AP Physics C: Electricity and Magnetism Practice Test 11 – 60 Questions with Answers

Test yourself with 60 original AP Physics C: Electricity and Magnetism practice questions covering all 6 units of the course: electric charges, fields, and Gauss’s law, electric potential, conductors and capacitors, electric circuits, magnetic fields and electromagnetism, and electromagnetic induction. Every answer includes the full working. All questions are calculus-based.

Unit 8: Electric Charges, Fields, and Gauss’s Law

Q1. Two small point charges of 3 μC and 6 μC are held 0.2 m apart in a vacuum. What is the magnitude of the electrostatic force exerted on either charge?

  • (A) 8.1 N
  • (B) 4.05 N
  • (C) 2.02 N
  • (D) 16.2 N
Show Answer

Answer: B – Use Coulomb’s law F = k q1 q2 / r2 = (9.0 × 109)(3 × 10-6)(6 × 10-6) / (0.2)2 = 4.05 N.

Q2. A point charge of 2 μC is fixed in a vacuum. What is the magnitude of the electric field at a point 0.4 m from the charge?

  • (A) 2.25 × 105 N/C
  • (B) 5.62 × 104 N/C
  • (C) 1.12 × 106 N/C
  • (D) 1.12 × 105 N/C
Show Answer

Answer: D – E = k q / r2 = (9.0 × 109)(2 × 10-6) / (0.4)2 = 1.12 × 105 N/C.

Q3. Two point charges, +8 μC and +4 μC, are fixed 0.6 m apart. What is the magnitude of the net electric field at the midpoint between them?

  • (A) 8.00 × 105 N/C
  • (B) 4.00 × 105 N/C
  • (C) 2.00 × 105 N/C
  • (D) 1.20 × 106 N/C
Show Answer

Answer: B – At the midpoint each field has magnitude k q / (0.3)2. Like charges give fields in opposite directions there, so Enet = k |q1 – q2| / (0.3)2 = 4.00 × 105 N/C.

Q4. A closed Gaussian surface encloses a net charge of 9 μC and no other charge. What is the net electric flux through the surface?

  • (A) 2.03 × 106 N·m2/C
  • (B) 5.08 × 105 N·m2/C
  • (C) 1.02 × 107 N·m2/C
  • (D) 1.02 × 106 N·m2/C
Show Answer

Answer: D – By Gauss’s law, Φ = Qenc / ε0 = (9 × 10-6) / (8.85 × 10-12) = 1.02 × 106 N·m2/C.

Q5. A solid insulating sphere of radius 0.1 m carries a total charge of 10 μC uniformly distributed through its volume. What is the magnitude of the electric field at a distance 0.4 m from the centre of the sphere?

  • (A) 1.12 × 106 N/C
  • (B) 2.81 × 105 N/C
  • (C) 2.25 × 106 N/C
  • (D) 5.62 × 105 N/C
Show Answer

Answer: D – Outside a spherically symmetric distribution, E = k Q / r2 = (9.0 × 109)(10 × 10-6) / (0.4)2 = 5.62 × 105 N/C.

Q6. A solid insulating sphere of radius 0.4 m carries a total charge of 10 μC uniformly distributed through its volume. What is the magnitude of the electric field inside the sphere at a distance 0.1 m from its centre?

  • (A) 1.41 × 105 N/C
  • (B) 2.81 × 105 N/C
  • (C) 7.03 × 104 N/C
  • (D) 5.62 × 105 N/C
Show Answer

Answer: A – Inside a uniformly charged solid sphere, Gauss’s law gives E = k Q r / R3 = (9.0 × 109)(10 × 10-6)(0.1) / (0.4)3 = 1.41 × 105 N/C.

Q7. An infinitely long straight line of charge has a uniform linear charge density of 8 nC/m. What is the magnitude of the electric field at a perpendicular distance of 0.25 m from the line?

