Original practice sets for Magnetism are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
1. Magnetic force on a moving charge is maximum when the velocity is:
Explanation: F = qvB sinθ is maximum when θ = 90°.
2. Magnetic force does no work on a moving charge because it is always:
Explanation: Power = F·v = 0 when F ⊥ v, so the magnetic force never changes the kinetic energy of a charge.
3. Radius of circular path of charge q, mass m, speed v in uniform field B is:
Explanation: Equate magnetic force to centripetal: qvB = mv²/r → r = mv/(qB).
4. The cyclotron frequency of a charged particle in a uniform magnetic field:
Explanation: f = qB/(2πm). Since it depends only on q, B, and m — and not on v — the cyclotron frequency is speed-independent.
5. Force on a straight wire of length L carrying current I perpendicular to field B is:
Explanation: F = IL × B; for perpendicular orientation the magnitude is ILB.
6. Torque on a rectangular current loop of area A, carrying current I, in field B at angle θ to its plane is:
Explanation: τ = NIA B sinθ where θ is the angle between the plane of the loop and B. (Equivalently τ = NIAB cosφ where φ is the angle between the magnetic moment and B.)
7. Magnetic moment of a current loop of area A carrying current I is:
Explanation: Magnetic moment m = IA (SI unit: A·m²).
8. Fleming's left-hand rule gives the direction of:
Explanation: Left-hand rule: forefinger → B, middle finger → I, thumb → force on conductor.
9. The magnetic field at the centre of a circular loop of radius R carrying current I is:
Explanation: From Biot-Savart law, B = μ₀I/(2R) at the centre of a single circular loop.
10. Magnetic field inside a long solenoid with n turns per unit length carrying current I is:
Explanation: Applying Ampere's circuital law to a solenoid gives B = μ₀nI inside, 0 outside (ideal solenoid).
11. Two long parallel wires carrying currents in the same direction:
Explanation: The field of wire 1 at wire 2 interacts with wire 2's current (F = IL × B); using the right-hand rule both forces point toward each other → attraction.
12. Magnetic field in the empty space inside the core of a toroid is:
Explanation: By Ampere's law, the net enclosed current for a path inside the hole of a toroid is zero → B = 0 in the hollow core region.
13. A charged particle enters a uniform magnetic field at an angle θ to the field lines. Its path will be:
Explanation: The component parallel to B is unchanged (no force); the component perpendicular to B causes circular motion → combined path is a helix.
14. In a velocity selector, a charged particle passes undeflected when:
Explanation: The particle is undeflected when electric force equals magnetic force: qE = qvB → v = E/B. The selector is speed-selective.
15. Magnetic field at distance r from a long straight wire carrying current I is:
Explanation: Applying Ampere's law along a circular path: B·2πr = μ₀I → B = μ₀I/(2πr).
16. A galvanometer is converted into an ammeter by connecting:
Explanation: The shunt diverts most of the current, allowing only the full-scale deflection current through the galvanometer. Low shunt resistance → large current range.
17. A galvanometer is converted into a voltmeter by connecting:
Explanation: The high series resistance limits current so the potential drop across the galvanometer represents the full measured voltage. High resistance → high voltage range.
18. In a cyclotron, the particle is accelerated:
Explanation: The magnetic field keeps the particle in a semicircular path inside each dee; the alternating electric field in the gap between the dees does work and accelerates the particle each crossing.
19. Energy of a magnetic dipole of moment m in a field B at angle θ is:
Explanation: U = −m·B = −mB cosθ. Minimum energy (stable equilibrium) at θ = 0 (m parallel to B); maximum (unstable) at θ = 180°.
20. In a moving coil galvanometer, the deflection is proportional to current because:
Explanation: At equilibrium: NIAB = kφ → φ = (NAB/k)I. The linear relationship between φ and I holds when the radial field keeps sinθ = 1 always.
21. Force per unit length between two parallel wires carrying currents I₁ and I₂ separated by distance d is:
Explanation: F/L = μ₀I₁I₂/(2πd). This is the definition of the SI ampere when F/L = 2×10⁻⁷ N/m for two wires 1 m apart carrying 1 A each.
22. Magnetic field on the axis of a circular loop of radius R at distance x from centre is:
Explanation: From Biot-Savart integration: B = μ₀IR²/[2(R²+x²)^(3/2)]. At x=0 this reduces to μ₀I/(2R).
23. In a tangent galvanometer, current I is proportional to:
Explanation: I = (2rBH/μ₀N) tanθ. This is why it is called the tangent galvanometer — deflection gives tan of angle, not angle itself.
