A wire of length L is bent round into (i) a square coil having N turns and (ii) a circular coil having N turns

Question

L

(i) a square coil having

N

turns and

(ii) a circular coil having

N

turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.

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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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The maximum torque on a current-carrying coil in a magnetic field depends on its area and number of turns. For the same wire length and same number of turns, the square coil has a larger area than the circular coil, so the torque ratio is $\frac{\pi}{4}$ .

The key idea here is that the maximum torque on a coil in a uniform magnetic field is given by $\tau_{\text{max}} = N I A B$ , where $N$ is the number of turns, $I$ is the current, $A$ is the area of the coil, and $B$ is the magnetic field strength. Since the wire length $L$ is fixed and the same current flows in both cases, the only difference comes from the area each shape can enclose.

Let’s work through it step by step.

Understand the geometry constraint.

The total length of wire is $L$ . When this wire is bent into a coil of $N$ turns, each turn uses a fraction of the total length. For the square coil, each turn is a square of side $a$ , so the perimeter of one turn is $4a$ . With $N$ turns, the total wire length is $L = N \times (4a)$ .

For the circular coil, each turn is a circle of radius $r$ , so the circumference of one turn is $2\pi r$ . Thus, $L = N \times (2\pi r)$ .
Find the side and radius in terms of $L$ and $N$ .

From the square:

4aN = L \quad \Rightarrow \quad a = \frac{L}{4N}

From the circle:

2\pi r N = L \quad \Rightarrow \quad r = \frac{L}{2\pi N}

Compute the area of one turn for each shape.

Area of one square turn: $A_{\text{square}} = a^2 = \left(\frac{L}{4N}\right)^2 = \frac{L^2}{16N^2}$

Area of one circular turn: $A_{\text{circle}} = \pi r^2 = \pi \left(\frac{L}{2\pi N}\right)^2 = \pi \cdot \frac{L^2}{4\pi^2 N^2} = \frac{L^2}{4\pi N^2}$
Write the maximum torque for each coil.

The torque on a coil in a magnetic field is $\tau = N I A B \sin\theta$ , where $\theta$ is the angle between the plane of the coil and the field. The maximum torque occurs when $\sin\theta = 1$ , i.e., when the plane of the coil is parallel to the field.

So: …

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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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Q.A wire of length LLL is bent round into

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Q.A wire of length $L$ is bent round into

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