  • (A) 576 N/C
  • (B) 1150 N/C
  • (C) 288 N/C
  • (D) 2300 N/C
Show Answer

Answer: A – For an infinite line, E = 2kλ / r = 2(9.0 × 109)(8 × 10-9) / (0.25) = 576 N/C.

Q8. An infinite nonconducting sheet carries a uniform surface charge density of 2 μC/m2. What is the magnitude of the electric field near the sheet, on either side of it?

  • (A) 226 kN/C
  • (B) 56.5 kN/C
  • (C) 113 kN/C
  • (D) 452 kN/C
Show Answer

Answer: C – For a single infinite sheet, E = σ / (2ε0) = (2 × 10-6) / (2 × 8.85 × 10-12) = 113 kN/C.

Q9. A thin ring of radius 0.3 m carries a total charge of 4 μC uniformly distributed around it. What is the magnitude of the electric field on the axis of the ring at a distance of 0.4 m from its centre?

  • (A) 1.15 × 105 N/C
  • (B) 2.30 × 105 N/C
  • (C) 5.76 × 104 N/C
  • (D) 1.15 × 106 N/C
Show Answer

Answer: A – On the axis, E = k Q x / (R2 + x2)3/2 = (9.0 × 109)(4 × 10-6)(0.4) / (0.125) = 1.15 × 105 N/C.

Q10. A uniform electric field of 500 N/C passes through a flat surface of area 0.02 m2. The angle between the field and the normal to the surface is 60°. What is the electric flux through the surface?

  • (A) 5 N·m2/C
  • (B) 10 N·m2/C
  • (C) 2.5 N·m2/C
  • (D) 20 N·m2/C
Show Answer

Answer: A – Φ = E A cos θ = (500)(0.02) cos 60° = 5 N·m2/C.

Q11. The electric field at the surface of a uniformly charged spherical shell has magnitude 3000 N/C, directed radially outward, and the shell has radius 0.1 m. What is the total charge on the shell?

  • (A) 6.67 × 10-3 μC
  • (B) 3.33 × 10-3 μC
  • (C) 1.67 × 10-3 μC
  • (D) 0.0133 μC
Show Answer

Answer: B – Outside the shell E = k Q / r2, so Q = E r2 / k = (3000)(0.1)2 / (9.0 × 109) = 3.33 × 10-3 μC.

Q12. A thin rod of length 0.5 m carries a uniform linear charge density λ. A point P lies on the axis of the rod at a distance 0.1 m from its nearer end. Which integral gives the magnitude of the electric field at P?

  • (A) ∪00.5 kλ dx / (0.1 + x)
  • (B) ∪00.5 kλ x dx / (0.1 + x)2
  • (C) kλ0.5 / (0.1)2
  • (D) ∪00.5 kλ dx / (0.1 + x)2
Show Answer

Answer: D – An element dq = λ dx at distance (0.1 + x) from P contributes dE = k dq / r2 = kλ dx / (0.1 + x)2. Integrating from x = 0 to x = 0.5 gives the field.

Unit 9: Electric Potential

Q13. A point charge of 6 μC is fixed in a vacuum. Taking the potential at infinity to be zero, what is the electric potential at a distance 0.3 m from the charge?

  • (A) 360 kV
  • (B) 180 kV
  • (C) 90 kV
  • (D) 720 kV
Show Answer

Answer: B – V = k q / r = (9.0 × 109)(6 × 10-6) / (0.3) = 180 kV.

Q14. A charge of 2 μC is moved slowly from a point at potential 0 V to a point at potential 4000 V by an external agent. How much work is done by the external agent?

  • (A) 8 mJ
  • (B) 16 mJ
  • (C) 4 mJ
  • (D) 32 mJ
Show Answer

Answer: A – W = q ΔV = (2 × 10-6)(4000) = 8 mJ.

Q15. Two point charges of 5 μC and 6 μC are 0.5 m apart. What is the electric potential energy of this pair, taking zero energy at infinite separation?

