A flat dielectric disc of radius $R$ carries an excess charge on its surface, with uniform surface charge density $\sigma$. The disc rotates about an axis perpendicular to its plane, passing through its centre, with angular velocity $\omega$. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field of strength $B$, directed perpendicular to the axis of rotation. (a) $\dfrac{1}{4}\sigma\omega\pi BR$ (b) $\dfrac{1}{2}\sigma\omega\pi BR^2$ (c) $\dfrac{1}{4}\sigma\omega\pi BR^3$ (d) $\dfrac{1}{4}\sigma\omega\pi BR^4$

A flat dielectric disc of radius R carries an excess charge on its surface, with uniform surface charge densit

I. Multiple Choice Questions · Q14

R

(a)

\dfrac{1}{4}\sigma\omega\pi BR

(b)

\dfrac{1}{2}\sigma\omega\pi BR^2

(c)

\dfrac{1}{4}\sigma\omega\pi BR^3

(d)

\dfrac{1}{4}\sigma\omega\pi BR^4

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Concept understanding — Torque on a Current Loop

Torque on a Current Loop

The Intuition First

Imagine a compass needle in the Earth's magnetic field. The needle always turns until it points north. Why? Because the needle itself is a tiny magnet, and the field exerts a twist — a torque — that tries to align it.

A current-carrying loop behaves exactly like that tiny magnet. It has a magnetic moment $\vec{m}$ , which is like its own internal compass arrow. When you place this loop in an external magnetic field $\vec{B}$ , the field pulls on one side of the loop and pushes on the other, creating a turning effect.

The loop doesn't feel a net force (if the field is uniform), but it does feel a torque. That torque always tries to rotate the loop so that its magnetic moment points along the field — just like a compass needle.

The Key Players

The magnetic moment of a planar current loop is:

\vec{m} = I \, A \, \hat{n}

where $I$ is the current, $A$ is the area of the loop, and $\hat{n}$ is a unit vector perpendicular to the plane of the loop (direction given by the right-hand rule: curl your fingers along the current, your thumb points along $\vec{m}$ ).

The external field $\vec{B}$ is uniform — same magnitude and direction everywhere in the region of the loop.

The Torque: Two Equivalent Forms

The torque on the loop is:

\tau = m B \sin\theta

where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$ . The torque is maximum when $\vec{m}$ is perpendicular to $\vec{B}$ ( $\theta = 90^\circ$ ), and zero when they are parallel or antiparallel ( $\theta = 0^\circ$ or $180^\circ$ ).

The vector form captures both magnitude and direction:

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

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

The cross product tells you: the torque is perpendicular to both $\vec{m}$ and $\vec{B}$ , and its direction is given by the right-hand rule. This torque always rotates $\vec{m}$ toward $\vec{B}$ .

Why It Happens (The Physics)

Consider a rectangular loop of sides $a$ and $b$ , carrying current $I$ , placed in a uniform field $\vec{B}$ . Let the plane of the loop make an angle $\theta$ with the field.

The two sides of length $a$ are perpendicular to $\vec{B}$ . On each of these sides, the magnetic force is $F = I a B$ , but the forces on opposite sides are in opposite directions. These two forces form a couple — equal and opposite, not along the same line — which produces a torque.

The lever arm for each force is $(b/2) \sin\theta$ , so the net torque is:

\tau = 2 \times (I a B) \times \frac{b}{2} \sin\theta = I (ab) B \sin\theta = I A B \sin\theta

Since $m = I A$ , we get $\tau = m B \sin\theta$ .

Tip

For a rectangular loop, the torque comes only from the sides perpendicular to the field. The sides parallel to the field experience forces that are either zero or along the axis — they contribute nothing to the torque.

