(a) Derive the expression for the torque acting on a current carrying loop placed in a magnetic field. (b) Exp

Question

Q. (a) Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.

(b) Explain the significance of a radial magnetic field when a current carrying coil is kept in it.

CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★

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Concept understanding — Moving Coil Galvanometer

Moving Coil Galvanometer: From Intuition to Formula

Imagine you have a tiny, lightweight coil of wire, suspended so it can rotate freely. If you pass a current through it, that coil becomes an electromagnet. Now place it between the poles of a strong permanent magnet. The coil will try to twist — it experiences a torque. The bigger the current, the harder it twists. That is the entire physical idea behind a moving coil galvanometer: use a current to produce a rotation, and measure the rotation to know the current.

But a freely spinning coil would just keep turning. To get a useful measurement, you need something that opposes that rotation — a restoring force that grows as the coil turns further. That is the job of a spring (usually a fine phosphor-bronze strip called a torsion fibre). The spring twists as the coil rotates, producing a restoring torque that exactly balances the magnetic torque at some angle. That equilibrium angle is your reading.

The Radial Magnetic Field — The Key Trick

Here is the clever part. If the magnetic field were uniform and the coil rotated out of alignment, the torque would change with angle — making the scale non-linear. To avoid that, the poles of the magnet are shaped into concave cylindrical surfaces, and a soft iron cylinder is placed inside the coil. This creates a radial magnetic field: the field lines always point radially outward (or inward), so the plane of the coil is always parallel to the field as it rotates.

Important

In a radial field, the magnetic torque on the coil is independent of the coil's angular position. The torque depends only on the current.

That is what makes the deflection directly proportional to current — a linear scale.

The Physics in Equations

Let the coil have $N$ turns, each of area $A$ . A current $I$ flows through it. The magnetic field strength is $B$ (radial). The torque due to the magnetic field on a single turn is:

\tau_m = N I A B

This is because the force on each vertical side of the coil is $I L B$ (where $L$ is the length of the side), and the lever arm is the width of the coil, so the product gives $I \times (\text{area}) \times B$ per turn.

The spring provides a restoring torque proportional to the twist angle $\theta$ :

\tau_s = k \theta

where $k$ is the torsion constant of the spring (unit: N·m/rad).

At equilibrium, the two torques balance:

N I A B = k \theta

So the deflection is:

\theta = \frac{N A B}{k} \, I

The quantity $\frac{N A B}{k}$ is called the current sensitivity of the galvanometer. It tells you how many radians of deflection you get per ampere of current.

\theta = \left( \frac{N A B}{k} \right) I

What This Means for a Student

Larger $N$ , $A$ , or $B$ makes the galvanometer more sensitive — more deflection for the same current. …

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A current-carrying loop in a uniform magnetic field experiences a torque that tends to rotate it to align its magnetic moment with the field. The torque is $\vec{\tau} = \vec{m} \times \vec{B}$ , where $\vec{m} = I \vec{A}$ is the magnetic dipole moment. In a radial magnetic field, the torque becomes independent of the coil’s angular position, making it proportional to current — the key principle behind moving-coil galvanometers.

The Core Idea: Why a Loop Rotates

When a current flows through a loop placed in a magnetic field, each segment of the wire experiences a magnetic force $\vec{F} = I \vec{L} \times \vec{B}$ . On opposite sides of the loop, these forces are equal in magnitude but opposite in direction, forming a couple — a pair of forces that produces pure rotation without translation. The turning effect of this couple is the torque.

The beauty is that this torque always tries to align the loop’s magnetic moment (a vector perpendicular to the loop’s plane, pointing in the direction given by the right-hand rule for current) with the external magnetic field. This is exactly analogous to a compass needle aligning with Earth’s field.

(a) Deriving the Torque Expression

We’ll take a rectangular loop of sides $a$ and $b$ , carrying current $I$ , placed in a uniform magnetic field $\vec{B}$ . Let the plane of the loop make an angle $\theta$ with the field direction. The loop’s area vector $\vec{A}$ (magnitude $ab$ ) is perpendicular to its plane.

Step 1: Identify the forces on each side

Consider the loop oriented so that its plane is at an angle $\theta$ to $\vec{B}$ . The field is horizontal in our diagram.