  • (A) 1080 mJ
  • (B) 270 mJ
  • (C) 540 mJ
  • (D) 2160 mJ
Show Answer

Answer: C – U = k q1 q2 / r = (9.0 × 109)(5 × 10-6)(6 × 10-6) / (0.5) = 540 mJ.

Q16. A thin ring of radius 0.2 m carries a total charge of 6 μC uniformly distributed around it. What is the electric potential on the axis of the ring at a distance of 0.3 m from its centre, with zero potential at infinity?

  • (A) 300 kV
  • (B) 74.9 kV
  • (C) 150 kV
  • (D) 599 kV
Show Answer

Answer: C – Every charge element is 0.361 m from the point, so V = k Q / √(R2 + x2) = (9.0 × 109)(6 × 10-6) / (0.361) = 150 kV.

Q17. A thin rod of length 0.4 m carries a uniform linear charge density λ. A point P lies on the axis of the rod at a distance 0.2 m from its nearer end. Which integral gives the electric potential at P, with zero potential at infinity?

  • (A) ∪00.4 kλ dx / (0.2 + x)2
  • (B) ∪00.4 kλ dx / (0.2 + x)
  • (C) ∪00.4 kλ x dx / (0.2 + x)
  • (D) kλ0.4 / (0.2)
Show Answer

Answer: B – An element dq = λ dx contributes dV = k dq / r = kλ dx / (0.2 + x). Integrating from x = 0 to x = 0.4 gives V at P.

Q18. In a region, the electric potential varies with position as V(x) = 3x2 + 4x + 5, where V is in volts and x is in metres. What is the x-component of the electric field at x = 1 m?

  • (A) -20 V/m
  • (B) -5 V/m
  • (C) -10 V/m
  • (D) 10 V/m
Show Answer

Answer: C – Ex = -dV/dx = -(6x + 4). At x = 1 m, Ex = -(6 × 1 + 4) = -10 V/m.

Q19. A conducting spherical shell of radius 0.1 m carries a total charge of 8 μC. What is the electric potential at a point inside the shell, 0.05 m from its centre, with zero potential at infinity?

  • (A) 720 kV
  • (B) 1440 kV
  • (C) 360 kV
  • (D) 2880 kV
Show Answer

Answer: A – Inside a conducting shell the potential is constant and equal to its surface value, V = k Q / R = (9.0 × 109)(8 × 10-6) / (0.1) = 720 kV.

Q20. A particle of charge 2 μC is released from rest at a point where the potential is 1000 V higher than at a second point. If no non-electric forces act, what kinetic energy does it gain in moving to the second point?

  • (A) 4 mJ
  • (B) 2 mJ
  • (C) 1 mJ
  • (D) 8 mJ
Show Answer

Answer: B – Loss of potential energy becomes kinetic energy: ΔK = q ΔV = (2 × 10-6)(1000) = 2 mJ.

Q21. An electric dipole consists of charges +3 μC and -3 μC separated by 0.02 m. Estimate the electric potential on the dipole axis at a distance 0.2 m from its centre, with zero potential at infinity, using the far-field dipole approximation.

  • (A) 27 kV
  • (B) 13.5 kV
  • (C) 6.75 kV
  • (D) 54 kV
Show Answer

Answer: B – p = q d = (3 × 10-6)(0.02) and V ≈ k p / r2 = (9.0 × 109)(6.00 × 10-8) / (0.2)2 = 13.5 kV.

Unit 10: Conductors and Capacitors

Q22. A parallel-plate capacitor has plates of area 0.02 m2 separated by 2 mm in a vacuum. What is its capacitance?

  • (A) 177 pF
  • (B) 44.2 pF
  • (C) 88.5 pF
  • (D) 354 pF
Show Answer

Answer: C – C = ε0 A / d = (8.85 × 10-12)(0.02) / (2 × 10-3) = 88.5 pF.

Q23. A parallel-plate capacitor has capacitance 50 pF in air. A dielectric slab of dielectric constant 4 is inserted to fill the space between the plates completely, with the plates isolated so charge stays fixed. What is the new capacitance?

