Moving Charges and Magnetism Assignment
Class 12 Physics • JEE Advanced / NEET Level
Q1. Derive the expression for magnetic field at a distance r from a long straight conductor carrying current I. Also calculate B at r = 5 cm when I = 10 A.
Solution:
For a long straight wire, using Ampere’s law: B(2πr) = μ₀I. Hence B = μ₀I/(2πr). For I = 10 A and r = 0.05 m, B = (4π×10⁻⁷×10)/(2π×0.05) = 4×10⁻⁵ T.Q2. A circular coil of radius 10 cm carries current 5 A. Find magnetic field at its centre. Also mention the direction by right hand thumb rule.
Solution:
Magnetic field at centre of a circular loop is B = μ₀I/(2R). R = 0.10 m, I = 5 A. B = (4π×10⁻⁷×5)/(0.20) = π×10⁻⁵ T ≈ 3.14×10⁻⁵ T. Direction is perpendicular to the plane of the coil according to right hand thumb rule.Q3. Explain why a stationary charge does not experience magnetic force, while a moving charge can experience it.
Solution:
Magnetic force on a charge is F = q(v × B) = qvB sinθ. If the charge is stationary, v = 0, so F = 0. Therefore magnetic force acts only on a moving charge when its velocity has a component perpendicular to magnetic field.Q4. A proton moves with speed 2×10⁶ m/s perpendicular to a magnetic field of 0.5 T. Find the magnetic force on it. Take e = 1.6×10⁻¹⁹ C.
Solution:
Since v ⟂ B, F = qvB = 1.6×10⁻¹⁹ × 2×10⁶ × 0.5 = 1.6×10⁻¹³ N.Q5. A charged particle enters a uniform magnetic field perpendicular to its velocity. Derive the radius of its circular path.
Solution:
Magnetic force provides centripetal force: qvB = mv²/r. Therefore r = mv/(qB). The path is circular because force remains perpendicular to velocity.Q6. An electron enters a magnetic field of 2×10⁻³ T perpendicular to it with speed 3.5×10⁶ m/s. Find radius of path. Given mₑ = 9.1×10⁻³¹ kg, e = 1.6×10⁻¹⁹ C.
Solution:
r = mv/(qB) = (9.1×10⁻³¹×3.5×10⁶)/(1.6×10⁻¹⁹×2×10⁻³) = 9.95×10⁻³ m ≈ 1.0 cm.Q7. State the condition for a charged particle to pass undeflected through crossed electric and magnetic fields. Find speed if E = 3×10⁴ N/C and B = 0.02 T.
Solution:
For no deflection, electric force equals magnetic force: qE = qvB. Hence v = E/B = 3×10⁴/0.02 = 1.5×10⁶ m/s.Q8. A wire of length 0.5 m carrying current 4 A is placed perpendicular to a magnetic field 0.2 T. Find force on the wire.
Solution:
Force on current-carrying conductor: F = BIL sinθ. Here θ = 90°. F = 0.2×4×0.5 = 0.4 N.Q9. Two long parallel wires carry currents 5 A and 10 A in the same direction and are separated by 20 cm. Find force per unit length between them and nature of force.
Solution:
Force per unit length: F/L = μ₀I₁I₂/(2πd). = (4π×10⁻⁷×5×10)/(2π×0.2) = 5×10⁻⁵ N/m. Since currents are in the same direction, force is attractive.Q10. Why do two parallel conductors carrying currents in opposite directions repel each other?
Solution:
Each conductor produces a magnetic field at the position of the other. Using Fleming’s left hand rule or vector form F = I(L × B), currents in opposite directions produce forces away from each other. Hence they repel.Q11. A rectangular coil of 20 turns has area 0.04 m² and carries current 2 A. It is placed in a uniform magnetic field 0.5 T. Find maximum torque.
Solution:
Maximum torque on a current loop is τ = NIAB. τ = 20×2×0.04×0.5 = 0.8 N m.Q12. Derive the torque on a rectangular current loop placed in a uniform magnetic field.
Solution:
For a rectangular coil, forces on opposite sides form a couple. Torque τ = NIAB sinθ, where N is turns, I current, A area and θ is angle between magnetic moment and magnetic field. Vector form: τ = M × B, where M = NIA.Q13. A moving coil galvanometer has N = 100 turns, area = 2×10⁻⁴ m², B = 0.5 T and current 1 mA. Find torque on coil in radial magnetic field.
Solution:
In radial field, sinθ = 1. τ = NIAB = 100×10⁻³×2×10⁻⁴×0.5 = 1×10⁻⁵ N m.Q14. What is the role of radial magnetic field in a moving coil galvanometer?
Solution:
Radial magnetic field keeps the plane of coil always parallel to the field, so the angle between magnetic moment and field remains 90°. Thus torque τ = NIAB is directly proportional to current, giving a linear scale.Q15. A galvanometer of resistance 50 Ω gives full scale deflection for 2 mA. Find shunt resistance required to convert it into an ammeter of range 1 A.
