Class 12 Physics Chapter 4 Moving Charges and Magnetism – Extra Questions with Answers

Extra practice questions for Class 12 Physics Chapter 4 (Moving Charges and Magnetism), beyond the textbook. These Class 12 Physics Chapter 4 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. Write the formula for the Lorentz magnetic force.
Ans: F=qv×B.

Q2. Write the radius of circular motion of a charged particle in a magnetic field.
Ans: r=mv/(qB).

Q3. State Ampere’s circuital law.
Ans: ∮B·dl=μ₀I_enclosed.

Q4. What is the magnetic field at the center of a current loop of radius R carrying current I?
Ans: B=μ₀I/2R.

Q5. Do like currents attract or repel?
Ans: Attract.

Short Answer Questions (2–3 marks)

Q6. Find the magnetic field at a distance 0.1m from a long straight wire carrying 5A current.
Ans: B=μ₀I/2πr=(4π×10⁻⁷)(5)/(2π×0.1)=10⁻⁵ T.

Q7. A proton moves with velocity 2×10⁶ m/s perpendicular to a field of 0.5T. Find the force.
Ans: F=qvB=(1.6×10⁻¹⁹)(2×10⁶)(0.5)=1.6×10⁻¹³ N.

Q8. Two parallel wires carry 3A and 4A separated by 0.2m. Find the force per unit length.
Ans: F/l=μ₀I₁I₂/2πd=(4π×10⁻⁷)(3)(4)/(2π×0.2)=1.2×10⁻⁵ N/m.

Higher-Order Thinking / Application Questions

Q9. A charged particle enters a region with perpendicular electric field E=2×10⁴ V/m and magnetic field B=0.5T (velocity selector configuration) and passes through undeflected. It then enters a region with only the magnetic field B=0.5T and moves in a circle of radius 0.02m. Find the charge-to-mass ratio of the particle.
Ans: In the velocity selector, undeflected motion means the electric force balances the magnetic force: qE=qvB, so v=E/B=(2×10⁴)/(0.5)=4×10⁴ m/s. In the region with only the magnetic field, the particle moves in a circle with radius r=mv/(qB), so rearranging: q/m=v/(rB)=(4×10⁴)/(0.02×0.5)=(4×10⁴)/(0.01)=4×10⁶ C/kg. This two-stage method (velocity selector to fix v, then circular motion to relate q/m) is exactly the technique used historically (e.g., by J.J. Thomson) to measure the charge-to-mass ratio of the electron.

Q10. A moving coil galvanometer has resistance 50Ω and gives full-scale deflection for a current of 5mA. Convert it into (a) an ammeter reading up to 5A, and (b) a voltmeter reading up to 10V, finding the required shunt/series resistance in each case.
Ans: (a) Ammeter conversion: We need a shunt S in parallel with the galvanometer so that most of the 5A current bypasses the galvanometer, with only 5mA=0.005A flowing through it. The remaining current through the shunt is I_s=5−0.005=4.995A. Since the galvanometer and shunt are in parallel, they have the same voltage: I_gG=I_sS, so 0.005×50=4.995×S, giving S=0.25/4.995≈0.05Ω (a very small shunt resistance). (b) Voltmeter conversion: We need a series resistance R such that at full-scale deflection current I_g=0.005A, the total voltage across (galvanometer+R) equals 10V. Using V=I_g(G+R): 10=0.005(50+R), so 50+R=10/0.005=2000, giving R=2000−50=1950Ω (a large series resistance). This illustrates the general principle: ammeters need a very small parallel shunt (low resistance, to divert most current), while voltmeters need a large series resistance (to limit current through the sensitive galvanometer while dropping most of the voltage across the added resistor).

Written by Satish

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