  • (A) 200 pF
  • (B) 400 pF
  • (C) 100 pF
  • (D) 800 pF
Show Answer

Answer: A – With a dielectric, C = κ C0 = (4)(50 pF) = 200 pF.

Q24. Capacitors of 15 μF and 10 μF are connected in series. What is their equivalent capacitance?

  • (A) 12 μF
  • (B) 6 μF
  • (C) 3 μF
  • (D) 18 μF
Show Answer

Answer: B – 1/Ceq = 1/15 + 1/10 = 1/6 μF-1, so Ceq = 6 μF.

Q25. A capacitor of 2 μF is charged to a potential difference of 200 V. How much energy is stored in it?

  • (A) 40 mJ
  • (B) 80 mJ
  • (C) 20 mJ
  • (D) 160 mJ
Show Answer

Answer: A – U = (1/2) C V2 = (1/2)(2 × 10-6)(200)2 = 40 mJ.

Q26. A capacitor of 5 μF is connected across a 50 V battery until fully charged. What charge is stored on either plate?

  • (A) 500 μC
  • (B) 125 μC
  • (C) 1000 μC
  • (D) 250 μC
Show Answer

Answer: D – Q = C V = (5 μF)(50 V) = 250 μC.

Q27. A spherical capacitor consists of concentric conducting spheres of radii 0.08 m and 0.25 m, with vacuum between them. What is its capacitance?

  • (A) 26.2 pF
  • (B) 6.54 pF
  • (C) 13.1 pF
  • (D) 52.3 pF
Show Answer

Answer: C – C = 4πε0 a b / (b – a) = 4π(8.85 × 10-12)(0.08)(0.25) / (0.17) = 13.1 pF.

Q28. A hollow conducting sphere with an empty cavity is given a net charge and allowed to reach electrostatic equilibrium. Which statement is correct?

  • (A) Charge resides on the inner wall of the cavity, and the field inside the cavity is uniform and nonzero.
  • (B) No charge resides on the inner wall of the cavity, and the electric field inside the cavity is zero.
  • (C) No charge resides on the inner wall, but the field inside the cavity equals the field outside the sphere.
  • (D) Charge splits between the inner and outer walls, and the field inside the cavity points radially outward.
Show Answer

Answer: B – Gauss’s law applied to a surface inside the conducting material gives zero enclosed charge for the cavity region, so the cavity field is zero.

Q29. The electric field between the plates of a capacitor is 10000 N/C. What is the energy density stored in the field?

  • (A) 0.885 mJ/m3
  • (B) 0.443 mJ/m3
  • (C) 0.221 mJ/m3
  • (D) 1.77 mJ/m3
Show Answer

Answer: B – u = (1/2) ε0 E2 = (1/2)(8.85 × 10-12)(10000)2 = 0.443 mJ/m3.

Unit 11: Electric Circuits

Q30. A wire of length 4 m and cross-sectional area 5 mm2 is made of a material of resistivity 2.8 × 10-8 Ω·m. What is the resistance of the wire?

  • (A) 0.0448 Ω
  • (B) 0.0224 Ω
  • (C) 0.0112 Ω
  • (D) 0.0896 Ω
Show Answer

Answer: B – R = ρ L / A = (2.8 × 10-8)(4) / (5 × 10-6) = 0.0224 Ω.


Q31. Resistors of 10 Ω and 10 Ω are connected in parallel. What is their equivalent resistance?

  • (A) 10 Ω
  • (B) 5 Ω
  • (C) 2.5 Ω
  • (D) 15 Ω
Show Answer

Answer: B – 1/Req = 1/10 + 1/10, so Req = 5 Ω.

Q32. A 36 V battery with negligible internal resistance is connected across a 8 Ω resistor. What current flows through the resistor?

  • (A) 4.5 A
  • (B) 9 A
  • (C) 2.25 A
  • (D) 18 A
Show Answer

Answer: A – I = V / R = 36 / 8 = 4.5 A.

