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Physics · Moving Charges and Magnetism

Torque on a Current Loop & Magnetic Dipole

A current loop placed in a uniform magnetic field feels no net force, yet it twists. NCERT Section 4.9 shows that the two halves of the loop carry an equal and opposite pair of forces — a couple — whose turning effect is the torque $\boldsymbol{\tau} = \mathbf{m}\times\mathbf{B}$. This single result defines the magnetic moment $\mathbf{m}=NI\mathbf{A}$, explains why a loop behaves at a distance exactly like a magnetic dipole, and underpins the moving-coil galvanometer. For NEET, the formulae for maximum and zero torque and the potential energy of orientation are tested almost every year.

Why the net force is zero

Consider a rectangular loop of sides $a$ and $b$ carrying a steady current $I$, held in a uniform magnetic field $\mathbf{B}$. NCERT first takes the simplest orientation: the field lies in the plane of the loop. The two arms parallel to $\mathbf{B}$ feel no force, while the arm AB (perpendicular to $\mathbf{B}$) feels a force $F_1 = IbB$ directed into the plane, and the opposite arm CD feels an equal force $F_2 = IbB = F_1$ directed out of the plane.

Because $F_1$ and $F_2$ are equal in magnitude and opposite in direction, the net force on the loop is zero. This is true in any uniform field and is the magnetic analogue of an electric dipole in a uniform electric field. What survives is a couple — the two forces are not collinear, so they exert a turning effect that tends to rotate the loop.

Figure 1 · The couple
B A B C D I F₁ (into page) F₂ (out of page) τ rotates loop
The field lies in the plane of the loop ($\theta = 90^\circ$). Forces $F_1 = F_2 = IbB$ on arms AB and CD are equal and opposite but not collinear, forming a couple. Net force is zero; the torque is maximum.

Torque on a rectangular loop

With the field in the plane, the two forces are separated by the full width $a$, so the torque is the sum of the moments about the central axis:

$$ \tau = F_1\,\frac{a}{2} + F_2\,\frac{a}{2} = IbB\cdot a = I(ab)B = IAB $$

where $A = ab$ is the area of the loop. Now tilt the loop so its normal makes an angle $\theta$ with $\mathbf{B}$ — the previous case is $\theta = \pi/2$. The forces on AB and CD remain $F_1 = F_2 = IbB$, but the perpendicular distance between them shrinks to $a\sin\theta$. NCERT Eq. (4.21) gives the general result:

$$ \tau = IAB\sin\theta $$

For a coil of $N$ closely wound turns, the same expression holds with the area replaced by the total turns, giving the form you use in problems:

$$ \boxed{\;\tau = NIAB\sin\theta\;}$$ with $\theta$ the angle between the magnetic moment $\mathbf{m}$ and the field $\mathbf{B}$.

Magnetic moment m = NIA

The torque can be written compactly by defining the magnetic dipole moment of the loop (NCERT Eq. 4.22, 4.24):

$$ \mathbf{m} = NI\mathbf{A} \qquad (m = NIA) $$

The direction of the area vector $\mathbf{A}$ — and therefore of $\mathbf{m}$ — is given by the right-hand thumb rule: curl the right-hand fingers along the current, and the thumb points along $\mathbf{m}$, perpendicular to the plane of the loop. Its dimensions are $[\text{A}][\text{L}^2]$ and its SI unit is $\text{A m}^2$. With this definition, both torque results collapse into the single vector equation NCERT writes as Eq. (4.23):

$$ \boldsymbol{\tau} = \mathbf{m}\times\mathbf{B} $$

NEET Trap

The angle θ is between m and B — not between the loop and B

In $\tau = NIAB\sin\theta$, the angle $\theta$ is measured between the magnetic moment (normal to the loop) and the field. Many students plug in the angle between the plane of the loop and the field. The two differ by $90^\circ$, so a careless substitution turns a maximum into a zero.