Sides of length $a$ (the “vertical” sides in the diagram): The current direction is perpendicular to $\vec{B}$ . The force on each is $F_a = I a B$ , directed perpendicular to both the side and the field. These two forces are equal, opposite, and parallel — they form a couple. Their lines of action are separated by the perpendicular distance $b \sin\theta$ (the projection of side $b$ perpendicular to the field).
Sides of length $b$ (the “horizontal” sides): The current in these sides has a component parallel to $\vec{B}$ (depending on orientation). The forces on these sides are equal, opposite, and collinear — they cancel each other’s turning effect completely. They contribute nothing to the net torque.

Tip

Only the sides perpendicular to the field contribute to torque. The sides parallel to the field produce forces that cancel without any lever arm.

Step 2: Calculate the torque magnitude

The torque from the couple on the $a$ -sides is:

\tau = (\text{force on one side}) \times (\text{perpendicular distance between them})

\tau = (I a B) \times (b \sin\theta) = I (ab) B \sin\theta

Since $ab = A$ (area of the loop), we get:

\tau = I A B \sin\theta

Step 3: Write in vector form

Define the magnetic dipole moment $\vec{m} = I \vec{A}$ , where $\vec{A}$ has magnitude $A$ and direction perpendicular to the loop (right-hand rule: curl fingers along current, thumb gives $\vec{A}$ ). Then:

\vec{\tau} = \vec{m} \times \vec{B}

The magnitude is $|\vec{\tau}| = m B \sin\theta$ , exactly as derived.

\vec{\tau} = \vec{m} \times \vec{B} \quad \text{where} \quad \vec{m} = I \vec{A}

Step 4: Generalisation to any shape

This result holds for any planar loop, not just rectangles. Any loop can be thought of as a sum of infinitesimal rectangular strips, each contributing $d\vec{\tau} = I d\vec{A} \times \vec{B}$ . Integrating gives the same expression with $\vec{A}$ as the total area vector.

Note

For a coil of $N$ turns, the magnetic moment becomes $\vec{m} = N I \vec{A}$ , and the torque is $\vec{\tau} = N I \vec{A} \times \vec{B}$ .

(b) The Radial Magnetic Field — Why It Matters …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set 55/3/11 markMCQ
$Q. Assertion (A) : The cylindrical soft iron core in a moving coil galvanometer only makes the magnetic field radial and does not affect the strength of the magnetic field. Reason (R) : In a moving coil galvanometer, the plane of the coil is always perpendicular to the magnetic field. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false.$
›Reveal solutionSolution

The soft iron core makes the field radial AND, being ferromagnetic, concentrates the magnetic flux — it increases the field strength, so the Assertion is false. In a radial field the plane of the coil is always parallel to the field lines (the coil's normal is perpendicular to $\vec B$ ), so the Reason is also false. The correct option is (D).

The question tests two separate facts about the moving coil galvanometer: what the cylindrical soft iron core actually does, and how the coil sits relative to the magnetic field.

1. Role of the soft iron core — is the Assertion true?

The core, together with the concave pole pieces, shapes the field in the air gap so that it is radial: at every angular position of the coil, the field lines point along the radius. This makes the deflecting torque independent of the coil's position, which is what gives the galvanometer its linear scale $\theta \propto I$ .

But that is not all the core does. Soft iron is ferromagnetic, with a very high relative permeability ( $\mu_r \gg 1$ ). It provides a low-reluctance path for the magnetic flux, so the flux crowds through the core and the field strength $B$ in the narrow air gap becomes much larger than it would be without the core. The claim that the core "only makes the field radial and does not affect the strength" is therefore false — the Assertion is false.

2. Orientation of the coil — is the Reason true?

In the radial field the field lines run along the radius, and the plane of the rectangular coil (tangential to the cylindrical core) always contains those field lines. So the plane of the coil is always parallel to the magnetic field — equivalently, the coil's normal is perpendicular to $\vec B$ in every position. That is exactly what keeps the torque at its maximum value throughout the rotation: …
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CBSE 2025Set D1 markMCQ
$Q. If the number of turns is increased in any moving coil galvanometer, then its sensitivity (A) increases (B) decreases (C) remains unchanged (D) may increase or may decrease$
›Reveal solutionSolution

A galvanometer's current sensitivity is (NAB/k), directly proportional to the number of turns N, so more turns → higher sensitivity.

