Comprehensive Numerical & Conceptual MCQs | WBBSE, CBSE & NCERT Standard (Moderate to Hard Level)
Q1. A circular coil of radius 2 cm has 100 turns and carries a current of 2 A. What is the magnetic field intensity at the center of the coil?
Answer: (B) — Using B = μ0NI / 2R = (4π × 10-7 × 100 × 2) / (2 × 0.02) = 0.0063 T.
Q2. Two long straight parallel conductors separated by distance 2 cm carry currents of 5 A and 10 A in opposite directions. The force per unit length between them is:
Answer: (A) — Using F/L = (μ0 I1 I2)/(2πd) = (2 × 10-7 × 50)/(0.02m) = 0.0005 N/m. Since currents flow in opposite directions, the force is repulsive.
Q3. At what distance along the axis of a circular coil of radius R = 3 cm is the magnetic field equal to 1/8th of its value at the center?
Answer: (A) — Formula: Baxis = Bcenter / (1 + x2/R2)3/2. Setting denominator = 8 gives (1 + x2/R2) = 4 ⇒ x = R√3 = 3√3 = 5.2 cm.
Q4. A current I flows along an infinitely long thin-walled straight tube. The magnetic field at any point inside the tube is:
Answer: (C) — Applying Ampere's Circuital Law with an Amperian loop inside the tube, the enclosed current Ienc = 0, hence B = 0.
Q5. An electron moves in a circle of radius 5 mm with speed 2 × 106 m/s. The equivalent magnetic dipole moment is:
Answer: (A) — Magnetic moment M = I · A = (ev / 2πr) · (πr2) = ½ evr = 0.5 × (1.6×10-19) × (2×106) × (5×10-3) = 8e-16 A·m².
Q6. A circular coil of radius 12 cm has 100 turns and carries a current of 2 A. What is the magnetic field intensity at the center of the coil?
Answer: (B) — Using B = μ0NI / 2R = (4π × 10-7 × 100 × 2) / (2 × 0.12) = 0.001 T.
Q7. Two long straight parallel conductors separated by distance 7 cm carry currents of 5 A and 10 A in opposite directions. The force per unit length between them is:
Answer: (A) — Using F/L = (μ0 I1 I2)/(2πd) = (2 × 10-7 × 50)/(0.07m) = 0.000143 N/m. Since currents flow in opposite directions, the force is repulsive.
Q8. At what distance along the axis of a circular coil of radius R = 8 cm is the magnetic field equal to 1/8th of its value at the center?
Answer: (A) — Formula: Baxis = Bcenter / (1 + x2/R2)3/2. Setting denominator = 8 gives (1 + x2/R2) = 4 ⇒ x = R√3 = 8√3 = 13.86 cm.
Q9. A current I flows along an infinitely long thin-walled straight tube. The magnetic field at any point inside the tube is:
Answer: (C) — Applying Ampere's Circuital Law with an Amperian loop inside the tube, the enclosed current Ienc = 0, hence B = 0.
Q10. An electron moves in a circle of radius 10 mm with speed 2 × 106 m/s. The equivalent magnetic dipole moment is:
Answer: (A) — Magnetic moment M = I · A = (ev / 2πr) · (πr2) = ½ evr = 0.5 × (1.6×10-19) × (2×106) × (10×10-3) = 1.6e-15 A·m².
Q11. A circular coil of radius 22 cm has 100 turns and carries a current of 2 A. What is the magnetic field intensity at the center of the coil?
Answer: (B) — Using B = μ0NI / 2R = (4π × 10-7 × 100 × 2) / (2 × 0.22) = 0.0006 T.
Q12. Two long straight parallel conductors separated by distance 12 cm carry currents of 5 A and 10 A in opposite directions. The force per unit length between them is:
Answer: (A) — Using F/L = (μ0 I1 I2)/(2πd) = (2 × 10-7 × 50)/(0.12m) = 8.3e-05 N/m. Since currents flow in opposite directions, the force is repulsive.
