NUST NET Physics chapter 6: Physics · Magnetism and alternating current MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

42 exam-style questions on Physics · Magnetism and alternating current, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

42 questions · 3 topics · 10 per test

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What chapter 6 covers

Ten Physics · Magnetism and alternating current questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. A charged particle moves parallel to a uniform magnetic field. The magnetic force on it is

  1. A
    qvB, at right angles to the field
  2. B
    qvB, along the field
  3. C
    zero
  4. D
    qvB/2
Show answer

Answer: C. F = qvB sin θ and θ = 0°, so the force is zero and the particle carries on in a straight line.

2. A series LC circuit resonates at frequency f. The inductance is made four times as large with the same capacitor. The new resonant frequency is

  1. A
    f/4
  2. B
    2f
  3. C
    f/2
  4. D
    4f
Show answer

Answer: C. f = 1/(2π√(LC)), so f ∝ 1/√L and multiplying L by 4 halves f. Forgetting the square root gives f/4.

3. An ideal step-down transformer changes 220 V to 11 V. Its secondary coil has 50 turns. The number of turns on its primary is

  1. A
    20
  2. B
    2.5
  3. C
    500
  4. D
    1000
Show answer

Answer: D. Np/Ns = Vp/Vs = 220/11 = 20, so Np = 20 × 50 = 1000. Inverting the ratio gives 2.5; 20 is the ratio itself.

4. A rod 0.20 m long slides at a steady 5.0 m s−1 along two parallel rails joined by a 2.0 Ω resistor, at right angles to a uniform field of 0.50 T. Ignoring other resistance and friction, the force needed to keep it moving is

  1. A
    0.125 N
  2. B
    0.025 N
  3. C
    0.25 N
  4. D
    0.50 N
Show answer

Answer: B. EMF = BLv = 0.50 × 0.20 × 5.0 = 0.50 V, so I = 0.25 A and F = BIL = 0.50 × 0.25 × 0.20 = 0.025 N. Leaving out L in BIL gives 0.125 N.

5. A coil of 200 turns and area 5.0 × 10−3 m2 is at right angles to a magnetic field that rises steadily from 0 to 0.40 T in 0.20 s. The induced EMF is

  1. A
    0.010 V
  2. B
    0.40 V
  3. C
    20 V
  4. D
    2.0 V
Show answer

Answer: D. EMF = NAΔB/Δt = 200 × 5.0 × 10−3 × 0.40/0.20 = 2.0 V. Leaving out the number of turns gives 0.010 V.

6. In a velocity selector, a uniform electric field of 2.0 × 104 V m−1 is at right angles to a uniform magnetic field of 0.10 T. Charged particles pass through undeflected if their speed is

  1. A
    2.0 × 103 m s−1
  2. B
    2.0 × 105 m s−1
  3. C
    5.0 × 10−6 m s−1
  4. D
    2.0 × 104 m s−1
Show answer

Answer: B. No deflection when qE = qvB, so v = E/B = 2.0 × 104/0.10 = 2.0 × 105 m s−1. Multiplying E by B gives 2.0 × 103.

7. A coil of 50 turns and area 0.040 m2 is at right angles to a field of 0.25 T. It is turned in 0.10 s until its plane is parallel to the field. The average induced EMF is

  1. A
    5.0 V
  2. B
    10 V
  3. C
    0.10 V
  4. D
    zero
Show answer

Answer: A. Flux per turn falls from BA = 0.010 Wb to zero, so EMF = 50 × 0.010/0.10 = 5.0 V. 10 V would need the flux to reverse (a 180° turn), and 0.10 V forgets the 50 turns.

8. A wire 0.50 m long carries a current of 4.0 A in a uniform magnetic field of 0.20 T. The wire makes an angle of 30° with the field. The force on the wire is

  1. A
    0.20 N
  2. B
    0.40 N
  3. C
    0.35 N
  4. D
    0.10 N
Show answer

Answer: A. F = BIL sin θ = 0.20 × 4.0 × 0.50 × 0.5 = 0.20 N. Ignoring the angle gives 0.40 N; using cos 30° gives about 0.35 N.

9. The coil of an AC generator has 100 turns of area 0.010 m2 and rotates at 50 revolutions per second in a uniform field of 0.20 T. The peak EMF is

  1. A
    20π V
  2. B
    10 V
  3. C
    10π V
  4. D
    0.2π V
Show answer

Answer: A. ω = 2πf = 100π rad s−1, so ε0 = NABω = 100 × 0.010 × 0.20 × 100π = 20π V (about 63 V). Using f in place of ω gives 10 V, and leaving out N gives 0.2π V.

10. A capacitor of capacitance 100/π μF is connected to a 50 Hz AC supply. Its reactance is

  1. A
    100 Ω
  2. B
    0.010 Ω
  3. C
    628 Ω
  4. D
    1.0 × 10−4 Ω
Show answer

Answer: A. XC = 1/(2πfC) = 1/(2π × 50 × 10−4/π) = 1/(100 × 10−4) = 100 Ω. 0.010 Ω is 2πfC, the formula turned upside down.

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Ways to practise this chapter

Physics · Magnetism and alternating current: questions students ask

How many Physics · Magnetism and alternating current MCQs are there?

42, each with the answer and a short explanation, across 3 topics: Electromagnetism, Electromagnetic induction, Alternating current. Every chapter test draws ten of them, so you can take it many times.

Is the NUST NET Physics chapter 6 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

Can I read the explanations in Urdu?

Yes. In practice mode and on your results you can switch the explanations to Urdu.

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