Q33. In a single loop, two batteries of emfs 12 V and 8 V oppose each other, and the total resistance in the loop is 3 Ω. What is the current in the loop, in the direction of the larger emf?

  • (A) 2.67 A
  • (B) 0.667 A
  • (C) 5.33 A
  • (D) 1.33 A
Show Answer

Answer: D – Kirchhoff’s loop rule gives I = (E1 – E2) / R = (12 – 8) / 3 = 1.33 A.

Q34. A current of 3 A flows through a 10 Ω resistor. At what rate is energy dissipated as heat in the resistor?

  • (A) 180 W
  • (B) 90 W
  • (C) 45 W
  • (D) 360 W
Show Answer

Answer: B – P = I2 R = (3)2(10) = 90 W.

Q35. A capacitor of 10 μF is connected in series with a 10 kΩ resistor. What is the time constant of this RC circuit?

  • (A) 0.2 s
  • (B) 0.05 s
  • (C) 0.1 s
  • (D) 0.4 s
Show Answer

Answer: C – τ = R C = (10 × 103)(10 × 10-6) = 0.1 s.

Q36. A capacitor in an RC circuit charges toward a maximum charge of 100 μC. After a time equal to 2 time constant(s), what charge is on the capacitor?

  • (A) 86.5 μC
  • (B) 173 μC
  • (C) 43.2 μC
  • (D) 346 μC
Show Answer

Answer: A – q(t) = Qmax(1 – e-t/τ) = 100(1 – e-2) = 86.5 μC.

Q37. A discharging RC circuit has an initial current of 15 mA. After a time equal to 2 time constant(s), what is the current?

  • (A) 4.06 mA
  • (B) 1.02 mA
  • (C) 8.12 mA
  • (D) 2.03 mA
Show Answer

Answer: D – I(t) = I0 e-t/τ = 15 e-2 = 2.03 mA.

Q38. An RC circuit with time constant 2 s is charging from zero charge. How long does it take for the charge to reach 50% of its maximum value?

  • (A) 2.77 s
  • (B) 0.693 s
  • (C) 1.39 s
  • (D) 5.54 s
Show Answer

Answer: C – q/Qmax = 1 – e-t/τ = 0.5, so t = -τ ln(1 – 0.5) = 2 × 0.693 = 1.39 s.

Q39. A battery of emf 24 V and internal resistance 2 Ω delivers a current of 3 A to an external circuit. What is the terminal potential difference of the battery?

  • (A) 36 V
  • (B) 9 V
  • (C) 18 V
  • (D) 72 V
Show Answer

Answer: C – V = ε – I r = 24 – (3)(2) = 18 V.

Q40. Resistors of 4 Ω and 12 Ω are connected in parallel across a 12 V battery. What is the current through the 4 Ω resistor?

  • (A) 6 A
  • (B) 1.5 A
  • (C) 12 A
  • (D) 3 A
Show Answer

Answer: D – Each parallel branch sees the full battery voltage, so I1 = V / R1 = 12 / 4 = 3 A.

Q41. In a series circuit, a 12 V battery is connected to a 6 Ω resistor in series with a 6 Ω resistor, and a capacitor is connected in parallel with the 6 Ω resistor. In steady state, what is the potential difference across the capacitor?

  • (A) 12 V
  • (B) 6 V
  • (C) 3 V
  • (D) 24 V
Show Answer

Answer: B – In steady state no current flows through the capacitor branch, so the capacitor voltage equals the voltage across the 6 Ω resistor: VC = V R2 / (R1 + R2) = 12 × 6 / 12 = 6 V.

Unit 12: Magnetic Fields and Electromagnetism

Q42. A particle of charge 4 μC moves at 2 × 104 m/s perpendicular to a uniform magnetic field of 0.2 T. What is the magnitude of the magnetic force on the particle?

  • (A) 32 mN
  • (B) 8 mN
  • (C) 16 mN
  • (D) 64 mN
Show Answer

Answer: C – F = q v B sin 90° = (4 × 10-6)(2 × 104)(0.2) = 16 mN.

