MUET / Sukkur IBA Engineering Physics Electromagnetic Induction — Set 3

MUET / Sukkur IBA Engineering Physics Electromagnetic Induction — Set 3

1. Question 1

   Q1. The magnetic flux through a coil is given by ϕ = 5t² + 3t + 2. What is the induced emf at t = 2 s?

   • A) 23 V
   • B) 20 V
   • C) 10 V
   • D) 26 V

   Answer: 23 V

   Explanation: ε = -dϕ/dt = -(10t + 3), at t = 2, ε = -(10*2 + 3) = -23 V, magnitude is 23 V.

2. Question 2

   Q2. A rectangular coil of 100 turns, area 0.01 m² is rotated at 50 Hz in a field of 0.1 T. What is the peak emf?

   • A) 31.4 V
   • B) 3.14 V
   • C) 314 V
   • D) 0.314 V

   Answer: 31.4 V

   Explanation: ε = NABω = 100 * 0.01 * 0.1 * 2 * π * 50 = 31.4 V.

3. Question 3

   Q3. A circular coil of radius 0.1 m, 100 turns is rotated at 100 rpm in a uniform magnetic field of 0.1 T. The maximum induced emf is

   • A) 1.047 V
   • B) 10.47 V
   • C) 0.1047 V
   • D) 104.7 V

   Answer: 10.47 V

   Explanation: ε = NABω = 100 * π * (0.1)² * 0.1 * (100/60) * 2π = 10.47 V.

4. Question 4

   Q4. The flux linked with a coil is given by ϕ = 2t² + 4t + 6. The induced emf at t = 1 / 2 s is

   • A) 6 V
   • B) 4 V
   • C) 8 V
   • D) 2 V

   Answer: 6 V

   Explanation: ε = -dϕ/dt = -(4t + 4), at t = 1/2, ε = -(4 * 1/2 + 4) = -6 V, magnitude is 6 V.

5. Question 5

   Q5. A solenoid of 1000 turns, area 0.01 m² has an inductance of 0.1 H. If the current changes at 100 A/s, the induced emf is

   • A) 10 V
   • B) -10 V
   • C) 100 V
   • D) -100 V

   Answer: 10 V

   Explanation: ε = -L(dI/dt) = -0.1 * 100 = -10 V, magnitude is 10 V.

6. Question 6

   Q6. A coil of inductance 0.1 H has a current of 1 A. The energy stored is

   • A) 0.05 J
   • B) 0.1 J
   • C) 0.5 J
   • D) 1 J

   Answer: 0.05 J

   Explanation: E = 1/2 * LI² = 1/2 * 0.1 * 1² = 0.05 J.

7. Question 7

   Q7. The self-inductance of a coil is 0.1 H. If the current changes from 1 A to 2 A in 0.1 s, the induced emf is

   • A) 1 V
   • B) -1 V
   • C) 10 V
   • D) -10 V

   Answer: 1 V

   Explanation: ε = -L(dI/dt) = -0.1 * (2-1)/0.1 = -1 V, magnitude is 1 V.

8. Question 8

   Q8. A conductor of length 1 m moves at 10 m/s in a field of 0.1 T. The induced emf is

   • A) 1 V
   • B) 0.1 V
   • C) 10 V
   • D) 0.01 V

   Answer: 1 V

   Explanation: ε = Blv = 0.1 * 1 * 10 = 1 V.

9. Question 9

   Q9. The magnetic flux through a coil is ϕ = 3t² + 2t + 1. The induced emf at t = 1 s is

   • A) 8 V
   • B) -8 V
   • C) 6 V
   • D) -6 V

   Answer: 8 V

   Explanation: ε = -dϕ/dt = -(6t + 2), at t = 1, ε = -(6 + 2) = -8 V, magnitude is 8 V.

10. Question 10

    Q10. A circular coil of 100 turns, radius 0.1 m is rotated at 1000 rpm in a field of 0.01 T. The maximum induced emf is

    • A) 10.47 V
    • B) 1.047 V
    • C) 104.7 V
    • D) 0.1047 V

    Answer: 10.47 V

    Explanation: ε = NABω = 100 * π * (0.1)² * 0.01 * (1000/60) * 2π = 10.47 V, simplified.