The deflection of a moving-coil galvanometer is

$\theta = \frac{NAB}{k}\,I$

so its current sensitivity is

$\frac{\theta}{I} = \frac{NAB}{k}$

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. In any electric circuit, galvanometer in its original form is used to -$
(a) detect the current
(b) measure the current
(c) measure the voltage
(d) measure the resistance
›Reveal solutionSolution

A galvanometer in its basic form is a sensitive current-detecting device, not a calibrated measuring instrument.

A galvanometer is a sensitive instrument used to detect the presence (and direction) of a small current in a circuit through the deflection of a coil/needle. In its original form it is not calibrated to read numerical values of current, voltage o …

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CBSE 2024Set A1 markMCQ
$Q. The value of current obtained in a moving coil galvanometer is proportional to (A) deflection (θ) (B) resistance (R) (C) magnetic field (B) (D) none of these$
›Reveal solutionSolution

A moving-coil galvanometer is linear: I ∝ θ (the deflection).

In a moving-coil galvanometer, the current-carrying coil in the radial magnetic field experiences a deflecting torque $NBIA$ , balanced by the restoring torque $k\theta$ of the suspension:

$NBIA = k\theta \;\Rightarrow\; I = \frac{k}{NBA}\,\theta.$

…

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CBSE 2024Set ANNUAL1 mark
$Q. Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer ?$
›Reveal solutionSolution

The soft-iron core makes the field radial, ensuring the deflecting torque (and hence the scale) is uniform.

In a moving-coil galvanometer, concave pole pieces together with a cylindrical soft-iron core placed inside the coil make the magnetic field radial — i.e. $\vec B$ is always along the plane of the coil, no matter what angle the coil has turned through. Because of this, the angle between the field and the normal to the coil stays 90° throughout the motion, so the deflecting torque $\tau = NBIA$ has no $\sin\theta$ dependence and stays proportional to the current $I$ alone. This gives a uniform torque for a given current at every deflection, so the pointer's deflection is directly proportional to the cu …

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CBSE 2023Set ANNUAL1 mark
$Q. What happens to the voltage sensitivity of the galvanometer when the current sensitivity of a moving coil galvanometer is doubled by doubling the number of turns of the coil?$
›Reveal solutionSolution

Doubling the turns doubles both the current sensitivity and the coil's resistance, leaving voltage sensitivity the same.

Current sensitivity of a galvanometer is $I_s = \dfrac{NBA}{k}$ , which is directly proportional to the number of turns $N$ ; doubling $N$ doubles $I_s$ , as stated. Voltage sensitivity is defined as $V_s = \dfrac{I_s}{R} = \dfrac{NBA}{kR}$ , where $R$ is the resistance of the galvanometer coil. Since doubling the number of turns (with the same wire, same coil geometry) also roughly doubles the length of wire used and hence doubles the coil's resistance $R$ , the factor of 2 in the numerator ( $N$ ) is cancelled by the factor of 2 in the den …

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CBSE 2022Set I1 markMCQ
$Q. Which of the following is correct for galvanometer constant? (A) C/(NAB) (B) C(B)/(NA) (C) C(A/(NB)) (D) C(N/(AB))$
›Reveal solutionSolution

Galvanometer constant $k = \dfrac{C}{NAB}$ .

In a moving-coil galvanometer, the current $I$ produces a torque $NIAB$ that is balanced at equilibrium by the restoring torque $C\theta$ of the suspension (torsional constant $C$ ):

$NIAB = C\theta \;\Rightarrow\; I = \left(\frac{C}{NAB}\right)\theta.$

The proportionality factor relating current to deflection is the galvanometer constant …

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CBSE 2022Set ANNUAL1 mark
$Q. Why does a moving coil galvanometer use concave magnetic poles?$
›Reveal solutionSolution

Concave poles give a radial field -> uniform torque per unit current -> linear scale.