24. The Hall effect is used to determine:
Explanation: The polarity of the Hall voltage reveals whether carriers are positive or negative. Hall voltage VH = IB/(ned) determines carrier density n.
25. Relative permeability of a paramagnetic material is:
Explanation: Paramagnetic materials are weakly attracted to magnetic fields; μr slightly > 1 (e.g., 1.00002 for O₂). Ferromagnetics have μr >> 1; diamagnetics have μr slightly
26. Diamagnetic materials when placed in a non-uniform magnetic field move:
Explanation: Diamagnets are repelled by magnetic fields — they migrate toward regions of lower field. Example: bismuth, copper, water.
27. Above the Curie temperature, a ferromagnetic material becomes:
Explanation: Above the Curie point, thermal agitation destroys long-range magnetic domain alignment; the material transitions from ferromagnetic to paramagnetic behaviour.
28. Retentivity (or remanence) of a ferromagnetic material is:
Explanation: When the magnetising field is removed (H = 0), the material retains a magnetic flux density equal to retentivity. Permanent magnets need high retentivity.
29. Coercivity of a ferromagnetic material is:
Explanation: Coercivity is the magnitude of reverse H required to bring B to zero after saturation. Permanent magnet materials need high coercivity so they aren't easily demagnetised.
30. Angle of dip is 0° at the:
Explanation: At the magnetic equator the Earth's field is horizontal; dip (inclination) = 0°. At the magnetic poles the field is vertical; dip = 90°.
31. If angle of dip is δ, horizontal component H and total field B are related by:
Explanation: The total field B has components BH = B cosδ (horizontal) and BV = B sinδ (vertical). BV/BH = tanδ.
32. Magnetic susceptibility χ is defined as:
Explanation: Susceptibility χ = M/H where M is magnetisation and H is the applied field. χ > 0 for paramagnetics and ferromagnetics; χ
33. Ampere's circuital law states that the line integral of B along a closed path equals:
Explanation: ∮ B·dl = μ₀I_enc. This is Ampere's law; the generalised form (Maxwell) adds the displacement current term μ₀ε₀ dΦE/dt.
34. Magnetic field at a point inside a solid cylindrical conductor of radius R at distance r < R from axis is:
Explanation: For r
35. Far from a small current loop, the magnetic field pattern is identical to that of:
Explanation: A small current loop has magnetic moment m = IA. Its far field is a dipole field exactly analogous to an electric dipole, with B ∝ 1/r³.
36. A magnetic dipole aligned with the external field B is in:
Explanation: At θ = 0° (dipole parallel to B), torque τ = mB sinθ = 0 and energy U = −mB cosθ = −mB (minimum). Any displacement restores the dipole → stable equilibrium.
37. At the equatorial point of a short bar magnet (magnetic moment m, distance d), field B is:
Explanation: Equatorial (broadside-on) field: B = μ₀m/(4πd³), directed anti-parallel to m. Axial field is twice this: 2μ₀m/(4πd³).
38. At the axial point of a short bar magnet (moment m, distance d), field B is:
Explanation: Axial (end-on) field: B = 2μ₀m/(4πd³) = μ₀m/(2πd³), directed along m (N to S outside the magnet).
39. When a bar magnet is placed along the magnetic meridian with north pole pointing north, neutral points occur:
Explanation: When the magnet's N points toward geographic north, its axial field aids Earth's field and its equatorial field opposes it. The net field is zero on the equatorial line (broadside) where equatorial field = Earth's horizontal component.
40. A circular coil of radius 5 cm, 20 turns, lies in the plane of a uniform field B = 0.1 T. Magnetic flux through the coil is:
Explanation: If the coil lies in the plane of B, then B is parallel to the plane — meaning B is perpendicular to the area vector. Flux = BA cosθ = BA cos90° = 0.
41. A proton moves with velocity 3×10⁶ m/s in a magnetic field of 0.2 T perpendicular to its velocity. The radius of its circular path is (m_p = 1.67×10⁻²⁷ kg, q = 1.6×10⁻¹⁹ C):
Explanation: r = mv/(qB) = (1.67×10⁻²⁷ × 3×10⁶)/(1.6×10⁻¹⁹ × 0.2) = 5.01×10⁻²¹/3.2×10⁻²⁰ ≈ 0.157 m.
42. A 10 cm wire carrying 5 A lies perpendicular to a field of 2 T. The force on the wire is:
Explanation: F = ILB sinθ = 5 × 0.10 × 2 × sin90° = 1 N.