11. Question 11

    Q11. The current in a coil changes from 1 A to 2 A in 0.1 s. If the induced emf is 10 V, the self-inductance is

    • A) 1 H
    • B) 0.1 H
    • C) 10 H
    • D) 0.01 H

    Answer: 1 H

    Explanation: ε = -L(dI/dt), 10 = L * (2-1)/0.1, L = 1 H.

12. Question 12

    Q12. A coil has an inductance of 0.1 H. The energy stored when the current is 2 A is

    • A) 0.2 J
    • B) 0.1 J
    • C) 0.4 J
    • D) 0.02 J

    Answer: 0.2 J

    Explanation: E = 1/2 * LI² = 1/2 * 0.1 * 2² = 0.2 J.

13. Question 13

    Q13. A rod of length 0.5 m moves at 20 m/s in a field of 0.2 T. The induced emf is

    • A) 2 V
    • B) 1 V
    • C) 0.2 V
    • D) 0.1 V

    Answer: 2 V

    Explanation: ε = Blv = 0.2 * 0.5 * 20 = 2 V.

14. Question 14

    Q14. The flux linked with a coil is ϕ = t² + 2t + 1. The induced emf at t = 2 s is

    • A) 6 V
    • B) -6 V
    • C) 4 V
    • D) -4 V

    Answer: 6 V

    Explanation: ε = -dϕ/dt = -(2t + 2), at t = 2, ε = -(2*2 + 2) = -6 V, magnitude is 6 V.

15. Question 15

    Q15. A coil of 500 turns, area 0.1 m² is rotated at 500 rpm in a field of 0.2 T. The maximum induced emf is

    • A) 104.7 V
    • B) 10.47 V
    • C) 52.35 V
    • D) 5.235 V

    Answer: 52.35 V

    Explanation: ε = NABω = 500 * 0.1 * 0.2 * (500/60) * 2π = 52.35 V.

16. Question 16

    Q16. The self-inductance of a coil is 0.01 H. If the current changes from 0 to 1 A in 0.01 s, the induced emf is

    • A) 1 V
    • B) -1 V
    • C) 10 V
    • D) -10 V

    Answer: 1 V

    Explanation: ε = -L(dI/dt) = -0.01 * (1-0)/0.01 = -1 V, magnitude is 1 V.

17. Question 17

    Q17. A conductor of length 0.2 m moves at 5 m/s in a field of 0.5 T. The induced emf is

    • A) 0.5 V
    • B) 0.1 V
    • C) 1 V
    • D) 0.05 V

    Answer: 0.5 V

    Explanation: ε = Blv = 0.5 * 0.2 * 5 = 0.5 V.

18. Question 18

    Q18. A coil of 100 turns is rotated at 1000 rpm in a magnetic field of 0.01 T. The maximum emf induced is

    • A) 10.47 V
    • B) 1.047 V
    • C) 104.7 V
    • D) 0.1047 V

    Answer: 104.7 V

    Explanation: ε = NABωsin(ωt), ε_max = NABω = 100 * 0.01 * (2π * 1000 / 60) = 104.7 V, using ω = 2πf.

19. Question 19

    Q19. A conductor of length 1 m moves at 10 m/s in a uniform magnetic field of 0.1 T. The induced emf is

    • A) 1 V
    • B) 0.1 V
    • C) 10 V
    • D) 0.01 V

    Answer: 1 V

    Explanation: ε = Blv = 0.1 * 1 * 10 = 1 V, using the formula for motional emf.

20. Question 20

    Q20. The magnetic flux linked with a coil is given by ϕ = 5t² + 2t + 3. The induced emf at t = 1 s is

    • A) 12 V
    • B) 10 V
    • C) 8 V
    • D) 6 V

    Answer: 12 V

    Explanation: ε = -dϕ/dt = -(10t + 2), at t = 1, ε = -(10 + 2) = 12 V, using Faraday's law.