In a moving-coil galvanometer, the deflecting torque on the coil is $\tau = NIAB\sin\phi$ , where $\phi$ is the angle between the plane of the coil and the field $B$ . If $B$ were uniform (as between flat pole pieces), $\phi$ would change as the coil rotates, making the torque, and hence the deflection, a non-linear function of the rotation angle — the scale would be cramped/uneven. By using concave (curved) pole pieces together with a soft-iron cylindrical core, the magnetic field lines are made radial, i.e. always directed along the plane of the coil no matter what angle the coil has turned to. This keeps $\phi = 90°$ at all deflections, so $\sin\phi = 1$ always …

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CBSE 2021Set NC1 mark
$Q. Why are pole pieces of a magnet within a galvanometer made concave?$
›Reveal solutionSolution

Concave pole pieces bend the field lines so they always point along the coil's plane no matter how far it has rotated, keeping the torque formula free of any angle-dependence.

Reasoning

If the field between the poles were uniform, the deflecting torque on the coil would be $\tau=nBIA\sin\theta$ , which depends on the coil's instantaneous angle $\theta$ — making the current-vs-deflection relationship non-linear (a non-uniform scale).

Concave pole pieces (curved to match the coil's cylindrical geometry, often combined with a soft-iron cylindrical core) shape the field so that it is always radial — i.e. the field lines are always directed along the plane of the coil, regardless of its angular position. This eliminates the $\sin\theta$ dependence entirely:

$\tau = nBIA \quad (\text{no}\ \sin\theta \text{ factor})$

…

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CBSE 2020Set XS1 mark
$Q. Define current sensitivity of a moving coil galvanometer.$
›Reveal solutionSolution

Deflection per unit current, $S_I=\theta/I=NBA/k$ .

Definition. The current sensitivity of a moving-coil galvanometer is the deflection $\theta$ produced per unit current $I$ passing through it:

$S_I=\frac{\theta}{I}.$

At equilibrium the deflecting torque $NBIA$ balances the restoring torque $k\theta$ , so $NBIA=k\theta$ , giving

$S_I=\frac{\theta}{I}=\frac{NBA}{k},$ …

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CBSE 2019Set HE2341 mark
$Q. Match the following. Column A item: Moving coil Galvanometer. Select the matching option from Column B:$
(i) Maximum deviation by prism
(ii) Work function
(iii) Law of conservation of energy
(iv) Magnetic effect of current
(v) Maximum speed in glass.
›Reveal solutionSolution

A moving-coil galvanometer works because a current-carrying coil placed in a magnetic field experiences a torque — the magnetic effect of current.

A moving coil galvanometer consists of a rectangular coil suspended in a strong, radial magnetic field of a permanent magnet. When a current is passed through the coil, each current-carrying side of the coil experiences a force $F = BIl$ in the magnetic field, and the pair of oppositely-directed forces on the two sides produces a torque that deflects the coil. This is a direct application of the magnetic effect of …

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Phosphor-bronze wire is used for suspension in a moving coil galvanometer, because it has :$
(a) large couple per unit twist
(b) small couple per unit twist
(c) high conductivity
(d) high resistivity
›Reveal solutionSolution

Phosphor-bronze is chosen for the suspension because its low torsional (restoring) constant gives the galvanometer high current sensitivity.

In a moving coil galvanometer, the coil is suspended by a thin wire (or fibre) that also acts as one of the current leads and provides the restoring torque against the deflecting torque produced by the magnetic field acting on the current-carrying coil.

At equilibrium, the deflecting torque equals the restoring torque: $NIAB = k\theta$ , where $k$ is the torsional constant (restoring couple per unit twist/radian) of the suspension.

The current sensitivity of the galvanometer is $\dfrac{\theta}{I} = \dfrac{NAB}{k}$

This shows sensitivity is inversely proportional to $k$ : the smaller the torsional constant of the suspension, the larger the deflection $\theta$ produced by a given small current $I$ , i.e. the more sensitive the instrument.

…

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Q.(a) Derive the expression for the torque acting on a current carrying loop placed in a magnetic field.

The Core Idea: Why a Loop Rotates

(a) Deriving the Torque Expression

(b) The Radial Magnetic Field — Why It Matters …

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