The Stable Equilibrium

When $\vec{m}$ is aligned with $\vec{B}$ ( $\theta = 0$ ), the torque is zero. This is a stable equilibrium — if you nudge the loop slightly, the torque brings it back. …

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Step 1. Split the disc into thin rings of radius $x$ and width $dx$ . A ring carries charge $dq=\sigma(2\pi x\,dx)$ , and rotating at $\omega$ it constitutes a current $dI = dq\times\dfrac{\omega}{2\pi} = \sigma\omega x\,dx$ .

Step 2. Each ring's magnetic moment is $dm = dI\times(\pi x^2) = \sigma\omega\pi x^3\,dx$ .

Step 3. Integrate over the whole disc, $x=0$ to $R$ :

$m = \sigma\omega\pi\int_0^R x^3\,dx = \sigma\omega\pi\frac{R^4}{4}$ …

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

CBSE 2026Set ANNUAL1 markMCQ
$Q. If vector m be the magnetic moment of a magnetic dipole placed in a magnetic field of induction vector B, the torque experienced by the dipole will be$
(a) m . B
(b) |m| / |B|
(c) m x B
(d) |m| |B|
›Reveal solutionSolution

The torque on a magnetic dipole is $\vec{\tau} = \vec{m}\times\vec{B}$ , analogous to torque on an electric dipole $\vec{p}\times\vec{E}$ .

When a magnetic dipole of moment $\vec{m}$ is placed in a uniform magnetic field $\vec{B}$ , it experiences a torque that tends to align it with the field. This torque is given by the vector product:

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

…

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CBSE 2026Set SEM31 markMCQ
$Q. The ratio of the radii of two circular loops is 1 : 2. The ratio of their magnetic moments is 1 : 2. The ratio of currents flowing through them is$
(a) 1 : 1
(b) 2 : 1
(c) 4 : 1
(d) 1 : 4
›Reveal solutionSolution

Magnetic moment M = I·(πr²), so I = M/(πr²) ∝ M/r². Substituting the given ratios gives I₁ : I₂ = 2 : 1. Option (b).

Step 1 — magnetic moment of a current loop (NCERT/CBSE Class 12 Physics, Moving Charges and Magnetism): M = I·A = I·πr².

Step 2 — so current I = M/(πr²), i.e. I ∝ M/r².

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Two circular loops having ratio of their radii 1 : 2 possess same magnetic moment. The ratio of their circulating currents will be$
(a) 4 : 1
(b) 1 : 4
(c) 2 : 1
(d) 1 : 2
›Reveal solutionSolution

Magnetic moment m = IA = Iπr²; equal m with r ratio 1:2 forces the current ratio to be 4:1 (inverse of the area ratio).

Magnetic moment of a current loop is $m = I\pi r^2$ . Let the radii be $r_1:r_2 = 1:2$ and the moments be equal, $m_1=m_2$ : …

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CBSE 2024Set A11 mark
$Q.The torque on a rectangular current loop in a uniform magnetic field increases by ———————— the area of the loop. Fill in the blank choosing the appropriate answer from the bracket: (decreasing, interference, helium, greater, diffraction, increasing)$
›Reveal solutionSolution

increasing (the torque is directly proportional to the area of the loop). …

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CBSE 2024Set A1 markMCQ
$Q. The value of torque (τ) experienced by current loop of magnetic moment (m) placed in magnetic field (B) is (A) τ = m \times B (B) τ = B \times m (C) τ = m/B (D) τ = B/m$
›Reveal solutionSolution

A current loop behaves like a magnetic dipole; the torque on it is the cross product of its magnetic moment and the field, τ = m × B.

A planar current loop carrying current I and enclosing area A has a magnetic (dipole) moment $\vec{m} = I\vec{A}$ , directed along the normal to the loop (right-hand rule).

When this dipole is placed in a uniform magnetic field $\vec{B}$ , the two sides of the loop carry equal and opposite forces that form a couple. The resulting torque is

$\vec{\tau} = \vec{m} \times \vec{B}, \qquad \tau = mB\sin\theta$

…

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CBSE 2024Set ANNUAL1 mark
$Q. Write the formula for torque acting on rotating current carrying coil in terms of magnetic dipole moment, in vector form.$
›Reveal solutionSolution

Torque on a magnetic dipole (current loop) in a magnetic field.

