Magnetic dipole moment and torque

3 marks

A 100-turn coil has area 3.0 × 10⁻² m² and carries a current of 0.50 A. It is placed in a 0.30 T magnetic field with its magnetic moment perpendicular to the field. Find (i) its magnetic dipole moment and (ii) the torque.

m = NIA = (100)(0.50)(3.0 × 10⁻²).

(i) m = 1.5 A m²

Since m is perpendicular to B, θ = 90°.

τ = mB sin90° = (1.5)(0.30).

(ii) τ = 0.45 N m

10. Converting a galvanometer into an ammeter

3 marks

A galvanometer has resistance 50 Ω and gives full-scale deflection at 2 mA. Calculate the shunt resistance required to convert it into an ammeter of range 2 A.

G = 50 Ω, I_g = 0.002 A and I = 2 A.

Shunt S = I_gG/(I − I_g).

S = (0.002 × 50)/(2 − 0.002).

S ≈ 0.0501 Ω

The shunt is connected in parallel with the galvanometer so that most of the current bypasses the delicate coil.

Section C — Mixed and Higher-Application Numericals

11. Converting a galvanometer into a voltmeter

3 marks

A galvanometer of resistance 50 Ω gives full-scale deflection at 2 mA. What resistance should be connected in series to convert it into a voltmeter of range 10 V?

R_total = V/I_g = 10/0.002 = 5000 Ω.

External series resistance R = R_total − G.

R = 5000 − 50.

R = 4950 Ω = 4.95 kΩ

12. Current sensitivity and galvanometer constant

3 marks

A moving-coil galvanometer has 50 turns, coil area 2.0 × 10⁻⁴ m² and is placed in a magnetic field of 0.20 T. Its current sensitivity is 0.050 rad mA⁻¹. Calculate the torsional constant k of its suspension.

Current sensitivity = θ/I = NBA/k.

0.050 rad mA⁻¹ = 50 rad A⁻¹.

k = NBA/(θ/I) = (50)(0.20)(2.0 × 10⁻⁴)/50.

k = 4.0 × 10⁻⁵ N m rad⁻¹

13. Two circular coils at the same centre

3 marks

Two circular coils are concentric and lie in the same plane. Coil 1 has 20 turns, radius 10 cm and carries 2 A. Coil 2 has 10 turns, radius 20 cm and carries 1 A. The currents produce magnetic fields in the same direction at the common centre. Find the resultant magnetic field.

For each coil, B = μ₀NI/(2R).

B₁ = (4π × 10⁻⁷ × 20 × 2)/(2 × 0.10) = 8π × 10⁻⁵ T.

B₂ = (4π × 10⁻⁷ × 10 × 1)/(2 × 0.20) = π × 10⁻⁵ T.

Same direction ⇒ B = B₁ + B₂.

B = 9π × 10⁻⁵ T ≈ 2.83 × 10⁻⁴ T

14. Changing the distance from a straight wire

2 marks

A long straight conductor carries a steady current. At a distance r from it, the magnetic field is B. At what distance from the conductor will the field become B/4?

For a long straight wire, B ∝ 1/r.

Therefore, B′/B = r/r′.

1/4 = r/r′ ⇒ r′ = 4r.

The required distance is 4r.

15. Force on a conductor when its angle changes

3 marks

A straight conductor of length 0.40 m carries a current of 3 A in a uniform magnetic field of 0.50 T. Initially it is perpendicular to the field. It is then rotated so that it makes 30° with the field. Find the force in both cases and the percentage decrease in force.

Initial: F₁ = BIL sin90° = (0.50)(3)(0.40) = 0.60 N.

After rotation: F₂ = BIL sin30° = (0.50)(3)(0.40)(0.5) = 0.30 N.

Percentage decrease = [(F₁ − F₂)/F₁] × 100.

= [(0.60 − 0.30)/0.60] × 100.

F₁ = 0.60 N, F₂ = 0.30 N; percentage decrease = 50%

Numerical-solving checklist

• Write the given quantities with SI units.
• Identify the exact physical situation before choosing a formula.
• Convert cm, kV, mA and other prefixes before substitution.
• For magnetic force, check the angle in sinθ.
• For current loops, distinguish area A from radius R.
• For parallel currents, state whether the force is attractive or repulsive.
• For galvanometer conversion, remember: ammeter → parallel shunt; voltmeter → series resistance.
• Always write the final SI unit and check the order of magnitude.

Common syllabus trap: Many older Chapter 4 resources still include cyclotron, toroid and detailed solenoid numericals. For CBSE 2026–27, the official syllabus specifically says the straight solenoid is qualitative treatment only, while the listed Chapter 4 scope does not include cyclotron. Do not let older question banks silently expand your current board-exam scope.

What to practise next

After completing these numericals, revise the verified Chapter 4 PYQs and then use the Formula Sheet + Quick Revision page and Chapter Test when they are published.

Chapter 4 PYQs Chapter 4 Notes Official CBSE Curriculum

Research and syllabus boundary used for this set

The 2026–27 CBSE curriculum was checked before editing this page. It lists Biot–Savart law for a circular loop, Ampere’s law for an infinitely long straight wire, force on moving charges, force on current-carrying conductors, parallel-current force, torque and magnetic dipole moment, and moving-coil galvanometer applications. It explicitly marks the straight solenoid as qualitative treatment only. The current official CBSE Class XII SQP/MS portal was also checked as the board’s current sample-paper reference point.

Content note: The numerical problems on this page are original practice questions prepared for Learn Revise Hub. They are not presented as verbatim CBSE board questions. The syllabus boundary and assessment references are based on the official CBSE 2026–27 Physics curriculum and sample-paper material.