Q13. At what distance along the axis of a circular coil of radius R = 13 cm is the magnetic field equal to 1/8th of its value at the center?
Answer: (A) — Formula: Baxis = Bcenter / (1 + x2/R2)3/2. Setting denominator = 8 gives (1 + x2/R2) = 4 ⇒ x = R√3 = 13√3 = 22.52 cm.
Q14. A current I flows along an infinitely long thin-walled straight tube. The magnetic field at any point inside the tube is:
Answer: (C) — Applying Ampere's Circuital Law with an Amperian loop inside the tube, the enclosed current Ienc = 0, hence B = 0.
Q15. An electron moves in a circle of radius 15 mm with speed 2 × 106 m/s. The equivalent magnetic dipole moment is:
Answer: (A) — Magnetic moment M = I · A = (ev / 2πr) · (πr2) = ½ evr = 0.5 × (1.6×10-19) × (2×106) × (15×10-3) = 2.4e-15 A·m².
Q16. A circular coil of radius 32 cm has 100 turns and carries a current of 2 A. What is the magnetic field intensity at the center of the coil?
Answer: (B) — Using B = μ0NI / 2R = (4π × 10-7 × 100 × 2) / (2 × 0.32) = 0.0004 T.
Q17. Two long straight parallel conductors separated by distance 17 cm carry currents of 5 A and 10 A in opposite directions. The force per unit length between them is:
Answer: (A) — Using F/L = (μ0 I1 I2)/(2πd) = (2 × 10-7 × 50)/(0.17m) = 5.9e-05 N/m. Since currents flow in opposite directions, the force is repulsive.
Q18. At what distance along the axis of a circular coil of radius R = 18 cm is the magnetic field equal to 1/8th of its value at the center?
Answer: (A) — Formula: Baxis = Bcenter / (1 + x2/R2)3/2. Setting denominator = 8 gives (1 + x2/R2) = 4 ⇒ x = R√3 = 18√3 = 31.18 cm.
Q19. A current I flows along an infinitely long thin-walled straight tube. The magnetic field at any point inside the tube is:
Answer: (C) — Applying Ampere's Circuital Law with an Amperian loop inside the tube, the enclosed current Ienc = 0, hence B = 0.
Q20. An electron moves in a circle of radius 20 mm with speed 2 × 106 m/s. The equivalent magnetic dipole moment is:
Answer: (A) — Magnetic moment M = I · A = (ev / 2πr) · (πr2) = ½ evr = 0.5 × (1.6×10-19) × (2×106) × (20×10-3) = 3.2e-15 A·m².
Q21. A circular coil of radius 42 cm has 100 turns and carries a current of 2 A. What is the magnetic field intensity at the center of the coil?
Answer: (B) — Using B = μ0NI / 2R = (4π × 10-7 × 100 × 2) / (2 × 0.42) = 0.0003 T.
Q22. Two long straight parallel conductors separated by distance 22 cm carry currents of 5 A and 10 A in opposite directions. The force per unit length between them is:
Answer: (A) — Using F/L = (μ0 I1 I2)/(2πd) = (2 × 10-7 × 50)/(0.22m) = 4.5e-05 N/m. Since currents flow in opposite directions, the force is repulsive.
Q23. At what distance along the axis of a circular coil of radius R = 23 cm is the magnetic field equal to 1/8th of its value at the center?
Answer: (A) — Formula: Baxis = Bcenter / (1 + x2/R2)3/2. Setting denominator = 8 gives (1 + x2/R2) = 4 ⇒ x = R√3 = 23√3 = 39.84 cm.
Q24. A current I flows along an infinitely long thin-walled straight tube. The magnetic field at any point inside the tube is:
Answer: (C) — Applying Ampere's Circuital Law with an Amperian loop inside the tube, the enclosed current Ienc = 0, hence B = 0.