A current-carrying coil of magnetic dipole moment $\vec{m}$ placed in a uniform magnetic field $\vec{B}$ experiences a torque that tends to align $\vec{m}$ with $\vec{B}$ :

$\vec{\tau} = \vec{m}\times\vec{B}$ …

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CBSE 2023Set 55/4/11 markMCQ
$L$
›Reveal solutionSolution
For a fixed wire length $L$ , a circle encloses the maximum area, so the square loop has less area than the circular loop. Since torque $\tau = NIAB\sin\theta$ depends directly on area, the square loop experiences less torque — making Assertion false and Reason false as well. The correct choice is (D).

The Core Idea: Magnetic Torque and Area

When a current-carrying loop sits in a uniform magnetic field, the net force on it is zero, but the field exerts a torque that tries to rotate the loop. The magnitude of this torque is given by:
$\tau = N I A B \sin\theta$
Here $N$ is the number of turns (1 for a single loop), $I$ is the current, $A$ is the area enclosed by the loop, $B$ is the magnetic field strength, and $\theta$ is the angle between the loop’s normal and the field. For a fixed $I$ , $B$ , and $\theta$ , the torque is directly proportional to the area $A$ .

So the question boils down to: for a given wire length $L$ , which shape — square or circle — gives a larger area?

Step-by-Step Reasoning

The wire length is fixed. Both loops are made from the same wire of total length $L$ . This length becomes the perimeter of each loop. For the square, each side is $L/4$ , so its area is:

$A_{\text{square}} = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16}$

For the circle, the circumference is $L = 2\pi r$ , so the radius is $r = L/(2\pi)$ . The area is:

$A_{\text{circle}} = \pi r^2 = \pi \left(\frac{L}{2\pi}\right)^2 = \frac{L^2}{4\pi}$

Compare the two areas. Which is larger? Compare $L^2/16$ and $L^2/(4\pi)$ . Since $L^2$ is positive, we compare the denominators: $16$ vs $4\pi \approx 12.57$ . A smaller denominator means a larger fraction, so:

$\frac{L^2}{4\pi} > \frac{L^2}{16}$
Therefore, $A_{\text{circle}} > A_{\text{square}}$ .

Tip

This is a classic isoperimetric result: for a given perimeter, the circle encloses the maximum possible area. Any other shape — square, rectangle, triangle — will have a smaller area. Memorising this saves you from recalculating every time. …
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CBSE 2019Set ANNUAL1 markMCQ
$Q. The torque (τ⃗) experienced by a current - loop of magnetic moment (M⃗) placed in magnetic field B⃗ is -$
(a) τ⃗ = M⃗ × B⃗
(b) τ⃗ = B⃗ × M⃗
(c) τ⃗ = M⃗ / B⃗
(d) τ⃗ = M⃗ . B⃗
›Reveal solutionSolution

Torque on a current loop: τ⃗ = M⃗ × B⃗.

A current loop of magnetic moment M⃗ = NIA n̂ placed in a uniform magnetic field B⃗ experiences a torque

$\vec{\tau} = \vec{M}\times\vec{B},$

…

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CBSE 2016Set ANNUAL1 markMCQ
$Q. A conducting circular loop of radius r carries a constant current I. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is$
(a) BIr
(b) 2πrIB
(c) zero
(d) πrIB
›Reveal solutionSolution

The net magnetic force on any closed current loop placed in a uniform magnetic field is always zero.

For a current loop in a UNIFORM magnetic field B, the net force is F = I∮dl×B = I(∮dl)×B. Since the vector sum of all the line elements around a closed loop is zero (∮dl = 0), the net force is always zero — regardless of the loop's shape or its orientation relative to B. (A net torque can still act if …

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