If a question states "the plane of the coil makes angle $\phi$ with $\mathbf{B}$", then $\theta = 90^\circ - \phi$, so $\tau = NIAB\cos\phi$.

How torque varies with angle

Because the torque carries a $\sin\theta$ factor, its behaviour is the mirror image of the everyday intuition about "facing" a field. The turning effect is strongest when the loop lies edge-on to the field and vanishes when it faces the field squarely.

Orientation θ (m to B) Plane vs B Torque τ = mB sinθ Energy U = −mB cosθ
Stable equilibrium 0° Plane ⟂ to B 0 (minimum) −mB (minimum)
Maximum twist 90° Plane ∥ to B mB (maximum) 0
Unstable equilibrium 180° Plane ⟂ to B 0 +mB (maximum)

At $\theta = 90^\circ$ the loop plane is parallel to $\mathbf{B}$ and the torque is at its peak $\tau_{\max} = NIAB$. As $\theta \to 0$ the forces of the couple become collinear, the perpendicular distance between them shrinks to zero, and both the torque and the net force vanish — a state of equilibrium.

Figure 2 · m perpendicular to the loop
B current loop m θ
The magnetic moment $\mathbf{m}=NI\mathbf{A}$ points along the loop's normal, set by the right-hand thumb rule. The torque depends on the angle $\theta$ between $\mathbf{m}$ and $\mathbf{B}$, vanishing when $\mathbf{m}$ aligns with $\mathbf{B}$.
→ Carry this forward

The same torque $\tau = NIAB$ drives the needle of an ammeter. See how a radial field linearises it in Moving Coil Galvanometer.

Potential energy U = −m·B

Because a torque acts to rotate the dipole, work must be done to turn it against the field, and this work is stored as orientation potential energy. Integrating the rotational work $\mathrm{d}W = \tau\,\mathrm{d}\theta = mB\sin\theta\,\mathrm{d}\theta$ gives the standard result:

$$ U = -\mathbf{m}\cdot\mathbf{B} = -mB\cos\theta $$

The minus sign captures the physics: energy is lowest ($U = -mB$) when $\mathbf{m}$ is parallel to $\mathbf{B}$ at $\theta = 0$, which is the stable equilibrium. Any small rotation produces a restoring torque that pulls the loop back. At $\theta = 180^\circ$ the energy is highest ($U = +mB$), the unstable equilibrium where a small disturbance grows. This is precisely why a small magnet — or any magnetic dipole — aligns itself with an external field.

The work to rotate a dipole from $\theta_1$ to $\theta_2$ is the change in potential energy, $W = U_2 - U_1 = mB(\cos\theta_1 - \cos\theta_2)$. A full flip from $\theta = 0$ to $\theta = 180^\circ$ therefore costs $W = mB(\cos 0 - \cos 180^\circ) = 2mB$ — a result NEET 2017 tested directly.

NEET Trap

Maximum torque ≠ maximum energy

Torque follows $\sin\theta$ and peaks at $\theta = 90^\circ$. Potential energy follows $-\cos\theta$ and peaks at $\theta = 180^\circ$. At $\theta = 90^\circ$ the energy is exactly zero, not maximum. Confusing the angle of maximum torque with the angle of maximum energy is a frequent slip.

$\tau_{\max}$ at $\theta = 90^\circ$ · $U_{\max}=+mB$ at $\theta = 180^\circ$ · $U_{\min}=-mB$ at $\theta = 0$.

Circular loop as a magnetic dipole

A circular loop of radius $R$ carrying current $I$ produces, on its axis at distance $x$, the field (from the axial field of a circular loop, NCERT Eq. 4.11):

$$ B = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}} $$

For points far from the loop, $x \gg R$, the $R^2$ term in the denominator is dropped. Using the area $A = \pi R^2$ and the magnetic moment $m = IA$, this simplifies to the axial field of a magnetic dipole (NCERT Eq. 4.25a):

$$ B = \frac{\mu_0}{4\pi}\,\frac{2m}{x^3} $$

So at large distances a planar current loop is equivalent to a magnetic dipole of moment $m = IA$. NCERT extends the analogy to the equatorial (perpendicular-bisector) point, where $B = \dfrac{\mu_0}{4\pi}\dfrac{m}{x^3}$ for $x \gg R$. The two results become exact for an idealised point magnetic dipole.

