Question 311 markMCQSection D

A moving coil galvanometer has a coil with area of cross-section 4.0×10−3 m24.0 \times 10^{-3} \text{ m}^2 and number of turns 50. The coil is rotating in a magnetic field of 0.25 T. The torque acting on the coil when a current of 5 A passes through it is


OR


A galvanometer coil has a resistance of 15 Ω and the meter shows full scale deflection for a current of 3 mA. The value of resistance required to convert it into a voltmeter of range (0 – 12 V) is

(A) 4015 Ω (B) 3985 Ω (C) 415 Ω (D) 385 Ω

Moving Charges and Magnetism
Torque on a moving coil galvanometer

Options

(A)1.0 N m
(B)2.0 N m
(C)0.50 N m
(D)0.25 N m

Official answer

Correct option: (D) 0.25 N m


Primary part — torque on the coil:


In a radial field the maximum deflecting torque on a moving-coil galvanometer is


  • τ=NBIA\tau = N B I A
  • τ=50×0.25 T×5 A×4.0×10−3 m2\tau = 50 \times 0.25 \text{ T} \times 5 \text{ A} \times 4.0 \times 10^{-3} \text{ m}^2
  • τ=0.25\tau = 0.25 N·m

OR alternative — converting to a voltmeter:


A high resistance R is added in series so full-scale current I_g gives 12 V:


  • R=V/Ig−G=12/(3×10−3)−15=4000−15R = V/I_g - G = 12 / (3 \times 10^{-3}) - 15 = 4000 - 15 = 3985 Ω → option (B)
Answer keywords:torque on galvanometer coilτ = NBIAmoving coil galvanometerradial magnetic fieldnumber of turnsvoltmeter conversion series resistancefull scale deflection0.25 N m

Marking scheme

  1. 1.1 mark: correct option (D) 0.25 N·m for the primary numerical part.
  2. 2.Full credit also for the OR alternative: option (B) 3985 Ω using R=V/Ig−GR = V/I_g - G.
  3. 3.No marks for merely writing the formula without the correct final value/option.

Hint

Radial field ⇒ maximum torque always: use τ=N⋅B⋅I⋅A\tau = N \cdot B \cdot I \cdot A with all values already in SI units.

Quick oral answer

Because the galvanometer uses a radial field, the coil plane is always parallel to B, so the torque is always the maximum value NBIANBIA — here 50×0.25×5×4×10−3=0.2550 \times 0.25 \times 5 \times 4 \times 10^{-3} = 0.25 newton-metre.

Common mistakes

  1. 1.Forgetting to multiply by the number of turns N=50N = 50, giving 0.005 N·m instead of 0.25 N·m.
  2. 2.Mishandling the area 4.0×10−3 m24.0 \times 10^{-3} \text{ m}^2 (using 4.0 or 4.0×10−24.0 \times 10^{-2} by mis-reading the exponent).
  3. 3.In the OR part, connecting the resistance in parallel (like an ammeter shunt) instead of in series for a voltmeter.

Analysis and explanation

Concept:


A current-carrying coil of N turns and area A in a magnetic field B experiences a torque τ=NBIAsin⁡θ\tau = NBIA \sin\theta. A moving-coil galvanometer uses a radial (cylindrical) magnetic field created by concave pole pieces and a soft-iron core, so the plane of the coil is always parallel to B and θ=90∘\theta = 90^\circ. Hence the torque stays at its maximum value NBIANBIA for every orientation.


Working:


  • All quantities are in SI units, so no conversion is needed.
  • τ=50×0.25×5×4.0×10−3=0.25\tau = 50 \times 0.25 \times 5 \times 4.0 \times 10^{-3} = 0.25 N·m.

Exam trap:


Do not treat 4.0×10−3 m24.0 \times 10^{-3} \text{ m}^2 as 4.0 m24.0 \text{ m}^2 or drop the number of turns N — both are common ways students land on the wrong distractor. The OR part is a standard voltmeter conversion where R goes in series, not parallel.


Real-world:


The radial-field design is exactly what makes analog ammeters and voltmeters give a linear, evenly spaced scale.

Interesting facts

The moving-coil galvanometer was refined by Jacques-Arsène d'Arsonval and Marcel Deprez in 1882; the 'd'Arsonval movement' is still the heart of most analog meters.

Because the radial field keeps torque constant, the deflection is directly proportional to current — this linearity is why analog meter dials have uniformly spaced markings.

A sensitive galvanometer can detect currents as small as 10−10 A10^{-10} \text{ A}, small enough to register the tiny thermoelectric currents in a thermocouple.

Question details

Q31 of 39 questions

Board
CBSE
Class
Class 12
Subject
Physics
Year
2026
Section
Section D
Marks
1
Type
MCQ
Chapter
Moving Charges and Magnetism
Topic
Torque on a moving coil galvanometer
Bloom's level
apply
Syllabus topic
Unit III: Magnetic Effects of Current and Magnetism — Moving Coil Galvanometer (torque on a current loop, τ = NBIA)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is the torque on the coil taken as NBIANBIA and not NBIAsin⁡θNBIA \sin\theta?

The galvanometer is built with a radial magnetic field using cylindrical pole pieces and a soft-iron core. This keeps the plane of the coil always parallel to the field, so the angle between the field and the coil plane makes sin⁡θ=1\sin\theta = 1 for every position of the coil. The torque therefore stays at its maximum value NBIANBIA throughout the rotation, which also makes the scale linear.

In the OR part, why is the resistance connected in series and not in parallel?

A voltmeter must draw very little current and read potential difference, so a high resistance R is added in series with the galvanometer. The full-scale current I_g then flows only when the applied voltage equals V, giving R=V/Ig−G=12/0.003−15=3985R = V/I_g - G = 12/0.003 - 15 = 3985 Ω. A parallel (shunt) resistance is used only to make an ammeter.