Solution:
Shunt resistance S = IgG/(I − Ig). Ig = 0.002 A, G = 50 Ω, I = 1 A. S = (0.002×50)/(0.998) ≈ 0.100 Ω.Q16. A galvanometer of resistance 100 Ω gives full scale deflection for 1 mA. Find series resistance to convert it into a voltmeter of range 10 V.
Solution:
Total resistance required = V/Ig = 10/0.001 = 10000 Ω. Series resistance R = 10000 − 100 = 9900 Ω.Q17. State Biot–Savart law and write its vector form.
Solution:
Biot–Savart law gives magnetic field due to a current element: dB = (μ₀/4π)(I dl sinθ/r²). Vector form: dB = (μ₀/4π) I (dl × r̂)/r².Q18. Using Ampere’s circuital law, derive magnetic field inside a long solenoid.
Solution:
For a long solenoid, choose an Amperian rectangle. Magnetic field outside is nearly zero and inside is uniform. Ampere’s law gives Bℓ = μ₀(nℓ)I. Hence B = μ₀nI.Q19. A long solenoid has 1000 turns per metre and carries current 2 A. Find magnetic field inside it.
Solution:
B = μ₀nI = 4π×10⁻⁷×1000×2 = 8π×10⁻⁴ T ≈ 2.51×10⁻³ T.Q20. Magnetic field inside a toroid is given by B = μ₀NI/(2πr). Explain why field outside an ideal toroid is nearly zero.
Solution:
In an ideal toroid magnetic field lines are confined within the core. For an Amperian loop outside the toroid, net enclosed current is zero, so ∮B·dl = 0 and B outside is nearly zero.Q21. A toroid has 500 turns, mean radius 20 cm and current 3 A. Find magnetic field inside the toroid.
Solution:
B = μ₀NI/(2πr) = (4π×10⁻⁷×500×3)/(2π×0.2) = 1.5×10⁻³ T.Q22. A wire carrying current I is bent into a semicircle of radius R. Find magnetic field at the centre due to the semicircular part.
Solution:
Magnetic field at centre due to arc angle θ is B = μ₀Iθ/(4πR). For semicircle θ = π, so B = μ₀I/(4R).Q23. A wire is bent into a quadrant of radius 10 cm carrying current 8 A. Find magnetic field at centre due to the arc.
Solution:
For arc, B = μ₀Iθ/(4πR). For quadrant θ = π/2. B = μ₀I/(8R) = (4π×10⁻⁷×8)/(8×0.10) = 4π×10⁻⁶ T ≈ 1.26×10⁻⁵ T.Q24. What is magnetic dipole moment of a current loop? Find it for a coil of 50 turns, area 0.01 m² and current 3 A.
Solution:
Magnetic dipole moment M = NIA. M = 50×3×0.01 = 1.5 A m².Q25. A current loop has magnetic moment 0.6 A m² and is placed in a field 0.4 T at 30° with the field. Find torque.
Solution:
Torque τ = MB sinθ = 0.6×0.4×sin30° = 0.12 N m.Q26. Explain the difference between electric field lines and magnetic field lines.
Solution:
Electric field lines start from positive charge and end on negative charge. Magnetic field lines are always closed loops; they do not begin or end because isolated magnetic monopoles are not observed.Q27. A charged particle moves parallel to a uniform magnetic field. What is its path and why?
Solution:
If velocity is parallel to magnetic field, θ = 0°. Magnetic force F = qvB sin0° = 0. Hence particle continues in a straight line with uniform velocity.Q28. A charged particle enters a uniform magnetic field at an angle θ with velocity. Describe its path.
Solution:
Velocity has two components: v cosθ parallel to B and v sinθ perpendicular to B. Parallel component remains unchanged while perpendicular component causes circular motion. Combined path is helical.Q29. Find pitch of helix for a particle of mass m and charge q entering magnetic field B with speed v at angle θ to the field.
Solution:
Time period T = 2πm/(qB). Component parallel to field = v cosθ. Pitch = v cosθ × T = (2πmv cosθ)/(qB).Q30. A proton enters B = 0.1 T with speed 10⁶ m/s at 60° to the field. Find pitch of helical path. Given mₚ = 1.67×10⁻²⁷ kg, q = 1.6×10⁻¹⁹ C.
Solution:
Pitch = (2πmv cosθ)/(qB). cos60° = 0.5. Pitch = 2π×1.67×10⁻²⁷×10⁶×0.5/(1.6×10⁻¹⁹×0.1) ≈ 0.328 m.Q31. State cyclotron frequency. Does it depend on speed of the particle?
Solution:
Cyclotron angular frequency ω = qB/m and frequency f = qB/(2πm). In non-relativistic case it does not depend on speed or radius.Q32. A cyclotron accelerates protons in magnetic field 1.5 T. Find cyclotron frequency. Given q = 1.6×10⁻¹⁹ C, m = 1.67×10⁻²⁷ kg.
Solution:
f = qB/(2πm) = (1.6×10⁻¹⁹×1.5)/(2π×1.67×10⁻²⁷) ≈ 2.29×10⁷ Hz.Q33. Why is a cyclotron not suitable for accelerating electrons to very high energies?