Figure 3 · τ versus θ
θ τ mB 0° 90° 180° τ = mB sinθ
Torque on the dipole rises from zero at $\theta = 0$, peaks at $\theta = 90^\circ$ when the loop plane is parallel to $\mathbf{B}$, and falls back to zero at $\theta = 180^\circ$.

Analogy with the electric dipole

The magnetic dipole formulae are obtained from the electric-dipole results by the substitutions $p_e \to m$, $\mathbf{E}\to\mathbf{B}$ and $1/\varepsilon_0 \to \mu_0$. The correspondence is exact term for term, which is why the two systems are taught back to back.

Quantity Electric dipole Magnetic dipole (current loop)
Dipole moment $p_e$ $m = NIA$
Torque in field $\boldsymbol{\tau} = \mathbf{p}_e\times\mathbf{E}$ $\boldsymbol{\tau} = \mathbf{m}\times\mathbf{B}$
Potential energy $U = -\mathbf{p}_e\cdot\mathbf{E}$ $U = -\mathbf{m}\cdot\mathbf{B}$
Axial field $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2p_e}{x^3}$ $B = \dfrac{\mu_0}{4\pi}\dfrac{2m}{x^3}$
Building block Two monopoles (charges) Loop itself — no monopoles exist

There is one deep difference NCERT stresses: an electric dipole is built from two elementary charges, but no isolated magnetic monopole has ever been found. The current loop is itself the most elementary magnetic element. This led Ampère to suggest that all magnetism arises from circulating currents — partly true, though intrinsic moments of electrons and protons go beyond it.

Quick Recap

Torque on a current loop & magnetic dipole — at a glance

  • Uniform field: net force on a loop is zero; the couple gives torque $\tau = NIAB\sin\theta = \mathbf{m}\times\mathbf{B}$.
  • Magnetic moment $\mathbf{m}=NI\mathbf{A}$, directed along the loop normal by the right-hand thumb rule; unit $\text{A m}^2$.
  • $\theta$ is the angle between $\mathbf{m}$ and $\mathbf{B}$. Torque is maximum ($NIAB$) when the loop plane is parallel to $\mathbf{B}$ ($\theta = 90^\circ$); zero when the plane is perpendicular to $\mathbf{B}$ ($\theta = 0$).
  • Potential energy $U = -mB\cos\theta$: minimum $-mB$ at $\theta = 0$ (stable), maximum $+mB$ at $\theta = 180^\circ$ (unstable). Work to flip = $2mB$.
  • Far from a circular loop, axial field $B = \dfrac{\mu_0}{4\pi}\dfrac{2m}{x^3}$ — the loop acts as a magnetic dipole, $m = IA$.

NEET PYQ Snapshot — Torque on a Current Loop & Magnetic Dipole

Real NEET questions on dipole moment and work done against torque.

NEET 2017

A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 µA and is subjected to a magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is:

  • (1) 1.15 µJ
  • (2) 9.1 µJ
  • (3) 4.55 µJ
  • (4) 2.3 µJ
Answer: (2) 9.1 µJ

$W = MB(\cos 0^\circ - \cos 180^\circ) = 2MB = 2(NIA)B$. With $A = 2.1\times1.25\times10^{-4}\,\text{m}^2$: $W = 2\times250\times85\times10^{-6}\times2.1\times1.25\times10^{-4}\times0.85 \approx 9.1\,\mu\text{J}$.

