Moving Charges and Magnetism — Previous Year Questions (Class 12 Physics)
This chapter explores magnetic forces on moving charges, magnetic fields, and electromagnetic induction. It explains how currents create magnetic fields an
TL;DR: This chapter explores magnetic forces on moving charges, magnetic fields, and electromagnetic induction. It explains how currents create magnetic fiel…
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This chapter explores magnetic forces on moving charges, magnetic fields, and electromagnetic induction. It explains how currents create magnetic fields an
Moving Charges and Magnetism — Previous Year Questions with Solutions
Q (2023, 2 marks): An electron moves with velocity 2 × 10^6 m/s perpendicular to a magnetic field of 0.01 T. Find the magnetic force on the electron. (e = 1.6 × 10^-19 C)
Answer: Magnetic force on moving charge: F = qvB sin(θ)
where θ = 90° (perpendicular), so sin(θ) = 1
F = evB
F = 1.6 × 10^-19 × 2 × 10^6 × 0.01
F = 1.6 × 10^-19 × 2 × 10^4
F = 3.2 × 10^-15 N
Q (2022, 2 marks): A straight wire of length 0.5 m carries a current of 2 A and is placed perpendicular to a magnetic field of 0.4 T. Calculate the force on the wire.
Answer: Force on current-carrying wire: F = BIL sin(θ)
where θ = 90° (perpendicular), so sin(θ) = 1
Given: B = 0.4 T, I = 2 A, L = 0.5 m
F = 0.4 × 2 × 0.5
F = 0.4 N
Q (2023, 2 marks): Calculate the magnetic field at the center of a circular coil of radius 10 cm carrying current 5 A. (μ0 = 4π × 10^-7 T·m/A)
Answer: Magnetic field at center of circular coil: B = (μ0 × I) / (2r)
where μ0 = 4π × 10^-7 T·m/A, I = 5 A, r = 0.1 m
B = (4π × 10^-7 × 5) / (2 × 0.1)
B = (20π × 10^-7) / 0.2
B = 100π × 10^-7
B = 100π × 10^-7 T ≈ 3.14 × 10^-5 T
Q (2021, 3 marks): A proton and an alpha particle (He nucleus) enter a magnetic field with the same kinetic energy. Compare their radii of circular paths.
Answer: Radius of circular path in magnetic field: r = mv / (qB)
For same kinetic energy: KE = (1/2)mv² = constant
For proton: m_p = m, q = e, KE = (1/2)m*v_p²
For alpha particle: m_α = 4m, q = 2e, KE = (1/2)(4m)*v_α²
Since KE is same:
(1/2)m*v_p² = (1/2)(4m)*v_α²
v_p² = 4v_α²
v_p = 2v_α
Radius for proton: r_p = m*v_p / (e*B) = m*(2v_α) / (e*B)
Radius for alpha: r_α = (4m)*v_α / (2e*B) = (4m)*v_α / (2e*B) = 2m*v_α / (e*B)
r_p / r_α = [m*2v_α/(e*B)] / [2m*v_α/(e*B)] = 1
Therefore, r_p = r_α (both have equal radii)
Q (2022, 2 marks): A conducting rod of length 1 m moves with velocity 5 m/s perpendicular to a magnetic field of 2 T. Find the induced EMF.
Answer: Induced EMF due to motional EMF: ε = BLv
where B = 2 T, L = 1 m, v = 5 m/s
ε = 2 × 1 × 5
ε = 10 V
Q (2023, 2 marks): Define magnetic flux. A magnetic field of 0.5 T is perpendicular to a surface of area 200 cm². Find the magnetic flux.
Answer: Magnetic flux (Φ) is the number of magnetic field lines passing through a surface.
Φ = B·A = BA cos(θ)
where θ is angle between field and normal to surface
Given: B = 0.5 T, A = 200 × 10^-4 m² = 0.02 m², θ = 0° (perpendicular)
Φ = 0.5 × 0.02 × cos(0°)
Φ = 0.5 × 0.02 × 1
Φ = 0.01 Wb (Weber)
Frequently Asked Questions
What is the right-hand rule and how is it used?
The right-hand rule is used to find the direction of magnetic field or force. For current: thumb points in direction of current, fingers curl in direction of magnetic field. For force: thumb = velocity, fingers = field, palm faces force direction.
Why does a charged particle move in a circle in a uniform magnetic field?
The magnetic force (F = qvB) is always perpendicular to velocity, providing centripetal force. Since direction changes but magnitude remains constant, the particle moves in a circular path.
More Class 12 Physics PYQs
- Current Electricity
- Ray Optics
- Electrostatics
- Electromagnetic Induction
- Alternating Current
- Atoms and Nuclei
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