Solution:
Electrons have very small mass, so they become relativistic quickly. Their mass effectively increases, cyclotron frequency changes, and they go out of phase with the alternating electric field.Q34. A current carrying circular coil behaves like a magnetic dipole. Which face acts as north pole?
Solution:
When viewed from a side, if current appears anticlockwise, that face acts as north pole. If current appears clockwise, that face acts as south pole.Q35. A square loop of side 20 cm carries current 5 A. Find magnetic field at its centre. Use formula B = 2√2 μ₀I/(πa).
Solution:
a = 0.20 m. B = 2√2 μ₀I/(πa) = 2√2×4π×10⁻⁷×5/(π×0.20) = 2.83×10⁻⁵ T.Q36. A straight wire carries current from south to north. Find direction of magnetic field at a point vertically above the wire.
Solution:
Using right hand thumb rule: thumb points north along current. At a point vertically above the wire, curled fingers point towards west. Hence magnetic field is towards west.Q37. Define one ampere using force between two parallel current carrying conductors.
Solution:
One ampere is the current which, when maintained in each of two long straight parallel conductors of negligible cross-section placed 1 m apart in vacuum, produces a force of 2×10⁻⁷ N per metre between them.Q38. A conductor of length 2 m carries current 3 A in a field 0.4 T. If force is 1.2 N, find angle between conductor and field.
Solution:
F = BIL sinθ. 1.2 = 0.4×3×2×sinθ = 2.4 sinθ. sinθ = 0.5, so θ = 30° or 150°. Usually smaller angle = 30°.Q39. Why does magnetic force do no work on a moving charged particle?
Solution:
Magnetic force F = q(v × B) is always perpendicular to velocity. Work done dW = F·ds. Since displacement is along velocity and force is perpendicular, F·ds = 0. Hence magnetic force does no work.Q40. A particle of charge 2 μC moves with velocity 3i + 4j m/s in magnetic field 5k T. Find magnetic force vector.
Solution:
F = q(v × B). v × B = (3i + 4j) × 5k = 20i − 15j. q = 2×10⁻⁶ C. F = (40i − 30j)×10⁻⁶ N = (4i − 3j)×10⁻⁵ N.Q41. A rectangular loop of dimensions 10 cm × 20 cm carries current 2 A in a magnetic field 0.5 T. Find magnetic moment and maximum torque.
Solution:
Area A = 0.10×0.20 = 0.02 m². Magnetic moment M = IA = 2×0.02 = 0.04 A m². Maximum torque = MB = 0.04×0.5 = 0.02 N m.Q42. A charged particle has q/m = 5×10⁷ C/kg and moves perpendicular to B = 0.02 T. Find angular frequency.
Solution:
Angular frequency ω = qB/m = (q/m)B = 5×10⁷×0.02 = 1×10⁶ rad/s.Q43. Find magnetic field at the centre of a coil with 20 turns, radius 5 cm, carrying current 2 A.
Solution:
B = μ₀NI/(2R) = 4π×10⁻⁷×20×2/(2×0.05) = 16π×10⁻⁵ T ≈ 5.03×10⁻⁴ T.Q44. A current of 4 A flows through a long wire. At what distance is magnetic field 8×10⁻⁶ T?
Solution:
For long wire B = μ₀I/(2πr). r = μ₀I/(2πB) = (4π×10⁻⁷×4)/(2π×8×10⁻⁶) = 0.10 m.Q45. A proton and an alpha particle enter the same magnetic field with the same speed perpendicular to field. Compare their radii.
Solution:
r = mv/(qB). For proton rₚ = mₚv/(eB). For alpha particle m = 4mₚ, q = 2e, so rα = 4mₚv/(2eB) = 2rₚ. Thus rα : rₚ = 2 : 1.Q46. An electron and proton enter the same magnetic field with same kinetic energy perpendicular to field. Which has larger radius?
Solution:
r = mv/(qB) = √(2mK)/(qB). For same K and same |q|, r ∝ √m. Proton has much larger mass, so proton has larger radius.Q47. Explain why magnetic field is called an axial vector.
Solution:
Magnetic field direction is associated with rotation/current loops and follows right-hand rule. Under mirror reflection it behaves differently from polar vectors, so it is an axial or pseudovector.Q48. A circular loop of radius R carries current I. If it is reshaped into a square loop using the same wire and same current, compare magnetic fields at centre approximately.
Solution:
Length of wire = 2πR. For square side a, 4a = 2πR so a = πR/2. Field at centre of square = 2√2 μ₀I/(πa) = 4√2 μ₀I/(π²R). Field at centre of circle = μ₀I/(2R). Ratio Bsquare/Bcircle = 8√2/π² ≈ 1.15.Q49. A wire loop is placed in a uniform magnetic field. Does it experience net force? Does it experience torque?
Solution:
In a uniform magnetic field, net force on a closed current loop is zero because forces on opposite elements cancel. But it may experience torque τ = MB sinθ unless magnetic moment is parallel or antiparallel to the field.Q50. In a region B = 0.4 T, a 10 cm long wire moves perpendicular to B with speed 5 m/s. Find motional emf induced across its ends.
Solution:
Motional emf ε = Bℓv = 0.4×0.10×5 = 0.20 V.