Q43. An alpha particle (mass 6.4 × 10-27 kg, charge 3.2 × 10-19 C) moves at 2 × 106 m/s perpendicular to a uniform magnetic field of 0.25 T. What is the radius of its circular path?

  • (A) 32 cm
  • (B) 8 cm
  • (C) 16 cm
  • (D) 64 cm
Show Answer

Answer: C – r = m v / (q B) = (6.4 × 10-27)(2 × 106) / ((3.2 × 10-19)(0.25)) = 16 cm.

Q44. A straight wire of length 1 m carries a current of 4 A perpendicular to a uniform magnetic field of 50 mT. What is the magnitude of the magnetic force on the wire?

  • (A) 400 mN
  • (B) 100 mN
  • (C) 800 mN
  • (D) 200 mN
Show Answer

Answer: D – F = I L B = (4)(1)(50 × 10-3) = 200 mN.

Q45. A long straight wire carries a current of 7 A. What is the magnitude of the magnetic field at a distance of 0.07 m from the wire?

  • (A) 20 μT
  • (B) 10 μT
  • (C) 40 μT
  • (D) 30 μT
Show Answer

Answer: A – B = μ0I/(2πr) = 20 μT.

Q46. A circular loop of radius 5 cm carries a current of 5 A. What is the magnitude of the magnetic field at the centre of the loop?

  • (A) 62.8 μT
  • (B) 126 μT
  • (C) 31.4 μT
  • (D) 251 μT
Show Answer

Answer: A – B = μ0 I / (2R) = (4π × 10-7)(5) / (2 × 0.05) = 62.8 μT.

Q47. A long solenoid has 1500 turns per metre and carries a current of 2 A. What is the magnitude of the magnetic field inside the solenoid, far from its ends?

  • (A) 7.54 mT
  • (B) 3.77 mT
  • (C) 1.88 mT
  • (D) 15.1 mT
Show Answer

Answer: B – B = μ0 n I = (4π × 10-7)(1500)(2) = 3.77 mT.

Q48. Two long parallel wires 10 cm apart carry currents of 4 A and 4 A in opposite directions. What is the magnitude of the net magnetic field at the midpoint between the wires?

  • (A) 32 μT
  • (B) 64 μT
  • (C) 16 μT
  • (D) 128 μT
Show Answer

Answer: A – At the midpoint each wire contributes B = μ0 I / (2π r) with r = 0.05 m, and for opposite currents the fields point the same way, so Bnet = μ0(I1 + I2) / (2π r) = 32 μT.

Q49. A rectangular coil of 10 turns and area 0.04 m2 carries a current of 3 A in a uniform magnetic field of 0.1 T. When the plane of the coil is parallel to the field, what is the magnitude of the torque on the coil?

  • (A) 240 mN·m
  • (B) 60 mN·m
  • (C) 120 mN·m
  • (D) 480 mN·m
Show Answer

Answer: C – Maximum torque τ = N I A B = (10)(3)(0.04)(0.1) = 120 mN·m.

Q50. In a velocity selector, a charged particle passes undeflected through crossed electric and magnetic fields of 4000 N/C and 50 mT. What is the speed of the particle?

  • (A) 1.60 × 105 m/s
  • (B) 4.00 × 104 m/s
  • (C) 3.20 × 105 m/s
  • (D) 8.00 × 104 m/s
Show Answer

Answer: D – For no deflection qE = qvB, so v = E / B = 4000 / (50 × 10-3) = 8.00 × 104 m/s.

Unit 13: Electromagnetic Induction

Q51. A flat loop of area 0.02 m2 is placed in a uniform magnetic field of 200 mT. The angle between the field and the normal to the loop is 30°. What is the magnetic flux through the loop?

  • (A) 3.46 mWb
  • (B) 6.93 mWb
  • (C) 1.73 mWb
  • (D) 13.9 mWb
Show Answer

Answer: A – Φ = B A cos θ = (200 × 10-3)(0.02) cos 30° = 3.46 mWb.