Q25. An electron moves in a circle of radius 25 mm with speed 2 × 106 m/s. The equivalent magnetic dipole moment is:
Answer: (A) — Magnetic moment M = I · A = (ev / 2πr) · (πr2) = ½ evr = 0.5 × (1.6×10-19) × (2×106) × (25×10-3) = 4e-15 A·m².
Q26. A circular coil of radius 52 cm has 100 turns and carries a current of 2 A. What is the magnetic field intensity at the center of the coil?
Answer: (B) — Using B = μ0NI / 2R = (4π × 10-7 × 100 × 2) / (2 × 0.52) = 0.0002 T.
Q27. Two long straight parallel conductors separated by distance 27 cm carry currents of 5 A and 10 A in opposite directions. The force per unit length between them is:
Answer: (A) — Using F/L = (μ0 I1 I2)/(2πd) = (2 × 10-7 × 50)/(0.27m) = 3.7e-05 N/m. Since currents flow in opposite directions, the force is repulsive.
Q28. At what distance along the axis of a circular coil of radius R = 28 cm is the magnetic field equal to 1/8th of its value at the center?
Answer: (A) — Formula: Baxis = Bcenter / (1 + x2/R2)3/2. Setting denominator = 8 gives (1 + x2/R2) = 4 ⇒ x = R√3 = 28√3 = 48.5 cm.
Q29. A current I flows along an infinitely long thin-walled straight tube. The magnetic field at any point inside the tube is:
Answer: (C) — Applying Ampere's Circuital Law with an Amperian loop inside the tube, the enclosed current Ienc = 0, hence B = 0.
Q30. An electron moves in a circle of radius 30 mm with speed 2 × 106 m/s. The equivalent magnetic dipole moment is:
Answer: (A) — Magnetic moment M = I · A = (ev / 2πr) · (πr2) = ½ evr = 0.5 × (1.6×10-19) × (2×106) × (30×10-3) = 4.8e-15 A·m².
Q31. A proton is projected with velocity 310 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q32. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q33. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q34. A proton is projected with velocity 340 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q35. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q36. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q37. A proton is projected with velocity 370 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q38. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q39. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q40. A proton is projected with velocity 400 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q41. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q42. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q43. A proton is projected with velocity 430 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q44. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q45. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q46. A proton is projected with velocity 460 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q47. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q48. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q49. A proton is projected with velocity 490 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q50. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q51. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q52. A proton is projected with velocity 520 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q53. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q54. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q55. A proton is projected with velocity 550 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q56. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q57. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q58. A proton is projected with velocity 580 m/s perpendicular to a uniform magnetic field of 0.5 T. If the kinetic energy is quadrupled, the new radius becomes:
Answer: (B) — Radius r = √(2mK) / (qB) ⇒ r ∝ √K. If K becomes 4K, radius doubles.
Q59. In a cyclotron, the maximum kinetic energy of an accelerated alpha particle is K. The maximum kinetic energy of a proton accelerated under the same magnetic field and radius will be:
Answer: (A) — Formula: Kmax = (q2 B2 R2) / (2m). For alpha particle qα = 2e, mα = 4m. Thus q2/m ratio is identical for proton (e2/m) and alpha particle (4e2/4m = e2/m). Hence Kmax is the same = K.
Q60. A charged particle enters a magnetic field at an angle of 30° to the field lines. The ratio of pitch of the helical path to the radius of helix is:
Answer: (A) — Pitch p = v cosθ · T = (2πm v cosθ) / (qB). Radius r = (m v sinθ) / (qB). Pitch/Radius ratio = 2π cotθ = 2π cot 30° = 2π√3.
Q61. A galvanometer of resistance 55 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 55 = 5000 - 55 = 4945.0 Ω.
Q62. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q63. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q64. A galvanometer of resistance 70 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 70 = 5000 - 70 = 4930.0 Ω.
