CEP 2
Chapter 4 – Moving Charges and Magnetism (2 Marks Questions with Answers)
Q1. Is any work done by a magnetic field on a moving charge? Justify your answer.
No, a magnetic field does no work on a moving charge because the magnetic force is always perpendicular to the direction of motion. Hence, it does not change the speed or kinetic energy of the charge.
Q2. Does a charged particle gain kinetic energy as it enters a magnetic field?
No, the magnetic force changes only the direction of the velocity, not its magnitude. Therefore, the kinetic energy remains constant.
Q3. Formulate the definition of magnetic field from F = qvB sinθ.
Magnetic field is the region around a moving charge or current where it experiences a magnetic force. Force is maximum when θ = 90° and zero when θ = 0° or 180°.
Q4. Write the expression for magnetic field at a point due to current element and explain the terms.
( dB = μ₀/4π × (I dl × r̂)/r² )
where μ₀ = permeability, I = current, dl = element length, r = distance.
Q5. Derive expression of magnetic field at the centre of a circular coil using Biot–Savart law.
( B = μ₀ NI / 2R ), where N = number of turns, I = current, R = radius.
Q6. Write the expression for magnetic field at a point on the axis of a circular current loop.
( B = μ₀ I R² / 2(R² + x²)³ᐟ² ), where x = distance of point from the centre.
Q7. Recall formula of magnetic dipole moment of a coil of N turns and area A.
( M = NIA ). The polarity is predicted by the right-hand rule (curling fingers in current direction, thumb gives north pole).
Q8. State and explain Ampere’s circuital law.
The line integral of magnetic field around a closed path is equal to μ₀ times the total current enclosed:
( ∮ B·dl = μ₀I ).
Q9. Write expression for magnetic field due to long straight current-carrying conductor.
( B = μ₀ I / 2πr ), where I = current, r = distance.
Q10. Sketch diagrams of magnetic field lines for (i) long straight wire, (ii) circular coil.
(i) Concentric circles around wire.
(ii) Circular loops with denser lines near centre.
Q11. Write expression for magnetic field inside a solenoid.
( B = μ₀ n I ), where n = number of turns per metre, I = current.
Q12. Write expression for force between two parallel conductors.
( F/L = μ₀ I₁ I₂ / 2πd ), where d = distance between wires.
Q13. Define ‘ampere’ using force between parallel currents.
One ampere is that current which, when flowing through each of two parallel wires 1 m apart, produces a force of 2×10⁻⁷ N/m.
Q14. Write formula for torque on current loop and find direction.
( τ = N I B A sinθ ) or ( τ⃗ = m⃗ × B⃗ ). Direction is given by the right-hand rule.
Q15. When is torque on a current loop (i) maximum and (ii) minimum?
(i) Maximum when θ = 90°, (ii) Minimum (zero) when θ = 0° or 180°.
Q16. Draw labeled diagram of moving coil galvanometer.
Diagram shows coil, pole pieces, pointer, scale, and radial magnetic field.
Q17. What is the significance of radial magnetic field in a galvanometer?
It ensures that the torque on the coil is proportional to current for all positions, giving a uniform scale.
Q18. How is galvanometer converted into (i) ammeter (ii) voltmeter?
(i) By connecting a low resistance (shunt) parallel to coil.
(ii) By connecting a high resistance (multiplier) in series.
Q19. A solenoid has 1000 turns/m, current = 5 A. Find magnetic field.
( B = μ₀ n I = 4π×10⁻⁷×1000×5 = 6.28×10⁻³ T ).
Q20. A galvanometer has n = 30, B = 0.25 T, A = 1.5×10⁻³ m², k = 10⁻³ N·m/deg. Find current sensitivity.
( Iₛ = n B A / k = (30×0.25×1.5×10⁻³)/10⁻³ = 11.25°/A ).
Chapter 5 – Magnetism and Matter (2 Marks Questions with Answers)
Q1. Can two magnetic field lines intersect each other?
No, because if they intersect, a point would have two directions of magnetic field, which is impossible.
Q2. Write expression for potential energy of a dipole in uniform magnetic field.
( U = -mB cosθ ), where m = magnetic moment, B = field strength.
Q3. Conditions for (i) stable and (ii) unstable equilibrium of a magnetic needle.
(i) Stable: θ = 0°, dipole aligns with field.
(ii) Unstable: θ = 180°, dipole opposite to field.
Q4. Write expression for torque on magnetic needle in uniform magnetic field.
( τ = mB sinθ ), where m = magnetic moment, θ = angle with field.
Q5. Sketch showing declination, inclination and horizontal component of earth’s field.
Diagram: ( B_H = B cosδ ), ( B_V = B sinδ ), declination (θ) between magnetic and geographic meridian.
Q6. Values of angle of dip and horizontal components at equator and pole.
At equator: δ = 0°, H = B.
At pole: δ = 90°, H = 0.
Q7. What is diamagnetism? Example.
Diamagnetism is weak repulsion from a magnetic field. Example: Bismuth.
Q8. What is paramagnetism? Example.
Paramagnetism is weak attraction toward a magnetic field. Example: Aluminium.
Q9. On what two factors does paramagnetic susceptibility depend?
It depends on (i) material nature and (ii) temperature (inversely proportional to T).
Q10. State Curie’s law for paramagnetic substances.
( χₘ = C / T ), where C = Curie constant, T = absolute temperature.
Q11. Draw field lines near (i) diamagnetic (ii) paramagnetic substance.
(i) Diamagnetic – lines are repelled.
(ii) Paramagnetic – lines are attracted inside.
Q12. What is ferromagnetism? Example.
Strong magnetism even without external field. Example: Iron.
Q13. Any two differences between paramagnetic and ferromagnetic substances.
| Property | Paramagnetic | Ferromagnetic |
|---|---|---|
| Magnetic strength | Weak | Very strong |
| Magnetic domains | Random | Well aligned |
Q14. What are domains? What happens at high temperature?
Small regions with aligned magnetic moments; at high temperature, domains lose alignment.
Q15. What are hard ferromagnets? Example.
Substances with strong retentivity and coercivity (used in permanent magnets). Example: Steel.
Q16. What are soft ferromagnets? Example.
Easily magnetized and demagnetized materials (used in electromagnets). Example: Soft iron.
Q17. The magnetic susceptibility of a ferromagnet is 1499. Find relative permeability and permeability.
( μᵣ = 1 + χ = 1500 ), ( μ = μ₀ μᵣ = 4π×10⁻⁷×1500 = 1.88×10⁻³ H/m ).
Q18. A bar magnet with axis 30° to 0.25 T field has torque 4.5×10⁻² Nm. Find m.
( τ = mB sin30° ⇒ m = τ / (B sin30°) = 4.5×10⁻² / (0.25×0.5) = 0.36 J/T ).
Q19. A bar magnet (m = 0.32 J/T) in 0.15 T field. Find potential energy in stable and unstable equilibrium.
( U = -mB cosθ )
Stable (θ = 0°): U = –48 mJ
Unstable (θ = 180°): U = +48 mJ
Q20. A solenoid of 800 turns, area = 2.5×10⁻⁴ m², current = 3 A. Find magnetic moment.
( M = NIA = 800×3×2.5×10⁻⁴ = 0.60 J/T )