Q52. The magnetic field through a single loop of area 0.1 m2 increases uniformly by 0.4 T in 0.2 s, with the field perpendicular to the loop. What is the magnitude of the induced emf in the loop?

  • (A) 400 mV
  • (B) 100 mV
  • (C) 800 mV
  • (D) 200 mV
Show Answer

Answer: D – |ε| = A ΔB / Δt = (0.1)(0.4) / (0.2) = 200 mV.

Q53. A conducting rod of length 0.5 m moves at 2 m/s perpendicular to a uniform magnetic field of 0.2 T, with the rod perpendicular to both its velocity and the field. What is the motional emf between the ends of the rod?

  • (A) 0.4 V
  • (B) 0.1 V
  • (C) 0.2 V
  • (D) 0.8 V
Show Answer

Answer: C – ε = B L v = (0.2)(0.5)(2) = 0.2 V.

Q54. A conducting loop lies in the plane of the page in a magnetic field directed into the page. The field strength decreases with time. Viewed from above the page, in which direction does the induced current flow?

  • (A) Counterclockwise, because the induced field must point into the page to oppose the decreasing into-page flux.
  • (B) Clockwise, because the induced field must point out of the page to speed the decrease.
  • (C) Counterclockwise, because a decreasing field always induces a counterclockwise current.
  • (D) Clockwise, because the induced field must point into the page to oppose the decreasing into-page flux.
Show Answer

Answer: D – Decreasing into-page flux requires an induced field in the original direction, into the page, which corresponds to a clockwise current.

Q55. A solenoid has 400 turns, a cross-sectional area of 10 cm2, and a length of 0.5 m. What is its self-inductance?

  • (A) 0.804 mH
  • (B) 0.201 mH
  • (C) 1.61 mH
  • (D) 0.402 mH
Show Answer

Answer: D – L = μ0 N2 A / l = (4π × 10-7)(400)2(10 × 10-4) / (0.5) = 0.402 mH.

Q56. An inductor of 20 mH is connected in series with a 10 Ω resistor. What is the time constant of this LR circuit?

  • (A) 4 ms
  • (B) 1 ms
  • (C) 2 ms
  • (D) 8 ms
Show Answer

Answer: C – τ = L / R = (20 × 10-3) / 10 = 2 ms.

Q57. A series LR circuit with battery voltage 24 V and resistance 6 Ω is switched on at t = 0. After a time equal to 2 time constant(s), what is the current in the circuit?

  • (A) 6.92 A
  • (B) 1.73 A
  • (C) 3.46 A
  • (D) 13.8 A
Show Answer

Answer: C – I(t) = (V/R)(1 – e-t/τ) = (24/6)(1 – e-2) = 3.46 A.

Q58. An LC circuit consists of an inductor of 10 mH and a capacitor of 10 μF. What is its natural frequency of oscillation?

  • (A) 1010 Hz
  • (B) 252 Hz
  • (C) 2010 Hz
  • (D) 503 Hz
Show Answer

Answer: D – f = 1 / (2π√(LC)) = 1 / (2π√((10 × 10-3)(10 × 10-6))) = 503 Hz.

Q59. In an ideal LC circuit, a capacitor of 40 μF is initially charged to 10 V and then connected to an inductor of 40 mH. What is the maximum current in the circuit?

  • (A) 0.316 A
  • (B) 0.632 A
  • (C) 0.158 A
  • (D) 1.26 A
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Answer: A – Energy conservation gives (1/2) C V2 = (1/2) L Imax2, so Imax = V √(C/L) = 10 √((40 × 10-6) / (40 × 10-3)) = 0.316 A.

Q60. A coil of 50 turns and area 0.02 m2 rotates in a uniform magnetic field of 0.2 T with angular speed 50 rad/s. What is the maximum induced emf in the coil?

  • (A) 20 V
  • (B) 5 V
  • (C) 40 V
  • (D) 10 V
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Answer: D – εmax = N B A ω = (50)(0.2)(0.02)(50) = 10 V.

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