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Answer the following. (Class 12 Physics 2024, Q24a)

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PhysicsCBSE Class 12 · Board exam 2024, set 55/5/1 · 3 marks · Short answer

CBSE Class 12 Physics 2024: question 24(a)

Set in the CBSE Class 12 board exam of 2024, paper 55/5/1, section C, question 24(a), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q24(a) · 3 marks · Short answerElectromagnetic Induction

Answer the following.

  1. (i)

    State Lenz's Law. In a closed circuit, the induced current opposes the change in magnetic flux that produced it as per the law of conservation of energy. Justify.

    0 marks
  2. (ii)

    A metal rod of length 2 m is rotated with a frequency 60 rev/s about an axis passing through its centre and perpendicular to its length. A uniform magnetic field of 2T perpendicular to its plane of rotation is switched-on in the region. Calculate the e.m.f. induced between the centre and the end of the rod.

    0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

What would happen to the flux and the current if the induced current helped the change?

Hint 2

For a rod rotating in a field, ε = ½Bωl²; what is l between the centre and an end?

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (i) Lenz's law: The induced emf has a polarity such that the current it drives opposes the change in magnetic flux that produced it.

    1 mark

  2. Justification: Suppose the induced current aided the change in flux. The flux would then increase further, which would induce a still larger current, and so on without limit. Current and energy would grow without any work being done by an external agent.

    This violates the law of conservation of energy. So the induced current must oppose the change, and the work done against this opposition appears as electrical energy.

    ½ mark

  3. (ii) The emf between the centre and one end equals that of a rod of length rotating about one end: .

    Here is the centre-to-end length m, and rad/s.

    ½ mark

  4. ½ mark

  5. ½ mark

Answer: (ii) emf between centre and end = ½Bωl² = 120π V ≈ 377 V (with l = 1 m)

Where the marks are

  • States Lenz's law (1 mark)
  • Justifies it using conservation of energy (½ mark)
  • Uses ε = ½Bωl² with ω = 2π × 60 rad/s (½ mark)
  • Substitutes l = 1 m, the centre-to-end length (½ mark)
  • ε ≈ 377 V (½ mark)

Revise the chapter

Practise this chapterBoard-style questions from Electromagnetic Induction, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 55/5/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.