43. A rectangular coil 4 cm × 6 cm, 50 turns, carries 2 A in field B = 0.5 T. The plane of the coil is parallel to B. Maximum torque is:
Explanation: τ = NIAB sinθ. Plane parallel to B means normal is perpendicular to B → sinθ = 1. τ = 50 × 2 × (0.04×0.06) × 0.5 = 50 × 2 × 0.0024 × 0.5 = 0.12 N·m.
44. A solenoid has 2000 turns, length 0.5 m, and carries 3 A. The field inside is (μ₀ = 4π×10⁻⁷):
Explanation: n = 2000/0.5 = 4000 turns/m. B = μ₀nI = 4π×10⁻⁷ × 4000 × 3 = 4.8π×10⁻³ T ≈ 1.508×10⁻² T.
45. Two parallel wires 20 cm apart carry 10 A each in opposite directions. Force per unit length between them is (attractive or repulsive?):
Explanation: F/L = μ₀I₁I₂/(2πd) = 4π×10⁻⁷ × 10 × 10/(2π × 0.2) = 1×10⁻⁴ N/m. Opposite currents → repulsion.
46. A circular coil of radius 0.1 m carries 4 A. The field at its centre is (μ₀ = 4π×10⁻⁷):
Explanation: B = μ₀I/(2R) = 4π×10⁻⁷ × 4/(2 × 0.1) = 8π×10⁻⁶ T.
47. A galvanometer of resistance 50 Ω gives full deflection at 1 mA. Shunt needed to convert it to an ammeter reading up to 1 A is:
Explanation: S = G·Ig/(I−Ig) = 50 × 0.001/(1 − 0.001) ≈ 50 × 0.001/0.999 ≈ 0.05 Ω.
48. The same galvanometer (G = 50 Ω, Ig = 1 mA) needs to be used as a 10 V voltmeter. Required series resistance is:
Explanation: R = V/Ig − G = 10/0.001 − 50 = 10000 − 50 = 9950 Ω.
49. A semiconductor strip of thickness 2 mm carries current 5 A in a field B = 0.3 T (perpendicular to strip). Hall voltage measured is 15 mV. Carrier density n is (q = 1.6×10⁻¹⁹):
Explanation: VH = IB/(ned) → n = IB/(VH×e×d) = (5×0.3)/(0.015×1.6×10⁻¹⁹×0.002) = 1.5/(4.8×10⁻²⁴) ≈ 3.1×10²³ m⁻³. (Check: this is a typical semiconductor value.)
50. In a cyclotron with dees of radius R and magnetic field B, maximum kinetic energy of the particle (charge q, mass m) is:
Explanation: Maximum speed at radius R: v = qBR/m. KE = ½mv² = ½m(qBR/m)² = q²B²R²/(2m).
51. A bar magnet of moment m and moment of inertia I oscillates in uniform field B. Its time period is:
Explanation: Restoring torque = −mB sinθ ≈ −mBθ for small θ. Angular frequency ω = √(mB/I), period T = 2π/ω = 2π√(I/mB).
52. Curie's law for paramagnetic materials states that susceptibility χ is proportional to:
Explanation: Curie's law: χ = C/T where C is the Curie constant and T is absolute temperature. Susceptibility decreases as temperature increases because thermal agitation opposes alignment.
53. A particle of mass m and charge q enters a uniform magnetic field B perpendicularly. The time taken to complete a semicircle is:
Explanation: Time for full circle T = 2πm/(qB). Half circle time = T/2 = πm/(qB).
54. Torque on a current loop in a uniform magnetic field is zero when:
Explanation: τ = NIAB sinθ = 0 when θ = 0 or 180°. When the plane is perpendicular to B, the normal is parallel to B → θ = 0 → τ = 0.
55. Magnetic field at perpendicular distance R from the end of a semi-infinite straight wire carrying current I is:
Explanation: For a semi-infinite wire, only half of the 2π integration contributes: B = μ₀I/(4πR). This is half the field of an infinite wire at the same distance.
56. A proton and deuteron enter a cyclotron with the same kinetic energy. The ratio of their radii (r_p/r_d) in the same field B is:
Explanation: r = mv/qB = √(2mKE)/(qB). r ∝ √m/q. For same KE and same q: r_p/r_d = √(m_p/m_d) = √(1/2) = 1/√2.
57. For a permanent magnet, the ideal material should have:
Explanation: A permanent magnet needs high retentivity (to maintain strong B when H = 0) and high coercivity (to resist demagnetisation). Steel and Alnico alloys qualify.
58. Soft iron is preferred for electromagnet cores because it has:
Explanation: Electromagnet cores must be easily magnetised (high permeability/low coercivity) and easily demagnetised when the current stops (low retentivity). Soft iron satisfies both.
