Moving Charges and Magnetism — Previous Year Questions | eduPhysics
eduPhysics / Notes / Class XII / Moving Charges and Magnetism / Previous Year Questions
Class XII · Chapter 04

Moving Charges and Magnetism — Exam Practice

Practice questions in CBSE, NEET, and JEE Main style covering the Lorentz force, Biot–Savart Law, force between currents, and magnetic moment.

📝 10 questions 🎯 CBSE · NEET · JEE Main
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.

1CBSE Board Exam Style

CBSE · 1 Mark
Q1. A charged particle moves parallel to a uniform magnetic field. Why does it experience no magnetic force?
Answer Magnetic force has magnitude F = qvB sinθ, where θ is the angle between velocity and field. When motion is parallel to B, θ = 0°, so sinθ = 0 and F = 0.
CBSE · 3 Marks
Q2. Using the Biot–Savart Law, derive the expression for the magnetic field at the centre of a circular current-carrying loop of radius R.
Answer By Biot–Savart Law, a current element I dl contributes dB = (μ₀/4π)(I dl × r̂)/r² at the centre. Every element is at the same distance R from the centre, and dl is always perpendicular to r̂ (radius direction), so dB = (μ₀/4π)(I dl)/R² for each element, all pointing in the same direction along the axis.

Integrating around the full loop (∮dl = 2πR): B = (μ₀/4π) × I(2πR)/R² = μ₀I/2R.
CBSE · 5 Marks
Q3. (a) Derive the expression for the force per unit length between two long, parallel current-carrying conductors, and use it to define the SI unit of current, the ampere. (b) Two long parallel wires, 10 cm apart, carry currents of 3 A and 5 A in the same direction. Find the force per unit length between them.
Answer (a) Wire 1 (current I₁) produces a field B₁ = μ₀I₁/2πd at the location of wire 2, a distance d away. The force per unit length on wire 2 is F/L = B₁I₂ = μ₀I₁I₂/(2πd).

Setting I₁ = I₂ = 1 A and d = 1 m gives F/L = 2×10⁻⁷ N/m — this defines one ampere as the current that, flowing in two infinite parallel wires 1 m apart, produces exactly this force per unit length between them.

(b) F/L = μ₀I₁I₂/(2πd) = (2×10⁻⁷ × 3 × 5) / 0.1 = 3×10⁻⁵ N/m, attractive (same-direction currents).

2NEET Style (Single-Correct MCQ)

NEET Style
Q4. A proton and an electron enter a magnetic field with equal speed, both perpendicular to the field. Which has the larger radius of circular path?
✓ The proton
The electron
Both equal
Neither moves in a circle
Solution r = mv/(qB). With equal v, q (magnitude), and B, radius is proportional to mass — the much heavier proton traces a larger circle than the electron.
NEET Style
Q5. The magnetic field inside a long, ideal solenoid is:
Zero
✓ Uniform
Maximum at the ends
Radially outward
Solution Inside a long, tightly-wound solenoid, B = μ₀nI is constant in magnitude and direction throughout the interior, essentially independent of position along the axis (away from the very ends).
NEET Style
Q6. The work done by a magnetic force on a moving charged particle is:
Positive
Negative
✓ Always zero
Depends on field strength
Solution Magnetic force is always perpendicular to velocity (F = qv×B), so it can never do work — it changes only the direction of motion, never speed or kinetic energy.
NEET Style
Q7. The SI unit of magnetic field is:
Weber
✓ Tesla
Henry
Gauss
Solution Tesla (T) is the SI unit; gauss is a related but non-SI (CGS) unit, still commonly seen in some contexts (1 T = 10⁴ G).

3JEE Main Style

JEE Main Style · Numerical
Q8. A proton moving at 1×10⁶ m/s enters a magnetic field of 0.5 T, perpendicular to the field. Find the radius of its circular path. (mp = 1.67×10⁻²⁷ kg, q = 1.6×10⁻¹⁹ C)
Solution r = mv/(qB) = (1.67×10⁻²⁷ × 1×10⁶) / (1.6×10⁻¹⁹ × 0.5) = (1.67×10⁻²¹) / (8×10⁻²⁰) ≈ 0.0209 m ≈ 2.09 cm.
JEE Main Style · MCQ
Q9. Two long, parallel wires carry currents in opposite directions. The force between them is:
Attractive
✓ Repulsive
Zero
Depends on wire length only
Solution Opposite-direction currents repel, same-direction currents attract — the reverse of the like/unlike charge rule, and a frequent point of confusion.
JEE Main Style · Numerical
Q10. A circular coil of 100 turns and radius 5 cm carries a current of 2 A. Find its magnetic moment.
Solution m = NIA = 100 × 2 × π × (0.05)² = 200 × π × 0.0025 ≈ 1.57 A·m².
← Previous · PYQ Set 03

Current Electricity

Back
Up Next · PYQ Set 05

Magnetism and Matter

Continue →

Share this:

Like this:

Like Loading…