Physics · Magnetism · chapter 4 of 14

Class 12 Physics Chapter 4: Moving Charges and Magnetism case study question with answers

One case-based question of the kind the board asks: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

In a school laboratory, Meera has a moving coil galvanometer of resistance 25 Ω that shows full-scale deflection of 30 divisions for a current of 2.0 mA. Its coil turns in a radial field between concave pole pieces, and a spring provides the restoring torque kφ. She wants to use it in two ways: as an ammeter to read currents up to 1.0 A, and as a voltmeter to read up to 5.0 V. The figure shows the two arrangements: (i) a small resistance S in parallel with the galvanometer, (ii) a large resistance R in series with it.

Part 1 · 1 mark

Because the field is radial, the deflection φ of the coil is

(a) proportional to the current(b) proportional to the square of the current(c) proportional to the sine of the current(d) independent of the current

Show answer

Answer: (a) proportional to the current

In a radial field the plane of the coil is always along the field, so τ = NIAB at every position; kφ = NIAB gives φ = (NAB/k) I.

Part 2 · 1 mark

The shunt S needed for the ammeter of range 0 to 1.0 A is about

(a) 0.050 Ω(b) 0.10 Ω(c) 2475 Ω(d) 25 Ω

Show answer

Answer: (a) 0.050 Ω

S and the coil are in parallel, so they share one potential difference, and S carries I − I_g. With V = IR, I_g·R_G = (I − I_g)·S, so S = (2.0 × 10⁻³ × 25)/(1.0 − 0.002) = 0.050 Ω.

Part 3 · 2 marks

Find the series resistance R for the voltmeter of range 0 to 5.0 V, and explain why a voltmeter must have a large resistance.

Answer and marking scheme

In series one current I_g flows through the coil and R, and their potential differences add: V = I_g(R_G + R), so R = V/I_g − R_G = 5.0/(2.0 × 10⁻³) − 25 = 2475 Ω. A voltmeter is connected in parallel with the element; it must draw very little current so that it does not disturb the voltage it measures.

V = I_g(R_G + R) gives R = 2475 Ω (1)
connected in parallel, must draw very little current so as not to change the voltage (1)

Find the current sensitivity of the galvanometer in divisions per mA and its voltage sensitivity in divisions per mV. If the number of turns were doubled using the same wire, how would each change?

Answer and marking scheme

Current sensitivity = 30/2.0 = 15 div mA⁻¹. Full-scale voltage = 2.0 mA × 25 Ω = 50 mV, so voltage sensitivity = 30/50 = 0.60 div mV⁻¹. Doubling N doubles the current sensitivity NAB/k, but it also roughly doubles the resistance, so the voltage sensitivity (NAB/k)(1/R) stays about the same.

15 div mA⁻¹ and 0.60 div mV⁻¹ (1)
current sensitivity doubles; voltage sensitivity about unchanged since R also doubles (1)

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