59. An electron moves east in a magnetic field directed north. Magnetic force on the electron points:
Explanation: For the electron (q negative): F = q(v × B). v = east (+x̂), B = north (+ŷ). v × B = x̂ × ŷ = ẑ (up). For negative charge, F = −eẑ = down? Wait — electron charge is negative: F = (−e)(v×B). v×B = east×north = up(+z). F = −e(+z) = downward. Actually rechecking: F on electron = −e(v×B) where e>0. v=east(+x), B=north(+y). v×B = x̂×ŷ = ẑ (up). F = −e·ẑ = downward (−z). So F is downward. Correction: Force is downward (−z direction). The answer 'Up' was listed in error. The correct answer should be 'Down'.
60. Unlike a solenoid, a toroid has no magnetic field:
Explanation: By Ampere's law, the total enclosed current for a path outside a toroid is zero (equal positive and negative contributions cancel) → B = 0 outside. This is the key advantage — no external field leakage.
61. Maxwell added the displacement current term to Ampere's law to account for:
Explanation: Between charging capacitor plates, no conduction current flows but a changing E field exists. Maxwell added id = ε₀(dΦE/dt) to resolve the inconsistency in Ampere's law and to predict electromagnetic waves.
62. Which property is TRUE about magnetic field lines?
Explanation: Magnetic monopoles don't exist — field lines always close on themselves (they emerge from N, re-enter at S externally, and continue N→S inside the magnet). Two field lines never cross (that would imply two directions at one point).
63. A positive charge q enters a region where E points upward and B points out of the page. For the charge to move undeflected in the horizontal direction, which condition must hold?
Explanation: The electric force on +q is upward (qE upward). For horizontal motion in +x direction and B out of page (+z): magnetic force = q(v×B) = q(vx̂×Bẑ) = qvB(x̂×ẑ) = −qvBŷ (downward). Equilibrium: qE = qvB → v = E/B. This is the velocity selector principle.
64. A semi-circular wire of radius R lies in a plane perpendicular to a uniform field B. The net magnetic force on the wire when current I flows through it equals:
Explanation: The net force on any curved current in a uniform field equals ILB where L is the vector from start to end. For a semicircle, the chord length (straight-line distance between endpoints) = 2R, so F = I(2R)B = 2IRB. The net force is along the chord direction.
65. The cyclotron resonance condition breaks down when:
Explanation: Cyclotron frequency f = qB/(2πm). At relativistic speeds, m increases → f decreases → the particle falls out of phase with the fixed-frequency alternating field. The synchrotron resolves this by adjusting the field frequency or B to maintain resonance.
66. An electron revolves in a circular orbit of radius r with speed v. Its orbital magnetic moment is:
Explanation: Current I = e/(period) = ev/(2πr). Magnetic moment m = IA = (ev/2πr)(πr²) = evr/2.
67. The area enclosed by the B-H hysteresis loop represents:
Explanation: ∮H·dB = energy per unit volume per cycle lost as heat in the magnetic material. Materials for transformer cores need small hysteresis loops (low loss); permanent magnets need large loops.
68. A proton enters a magnetic field B at angle 30° to the field direction with speed v. The pitch of the resulting helix is:
Explanation: The component parallel to B is v‖ = v cos30°. Pitch = v‖ × T = v cos30° × (2πm/qB) = 2πmv cos30°/(qB).
69. Two identical bar magnets are placed coaxially end-to-end (north of one facing south of the other) at separation r. The force between them varies as:
Explanation: Force between magnetic dipoles = gradient of energy. Energy U ∝ 1/r³ (dipole field), so F = −dU/dr ∝ 1/r⁴.
70. Gauss's law for magnetism (∮ B·dA = 0) implies:
Explanation: ∮ B·dA = 0 means net magnetic flux through any closed surface is zero — i.e., every field line leaving a volume must re-enter it. This is equivalent to saying isolated magnetic poles (monopoles) don't exist; every N pole is accompanied by a S pole.
71. Inside a magnetic material, the relationship between B, H, and magnetisation M is:
Explanation: B = μ₀(H + M). For vacuum M = 0 so B = μ₀H. Inside a magnetic material, M is the magnetisation (magnetic moment per unit volume) and adds to the external H field.
72. A current loop is placed in a non-uniform magnetic field. It will move toward:
Explanation: Energy U = −m·B. The loop moves to minimise U. If m is aligned (parallel to B), U = −mB is most negative in high-B regions → moves to stronger field. If m is anti-aligned, U = +mB is minimised in low-B regions → moves toward weaker field.
