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(i) With the help of a… (Class 12 Physics 2025, Q32a)

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PhysicsCBSE Class 12 · Board exam 2025, set 55/7/1 · 5 marks · Long answer

CBSE Class 12 Physics 2025: question 32(a)

Set in the CBSE Class 12 board exam of 2025, paper 55/7/1, section E, question 32(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(a) · 5 marks · Long answerMoving Charges and Magnetism

(i) With the help of a labelled diagram, explain the principle of working of a moving coil galvanometer. Write the purpose of using (i) radial magnetic field, and (ii) soft iron core, in it.
(ii) Define current sensitivity of a galvanometer. "Increasing the current sensitivity may not necessarily increase the voltage sensitivity." Give reason.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 55/7/1.

Hints

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

Hint 1

Equate the magnetic torque NIAB with the spring's restoring torque kφ.

Hint 2

Write voltage sensitivity as φ/V with V = IR, and think about what happens to R when you add more turns.

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. Labelled diagram: a rectangular coil of many turns wound on a soft iron cylindrical core, placed between concave pole pieces N and S of a permanent magnet. The coil is suspended or pivoted, with a spring for restoring torque and a pointer moving over a scale.

    1 mark

  2. Principle and working: a current-carrying coil in a magnetic field experiences a torque . This torque rotates the coil until it is balanced by the restoring torque of the spring, : . So the deflection is proportional to the current.

    1 mark

  3. (i) Radial magnetic field: the plane of the coil is always parallel to the field, so the torque is maximum and independent of the coil's position. This makes and the scale linear.

    ½ mark

  4. (ii) Soft iron core: it has high permeability, so it concentrates and strengthens the magnetic field (and helps make it radial). This increases the sensitivity of the galvanometer.

    ½ mark

  5. (ii) Current sensitivity is the deflection per unit current: .

    1 mark

  6. Voltage sensitivity is .

    ½ mark

  7. If N is doubled to increase , the length of wire, and hence the coil resistance R, also roughly doubles. therefore stays essentially unchanged. So increasing current sensitivity does not necessarily increase voltage sensitivity.

    ½ mark

Answer: φ = (NAB/k) I. A radial field gives a linear scale; the soft iron core strengthens B and raises sensitivity. Current sensitivity = φ/I = NAB/k. Increasing N raises I_s, but R rises in the same proportion, so V_s = NAB/(kR) is unchanged.

Where the marks are

  • Labelled diagram of the galvanometer (1 mark)
  • Principle: torque on a current-carrying coil balanced by spring torque, φ ∝ I (1 mark)
  • Purpose of the radial field (½ mark)
  • Purpose of the soft iron core (½ mark)
  • Definition of current sensitivity, NAB/k (1 mark)
  • Voltage sensitivity expression (½ mark)
  • Reason: R increases in the same proportion as N, so V_s is unchanged (½ mark)

Revise the chapter

Practise this chapterBoard-style questions from Moving Charges and Magnetism, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 55/7/1: 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.