Q31

1 markMCQSection D

A moving coil galvanometer has a coil with area of cross-section $4.0 \times 10^{-3} \text{ m}^2$ 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

Q: A moving coil galvanometer has a coil with area of cross-section $4.0 \times 10^{-3} \text{ m}^2$ 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

**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 - $\tau = N B I A$ - $\tau = 50 \times 0.25 \text{ T} \times 5 \text{ A} \times 4.0 \times 10^{-3} \text{ m}^2$ - **$\tau = 0.25$ 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: - $R = V/I_g - G = 12 / (3 \times 10^{-3}) - 15 = 4000 - 15$ = **3985 Ω → option (B)**

Moving Charges and Magnetism

Torque on a moving coil galvanometer

Options

(A)1.0 N m

(B)2.0 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

$\tau = N B I A$
$\tau = 50 \times 0.25 \text{ T} \times 5 \text{ A} \times 4.0 \times 10^{-3} \text{ m}^2$
$\tau = 0.25$ 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:

$R = V/I_g - G = 12 / (3 \times 10^{-3}) - 15 = 4000 - 15$ = 3985 Ω → option (B)

torque on galvanometer coilτ = NBIAmoving coil galvanometerradial magnetic fieldnumber of turnsvoltmeter conversion series resistancefull scale deflection0.25 N m

Marking Scheme

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

Hint

Radial field ⇒ maximum torque always: use $\tau = N \cdot B \cdot I \cdot A$ 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 $NBIA$ — here $50 \times 0.25 \times 5 \times 4 \times 10^{-3} = 0.25$ newton-metre.

Analysis & Explanation

Concept:

A current-carrying coil of N turns and area A in a magnetic field B experiences a torque $\tau = NBIA \sin\theta$ . 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 $\theta = 90^\circ$ . Hence the torque stays at its maximum value $NBIA$ for every orientation.

Working:

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

Exam trap:

Do not treat $4.0 \times 10^{-3} \text{ m}^2$ as $4.0 \text{ m}^2$ 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.

Common Mistakes

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

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 $10^{-10} \text{ A}$ , small enough to register the tiny thermoelectric currents in a thermocouple.

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Frequently Asked Questions

Why is the torque on the coil taken as $NBIA$ and not $NBIA \sin\theta$ ?

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 $\sin\theta = 1$ for every position of the coil. The torque therefore stays at its maximum value $NBIA$ 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 $R = V/I_g - G = 12/0.003 - 15 = 3985$ Ω. A parallel (shunt) resistance is used only to make an ammeter.