73. A long cylindrical conductor of inner radius a and outer radius b carries current I uniformly distributed through the annular cross-section. At a point r where a < r < b, B is:
Explanation: Current density J = I/[π(b²−a²)]. Current enclosed by circle of radius r: I_enc = J·π(r²−a²) = I(r²−a²)/(b²−a²). Ampere's law: B(2πr) = μ₀I_enc → B = μ₀I(r²−a²)/[2πr(b²−a²)].
74. If a negative charge moves anti-parallel to a magnetic field B, the magnetic force on it is:
Explanation: F = q(v × B). If v is anti-parallel to B, then v × B = 0 → force = 0.
75. A solenoid has 500 turns, length 25 cm, and carries 2 A. Field inside (μ₀ = 4π×10⁻⁷) is:
Explanation: n = 500/0.25 = 2000 turns/m. B = μ₀nI = 4π×10⁻⁷ × 2000 × 2 = 16π×10⁻⁴ T = 4π×10⁻³ T.
76. A bar magnet with moment 0.5 A·m² and moment of inertia 2×10⁻⁵ kg·m² oscillates in a 0.2 T field. Time period is:
Explanation: T = 2π√(I/mB) = 2π√(2×10⁻⁵/(0.5×0.2)) = 2π√(2×10⁻⁵/0.1) = 2π√(2×10⁻⁴) = 2π×√2×10⁻² ≈ 2π×1.414×10⁻² ≈ 2π×0.01414 ≈ 0.0888 s ≈ 2π×10⁻² s (for cleaner number: if I=10⁻⁴, T=2π×10⁻² s).
77. A galvanometer of resistance 20 Ω shows full deflection at 5 mA. To measure up to 2 A, the shunt resistance required is:
Explanation: S = GIg/(I−Ig) = 20×0.005/(2−0.005) ≈ 0.1/1.995 ≈ 0.0501 Ω ≈ 0.05 Ω.
78. A good transformer core material should have a B-H loop that is:
Explanation: Transformer cores must undergo millions of B-H cycles. A narrow loop means low energy loss per cycle → low heating. Soft iron and silicon steel have narrow loops.
79. For a paramagnetic substance, if temperature is doubled, susceptibility:
Explanation: Curie's law: χ = C/T. If T doubles → χ halves.
80. An α-particle (charge 2e, mass 4u) moves at 10⁶ m/s perpendicular to B = 0.5 T. Its radius in meters (e = 1.6×10⁻¹⁹, u = 1.67×10⁻²⁷) is:
Explanation: r = mv/(qB) = (4×1.67×10⁻²⁷×10⁶)/(2×1.6×10⁻¹⁹×0.5) = 6.68×10⁻²¹/(1.6×10⁻¹⁹) ≈ 0.0418 m ≈ 0.042 m. (r ≈ 0.042 m is correct; 0.084 m would be for full circle diameter.)
81. Far from a small current loop with magnetic moment m, the field at axial distance d is:
Explanation: B_axial = 2μ₀m/(4πd³) = μ₀m/(2πd³) — exactly like the axial field of a magnetic dipole (bar magnet).
82. A toroid has N = 800 turns, mean circumference 40 cm, and carries current 5 A. Field inside the core is:
Explanation: B = μ₀NI/L = 4π×10⁻⁷ × 800 × 5 / 0.40 = 4π×10⁻⁷ × 10000 = 4π×10⁻³ T... Let me recalculate: 4π×10⁻⁷ × 800 × 5 / 0.4 = 4π×10⁻⁷ × 4000/0.4... Wait: N/L = 800/0.4 = 2000 turns/m. B = μ₀nI = 4π×10⁻⁷ × 2000 × 5 = 4π×10⁻³ T. Answer: 4π×10⁻³ T.
83. Net force on a closed current loop placed in a uniform magnetic field is:
Explanation: In a uniform field, the forces on each element of the loop are equal and opposite for opposite sides → net force = 0. Only torque acts on the loop in a uniform field, not net translational force.
84. At a boundary between two magnetic media, which component of B is continuous?
Explanation: From ∇·B = 0 (Gauss's law for magnetism), the normal component of B is continuous at an interface: B₁ₙ = B₂ₙ. The tangential component of H (not B) is continuous at a boundary with no surface current.
85. When a metal plate is swung into a magnetic field, it slows down due to:
Explanation: Moving the conductor through a magnetic field induces eddy currents (by Faraday's law). By Lenz's law, these currents create a field that opposes the motion — an electromagnetic braking effect used in braking systems and damping.