Moving Charges and Magnetism — Class 12 MCQs with Answers

Moving Charges and Magnetism — important questions & MCQs with answers (Class 12 Physics)

14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.

1. Q1Easy

   Force on charge q moving with v in field B:

   A.qvB²

   B.q/v

   C.qB/v

   D.qv × B✓ Correct

   Answer: D. qv × B

   Explanation: Lorentz force law: F = qv × B, a vector (cross) product — force is perpendicular to both velocity and field, unlike a scalar mix like qv/B or qB/v that ignores direction.

2. Q2Easy

   SI unit of magnetic field B:

   A.Volt

   B.Tesla✓ Correct

   C.Newton

   D.Ohm

   Answer: B. Tesla

   Explanation: SI unit of B is the tesla, defined via F = qvB so 1 T = 1 N/(A·m). Confusing it with volt or ohm mixes up electric-field and resistance units, which carry different dimensions.

3. Q3Easy

   What is the magnetic force on a stationary charge placed in a magnetic field?

   A.qB/2

   B.qB

   C.Zero✓ Correct

   D.Infinite

   Answer: C. Zero

   Explanation: Magnetic force F = qv × B needs motion — with v = 0 the cross product vanishes, so a stationary charge feels zero magnetic force even in a strong field, unlike an electric field.

4. Q4Easy

   The magnetic field at a distance r from a long straight wire carrying current I is:

   A.μ₀I/(4πr)

   B.μ₀I/(2πr)✓ Correct

   C.μ₀I/r

   D.2μ₀I/r

   Answer: B. μ₀I/(2πr)

   Explanation: Ampere's law for an infinite straight wire gives B = μ₀I/(2πr) — note the 2π and the 1/r fall-off; forgetting the 2π gives the tempting but wrong μ₀I/r distractor.

5. Q5Easy

   The work done by a magnetic force on a moving charged particle is:

   A.Zero✓ Correct

   B.qvB

   C.½mv²

   D.qE·d

   Answer: A. Zero

   Explanation: Magnetic force F = qv × B is always perpendicular to v, so it can never do work (W = F·d = 0) — it changes the direction of motion but never the speed or kinetic energy.

6. Q6Easy

   Which law gives the magnetic field due to a current element?

   A.Lenz's law

   B.Ohm's law

   C.Coulomb's law

   D.Biot-Savart law✓ Correct

   Answer: D. Biot-Savart law

   Explanation: The Biot–Savart law, dB = (μ₀/4π)(I dl × r̂)/r², gives the field from a tiny current element — the magnetic analogue of Coulomb's law for a point charge.

7. Q7Medium

   Force per unit length on current-carrying wire in B:

   A.I/B

   B.IB only

   C.IL × B (per length: I × B)✓ Correct

   D.L/B

   Answer: C. IL × B (per length: I × B)

   Explanation: Dividing the wire force F = IL × B by length L leaves force per unit length as I × B — a vector product, so it depends on the angle between current and field too.

8. Q8Medium

   Field at centre of circular loop (radius R, current I):

   A.μ₀I/(2R)✓ Correct

   B.μ₀I/R

   C.2μ₀I/R

   D.μ₀IR

   Answer: A. μ₀I/(2R)

   Explanation: Integrating Biot–Savart around a circular loop of radius R gives centre field B = μ₀I/(2R) — mixing this up with the straight-wire result μ₀I/(2πR) is the usual trap.

9. Q9Medium

   An electron (m = 9.1×10⁻³¹ kg) moves with v = 1.0×10⁷ m/s perpendicular to B = 1.0 mT. The radius of its circular path is approximately:

   A.5.7 m

   B.5.7 mm

   C.0.57 m

   D.5.7 cm✓ Correct

   Answer: D. 5.7 cm

   Explanation: Radius r = mv/(qB) = (9.1×10⁻³¹ × 1.0×10⁷)/(1.6×10⁻¹⁹ × 1.0×10⁻³) ≈ 5.7 cm — mixing up mT with T in the denominator throws the answer off by 1000×.

10. Q10Medium

    A long straight wire carries 5 A. What is B at 10 cm from it?

    A.1.0 × 10⁻⁵ T✓ Correct

    B.5.0 × 10⁻⁶ T

    C.2.0 × 10⁻⁴ T

    D.1.0 × 10⁻³ T

    Answer: A. 1.0 × 10⁻⁵ T

    Explanation: B = μ₀I/(2πr) = (4π×10⁻⁷ × 5)/(2π × 0.1) = 1.0×10⁻⁵ T. Forgetting to convert 10 cm to 0.1 m before dividing is the usual arithmetic slip here.

11. Q11Medium

    A proton moves in a magnetic field of 0.5 T. The cyclotron frequency is approximately:

    A.7.6 kHz

    B.7.6 MHz✓ Correct

    C.76 Hz

    D.7.6 GHz

    Answer: B. 7.6 MHz

    Explanation: Cyclotron frequency f = qB/(2πm) = (1.6×10⁻¹⁹ × 0.5)/(2π × 1.67×10⁻²⁷) ≈ 7.6×10⁶ Hz — using the electron mass instead of the proton mass gives a wildly different order of magnitude.

12. Q12Hard

    Cyclotron frequency:

    A.qB/m

    B.qB/(2πm)✓ Correct

    C.mB/q

    D.qBm

    Answer: B. qB/(2πm)

    Explanation: Setting magnetic force equal to centripetal force and using v = 2πrf gives cyclotron frequency f = qB/(2πm) — independent of the particle's speed or orbit radius.

13. Q13Hard

    Field inside long solenoid (n turns/length):

    A.μ₀I/n

    B.μ₀I

    C.μ₀nI✓ Correct

    D.0

    Answer: C. μ₀nI

    Explanation: Inside a long solenoid the field is uniform along the axis: B = μ₀nI, where n is turns per unit length — using total turns N alone without dividing by length is the usual slip.

14. Q14Hard

    Maximum kinetic energy of a particle in a cyclotron of radius R, field B, charge q, mass m is:

    A.q²B²R²/(2m)✓ Correct

    B.qBR/m

    C.mB²R²/q

    D.qBR²

    Answer: A. q²B²R²/(2m)

    Explanation: Maximum speed at radius R is v_max = qBR/m, so K_max = ½mv² = q²B²R²/(2m) — squaring both q and B (not just one of them) is the step most often missed.

Moving Charges and Magnetism — FAQs

More Class 12 Physics chapters

• 01Electric Charges and Fields
• 02Electrostatic Potential and Capacitance
• 03Current Electricity
• 05Magnetism and Matter
• 06Electromagnetic Induction
• 07Alternating Current

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