Q65. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q66. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q67. A galvanometer of resistance 85 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 85 = 5000 - 85 = 4915.0 Ω.
Q68. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q69. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q70. A galvanometer of resistance 50 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 50 = 5000 - 50 = 4950.0 Ω.
Q71. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q72. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q73. A galvanometer of resistance 65 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 65 = 5000 - 65 = 4935.0 Ω.
Q74. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q75. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q76. A galvanometer of resistance 80 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 80 = 5000 - 80 = 4920.0 Ω.
Q77. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q78. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q79. A galvanometer of resistance 95 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 95 = 5000 - 95 = 4905.0 Ω.
Q80. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q81. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q82. A galvanometer of resistance 60 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 60 = 5000 - 60 = 4940.0 Ω.
Q83. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q84. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q85. A galvanometer of resistance 75 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 75 = 5000 - 75 = 4925.0 Ω.
Q86. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q87. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q88. A galvanometer of resistance 90 Ω gives full scale deflection for 2 mA current. To convert it into a voltmeter of range 0 - 10 V, the required series resistance is:
Answer: (A) — Formula: R = (V / Ig) - G = (10 / 0.002) - 90 = 5000 - 90 = 4910.0 Ω.
Q89. If the current sensitivity of a moving coil galvanometer is increased by 20%, its resistance increases by 50%. The percentage change in its voltage sensitivity is:
Answer: (A) — Voltage sensitivity Vs = Is / R. New Vs' = (1.2 Is) / (1.5 R) = 0.8 Vs. Decrease = 20%.
Q90. A wire carrying current I is bent into an equilateral triangle of side L. The magnetic dipole moment of the loop is:
Answer: (A) — Magnetic moment M = I · A. Area of equilateral triangle A = (√3/4) L2, so M = (√3/4) I L2.
Q91. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q92. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q93. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q94. A paramagnetic material has a magnetic susceptibility of 0.002 at 388 K. Its susceptibility at 488 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (388 / 488) = 0.00159.
Q95. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q96. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q97. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q98. A paramagnetic material has a magnetic susceptibility of 0.002 at 396 K. Its susceptibility at 496 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (396 / 496) = 0.0016.
Q99. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q100. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q101. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q102. A paramagnetic material has a magnetic susceptibility of 0.002 at 404 K. Its susceptibility at 504 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (404 / 504) = 0.0016.
Q103. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q104. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q105. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q106. A paramagnetic material has a magnetic susceptibility of 0.002 at 412 K. Its susceptibility at 512 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (412 / 512) = 0.00161.
Q107. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q108. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q109. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q110. A paramagnetic material has a magnetic susceptibility of 0.002 at 420 K. Its susceptibility at 520 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (420 / 520) = 0.00162.
Q111. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q112. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q113. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q114. A paramagnetic material has a magnetic susceptibility of 0.002 at 428 K. Its susceptibility at 528 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (428 / 528) = 0.00162.
Q115. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q116. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q117. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q118. A paramagnetic material has a magnetic susceptibility of 0.002 at 436 K. Its susceptibility at 536 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (436 / 536) = 0.00163.
Q119. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q120. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q121. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q122. A paramagnetic material has a magnetic susceptibility of 0.002 at 444 K. Its susceptibility at 544 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (444 / 544) = 0.00163.
Q123. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q124. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q125. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q126. A paramagnetic material has a magnetic susceptibility of 0.002 at 452 K. Its susceptibility at 552 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (452 / 552) = 0.00164.
Q127. Which of the following materials is preferred for making permanent magnets?
Answer: (B) — Permanent magnets require material with high retentivity (to retain magnetism) and high coercivity (to resist demagnetization). Alnico is ideal.
Q128. The relation between relative permeability μr and magnetic susceptibility χm is:
Answer: (A) — Standard constitutive relation in magnetism: μr = 1 + χm.
