Torque on a current loop

This sits at the close of the CBSE Class 12 unit Moving Charges and Magnetism, directly after the force on a current-carrying conductor and the force between two parallel wires — both of which this derivation reuses. The board syllabus explicitly lists "torque experienced by a current loop in uniform magnetic field" with the relation $\vec\tau = \vec m\times\vec B$, so both the formula and a working derivation for a rectangular loop are fair game in a five-mark question. It is deliberately positioned right before the current loop is reintroduced as a magnetic dipole and before the moving-coil galvanometer, so mastering the force-pair argument here pays off twice more within the same chapter.

Picture a rectangular loop of wire, sides of length $a$ (horizontal) and $b$ (vertical), carrying a steady current $I$, free to rotate about a vertical axis through its centre. A uniform magnetic field $\vec B$ points horizontally, in the plane containing the rotation axis. Let $\theta$ be the angle between the loop's normal $\hat n$ — given by curling the right-hand fingers along the current and pointing the thumb — and $\vec B$; equivalently, $\theta$ is also the angle between the field and the plane of the two vertical sides. Each straight side carries current in a magnetic field, so each feels a force $\vec F = I\vec l\times\vec B$. The two horizontal sides (length $a$) make the same angle with $\vec B$ but carry current in opposite senses, so their forces are equal, opposite, and — crucially — collinear, acting along the same vertical line through the centre. Equal, opposite, collinear forces give zero net force and zero net torque from this pair; they merely try to stretch or compress the loop. The two vertical sides (length $b$) are the ones that matter. Each feels a force of magnitude $F = IbB$, directed horizontally, and because the currents in them run opposite ways, the two forces are equal and opposite too — so once again the net force is zero. But they do not act along the same line: they are separated, perpendicular to the force direction, by an arm of length $a\sin\theta$. An equal-and-opposite pair separated by a perpendicular arm is a couple, and a couple exerts a torque even with zero net force.

where: $F = IbB$ is the magnitude of each of the two equal-and-opposite forces on the vertical sides, $a\sin\theta$ is the perpendicular distance between their lines of action, and $A = ab$ is the area enclosed by the loop. For a coil of $N$ identical turns instead of a single loop, every turn contributes this same torque, so the total is $N$ times as large:

This scalar result is the magnitude of a vector relation. Define the loop's magnetic moment $\vec m = NI\vec A$, a vector of magnitude $NIA$ pointing along the normal $\hat n$ (right-hand rule on the current direction). Then the torque on the loop in a uniform field is

where: $\vec m$ is the magnetic moment (unit $\text{A}\,\cdot\,\text{m}^2$), $\vec B$ is the uniform external field (tesla, T), and $\vec\tau$ is the resulting torque (N·m), directed to rotate the loop toward smaller $\theta$ — toward aligning $\vec m$ with $\vec B$. This is structurally identical to $\vec\tau = \vec p\times\vec E$, the torque an electric dipole feels in a uniform field, with $\vec m$ playing the role of $\vec p$; a current loop is, for external-field purposes, a magnetic dipole, exactly the next topic in this chapter.

Board level - rectangular coil at maximum-torque orientation

A rectangular coil of $100$ turns, each of area $0.02\ \text{m}^2$, carries a current of $2\ \text{A}$ and is placed in a uniform magnetic field of $0.5\ \text{T}$ such that the plane of the coil is parallel to the field. Find the torque on the coil, and state whether this is the largest or smallest torque the coil can experience in this field.

Step 1. Identify $\theta$. The plane of the coil is parallel to $\vec B$, so the normal $\hat n$ is perpendicular to $\vec B$ — a common point of confusion, since the formula's angle is always between the normal and the field, not the plane and the field. Here $\theta = 90°$, so $\sin\theta = 1$.

Step 2. Write the values. $N = 100$, $I = 2\ \text{A}$, $A = 0.02\ \text{m}^2$, $B = 0.5\ \text{T}$, $\sin\theta = 1$.

Step 3. Apply $\tau = NIAB\sin\theta$ in stages: $N\times I = 100\times 2 = 200$; $200\times A = 200\times 0.02 = 4$; $4\times B = 4\times 0.5 = 2$; $2\times\sin\theta = 2$.

Step 4. State the result and its significance. $\tau = 2\ \text{N}\,\cdot\,\text{m}$. Since $\theta = 90°$ makes $\sin\theta$ take its largest value, this is the maximum torque this coil can experience while carrying $2\ \text{A}$ in a $0.5\ \text{T}$ field — every other orientation gives less, falling to zero when the plane turns perpendicular to $\vec B$. The torque on the coil is $2\ \text{N}\,\cdot\,\text{m}$, the maximum possible for these values of $N$, $I$, $A$, and $B$.

Competitive level - circular coil at an oblique angle, plus the restoring-torque check

A circular coil of $50$ turns and radius $0.1\ \text{m}$ carries a current of $3\ \text{A}$ in a magnetic field of $0.4\ \text{T}$. The normal to the coil makes an angle of $60°$ with the field. (a) Find the torque on the coil. (b) If the coil is now released from this orientation and free to rotate, in which sense does it turn, and what is the torque when it has rotated to $\theta = 30°$ from the field?