NEET 2021

A uniform conducting wire of length 12a is wound as a current-carrying coil in the shape of (i) an equilateral triangle of side a, and (ii) a square of side a. The magnetic dipole moments in each case respectively are:

  • (1) $4Ia^2$ and $3Ia^2$
  • (2) $\sqrt{3}\,Ia^2$ and $3Ia^2$
  • (3) $3Ia^2$ and $Ia^2$
  • (4) $3Ia^2$ and $4Ia^2$
Answer: (2)

$m = NIA$. Triangle: $N = 12a/3a = 4$ turns, $A = \tfrac{\sqrt3}{4}a^2$, so $m_1 = 4\cdot I\cdot\tfrac{\sqrt3}{4}a^2 = \sqrt3\,Ia^2$. Square: $N = 12a/4a = 3$ turns, $A = a^2$, so $m_2 = 3Ia^2$.

NEET 2025

A 2 A current flows through two different small circular copper coils having radii ratio 1 : 2. The ratio of their respective magnetic moments will be:

  • (1) 4 : 1
  • (2) 1 : 4
  • (3) 1 : 2
  • (4) 2 : 1
Answer: (2) 1 : 4

$m = IA = I\pi r^2$. With the same current, $m \propto r^2$, so $m_1 : m_2 = r_1^2 : r_2^2 = 1^2 : 2^2 = 1 : 4$.

FAQs — Torque on a Current Loop & Magnetic Dipole

Common doubts NEET aspirants raise on torque, magnetic moment and dipole energy.

What is the formula for torque on a current loop in a uniform magnetic field?
For a coil of N turns carrying current I, area A, in a uniform field B, the torque is τ = NIAB sinθ, where θ is the angle between the magnetic moment m (the normal to the loop) and B. In vector form τ = m × B, with m = NIA directed along the area vector by the right-hand thumb rule.
When is the torque on a current loop maximum and when is it zero?
Torque is maximum (τ = NIAB) when the plane of the loop is parallel to B, which means the angle θ between m and B is 90°. Torque is zero when the plane of the loop is perpendicular to B, i.e. θ = 0 (m parallel to B) or θ = 180° (m antiparallel to B).
What is the magnetic dipole moment of a current loop?
The magnetic moment is m = NIA, where N is the number of turns, I the current and A the area. Its direction is along the normal to the loop plane, fixed by the right-hand thumb rule. The SI unit is A m². A planar current loop is therefore equivalent to a magnetic dipole of moment m = IA.
What is the potential energy of a magnetic dipole in a magnetic field?
The potential energy is U = −m·B = −mB cosθ. It is minimum (U = −mB) at θ = 0, the stable equilibrium where m is parallel to B, and maximum (U = +mB) at θ = 180°, the unstable equilibrium where m is antiparallel to B. At θ = 90° the energy is zero.
How is a current loop analogous to an electric dipole?
The torque τ = m × B mirrors the electric-dipole torque τ = pₑ × E, and the axial field of a current loop at large distance, B = μ₀(2m)/(4πx³), mirrors the electric dipole field with the substitutions pₑ → m and 1/ε₀ → μ₀. The key difference is that magnetic monopoles do not exist, so a current loop is itself the most elementary magnetic dipole.
Does a current loop experience a net force in a uniform magnetic field?
No. In a uniform field the forces on opposite arms are equal and opposite, so the net force on the loop is zero. What remains is a couple, producing a torque that tends to rotate the loop until its magnetic moment aligns with the field.
Related Topics
Moving Charges and Magnetism Back to the full chapter map and all subtopics. Field on Axis of a Circular Loop The axial field that becomes the dipole field at large x. Moving Coil Galvanometer The torque τ = NIAB put to work in a measuring instrument. Magnetic Force & Lorentz Force The force on the loop's arms that creates the couple. Force Between Parallel Currents Forces on current-carrying conductors in a field. Biot–Savart Law The source law behind the loop's own magnetic field.