Q129. The magnetic susceptibility χm of a diamagnetic material is:
Answer: (C) — Diamagnetic materials have small, negative magnetic susceptibility (-1 ≤ χm < 0) and it is temperature independent.
Q130. A paramagnetic material has a magnetic susceptibility of 0.002 at 460 K. Its susceptibility at 560 K will be:
Answer: (A) — According to Curie's Law, χ ∝ 1/T ⇒ χ2 = χ1 (T1 / T2) = 0.002 × (460 / 560) = 0.00164.
Q131. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q132. If the horizontal component of Earth's field is B_H = 0.4 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.4 / 0.5 = 0.8 G.
Q133. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q134. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q135. If the horizontal component of Earth's field is B_H = 0.3 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.3 / 0.5 = 0.6 G.
Q136. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q137. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q138. If the horizontal component of Earth's field is B_H = 0.45 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.45 / 0.5 = 0.9 G.
Q139. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q140. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q141. If the horizontal component of Earth's field is B_H = 0.35 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.35 / 0.5 = 0.7 G.
Q142. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q143. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q144. If the horizontal component of Earth's field is B_H = 0.5 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.5 / 0.5 = 1.0 G.
Q145. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q146. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q147. If the horizontal component of Earth's field is B_H = 0.4 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.4 / 0.5 = 0.8 G.
Q148. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q149. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q150. If the horizontal component of Earth's field is B_H = 0.3 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.3 / 0.5 = 0.6 G.
Q151. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q152. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q153. If the horizontal component of Earth's field is B_H = 0.45 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.45 / 0.5 = 0.9 G.
Q154. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q155. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q156. If the horizontal component of Earth's field is B_H = 0.35 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.35 / 0.5 = 0.7 G.
Q157. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q158. At the magnetic poles of the Earth, the angle of dip is:
Answer: (C) — At the magnetic poles, Earth's magnetic field lines are perpendicular to the surface, so BH = 0 and dip angle δ = 90°.
Q159. If the horizontal component of Earth's field is B_H = 0.5 G and angle of dip is 60°, the total intensity of Earth's magnetic field is:
Answer: (A) — Formula: BH = B cos δ ⇒ B = BH / cos 60° = 0.5 / 0.5 = 1.0 G.
Q160. At a certain place, the horizontal component of Earth's magnetic field is equal to the vertical component. The angle of dip at this place is:
Answer: (C) — Formula: tan δ = BV / BH. Since BV = BH, tan δ = 1 ⇒ δ = 45°.
Q161. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q162. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q163. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q164. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Q165. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q166. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q167. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q168. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Q169. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q170. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q171. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q172. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Q173. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q174. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q175. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q176. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Q177. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q178. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q179. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q180. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Q181. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q182. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q183. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q184. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Q185. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q186. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q187. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q188. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Q189. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q190. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q191. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q192. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Q193. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q194. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q195. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q196. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Q197. A solenoid of length 50 cm has 500 turns and carries current of 2 A. The magnetic field inside the solenoid is:
Answer: (A) — Formula: B = μ0 n I where n = N/L = 500/0.5 = 1000 turns/m. B = (4π×10-7) × 1000 × 2 = 0.00251 T.
Q198. The magnetic torque acting on a bar magnet of dipole moment M placed at angle θ in uniform field B is maximum when θ is:
Answer: (C) — Torque τ = M B sinθ. Maximum when sinθ = 1 ⇒ θ = 90°.
Q199. The work done in rotating a magnetic dipole of moment M through 180° starting from stable equilibrium direction in uniform field B is:
Answer: (B) — Work done W = MB(cos θ1 - cos θ2) = MB(cos 0° - cos 180°) = 2 MB.
Q200. Electromagnets are made of soft iron because soft iron has:
Answer: (C) — Soft iron rapidly magnetizes and demagnetizes due to high magnetic permeability and low coercivity/hysteresis loss.
Tags:
FREECOURSES