Step 1. Find the enclosed area of the circular coil, since $A$ means the area of one turn regardless of shape: $A = \pi R^2 = \pi(0.1)^2 = 0.01\pi \approx 0.0314\ \text{m}^2$.

Step 2. Apply $\tau = NIAB\sin\theta$ with $\theta = 60°$, $\sin 60° \approx 0.866$: $N\times I = 50\times 3 = 150$; $150\times A = 150\times 0.0314 = 4.71$; $4.71\times B = 4.71\times 0.4 = 1.884$; $1.884\times 0.866 \approx 1.632$.

Step 3. State part (a). $\tau \approx 1.63\ \text{N}\,\cdot\,\text{m}$ at $\theta = 60°$.

Step 4. Reason about the sense of rotation for part (b). The torque $\vec\tau = \vec m\times\vec B$ always acts to reduce the angle between $\vec m$ and $\vec B$ — a restoring torque toward the stable equilibrium $\theta = 0°$, just as an electric or magnetic dipole always turns to align with its field. So released from $\theta = 60°$, the coil rotates so $\theta$ decreases.

Step 5. Recompute the torque at $\theta = 30°$, reusing the fixed product $4.71\times 0.4 = 1.884$ from Step 2 (unchanged, since $N$, $I$, $A$, $B$ are fixed) and only replacing $\sin\theta$: $\sin 30° = 0.5$, so $\tau = 1.884\times 0.5 = 0.942\ \text{N}\,\cdot\,\text{m}$. Answers: (a) $\tau \approx 1.63\ \text{N}\,\cdot\,\text{m}$ at $\theta = 60°$; (b) the coil rotates toward smaller $\theta$, and the torque falls to about $0.94\ \text{N}\,\cdot\,\text{m}$ at $\theta = 30°$ — smaller because $\sin\theta$ is smaller, even though the coil is "closer" to aligning. Timing note: keep this under three minutes. Compute the fixed product $NIAB$ once — here $1.884\ \text{N}\,\cdot\,\text{m}$ — and reuse it for every $\theta$ the question asks about, multiplying only by the new $\sin\theta$; do not recompute $N$, $I$, $A$, $B$ from scratch each sub-part. Keep the direction reasoning verbal: torque always pushes toward $\theta = 0°$, so there is no need to redraw a force diagram just to answer "which way."

Confusing the angle in the formula. The single most common error is treating $\theta$ as the angle between the loop's plane and $\vec B$, when it is actually between the loop's normal and $\vec B$. If a question gives the plane angle, convert first: angle with normal equals $90°$ minus angle with plane. Assuming a net force exists. Many students, having just learned $\vec F = I\vec l\times\vec B$ for a straight wire, expect a loop to be pushed sideways by a uniform field. It is not — forces on opposite sides always cancel, leaving torque only. (A non-uniform field can produce a small net force, but that is outside this syllabus.) Forgetting the number of turns $N$. A coil of $100$ turns produces $100$ times the torque of a single loop with the same current, field, and area; dropping $N$ is a routine one-mark loss. Treating maximum torque as the default. Some answers quietly assume $\sin\theta = 1$ even when a different angle is given. Torque is zero at $\theta = 0°$ (stable equilibrium) and rises to a maximum only at $\theta = 90°$ — compute it for whatever $\theta$ is actually stated. Mixing up stable and unstable equilibrium. $\theta = 0°$ ($\vec m$ parallel to $\vec B$) is stable; a small displacement produces a restoring torque back to it. $\theta = 180°$ ($\vec m$ antiparallel to $\vec B$) also gives zero torque but is unstable — the tiniest disturbance grows. Both give $\tau = 0$, but only one is a genuine resting configuration.

• State and apply $\vec\tau = \vec m\times\vec B$ (magnitude $\tau = NIAB\sin\theta$) for a current loop of $N$ turns, current $I$, area $A$, in a uniform field $B$.
• Derive the torque on a rectangular current loop from the forces on its four sides, identifying which pair cancels completely and which pair forms the couple.
• Explain, in words, why the net force on any current loop in a uniform field is always zero even when the net torque is not.
• Identify the orientation of maximum torque ($\theta = 90°$), zero torque at stable equilibrium ($\theta = 0°$), and zero torque at unstable equilibrium ($\theta = 180°$).
• Compute torque for loops of different shapes (rectangular, circular) given $N$, $I$, $A$, $B$ and $\theta$, and recompute it efficiently when only $\theta$ changes.
• Connect this torque expression to the current loop's identity as a magnetic dipole ($m = NIA$) and to the radial-field design of a moving-coil galvanometer.

Ask the tutor: Want to see how this same torque formula, kept constant by a radial field, turns into the linear current scale of a moving-coil galvanometer — ask the AI tutor to walk through that next step.