Torque on a Current Loop & Magnetic Dipole Moment

Forces on the Four Sides of a Rectangular Loop

A current loop sitting in an external magnetic field is the basic building block of motors, galvanometers, and even the orbiting electron in an atom. Let us see what an external $\vec{B}$ does to it.

Setup

Consider a rectangular loop of sides $a$ (along the axis of rotation) and $b$ (perpendicular to that axis), $N$ turns, carrying current $I$, placed in a uniform external magnetic field $\vec{B}$. Let the loop's plane make some angle with $\vec{B}$ — equivalently, the area vector $\hat{n}$ (perpendicular to the loop face) makes angle $\alpha$ with $\vec{B}$.

Forces on the four sides

Apply $\vec{F} = I\vec{L}\times\vec{B}$ to each side:

• Two sides parallel to $\vec{B}$ (the sides of length $a$ when the loop is oriented as shown): the current is along (or against) $\vec{B}$, so $\vec{L}\times\vec{B} = 0$ → zero force.

• Two sides perpendicular to $\vec{B}$ (the sides of length $b$): each experiences a force of magnitude

  $$F = NI\,b\,B.$$

  By geometry, these forces point in opposite directions along the loop's axis of rotation — equal in magnitude, opposite in direction, applied at different points.

So the net force on the loop is zero — the loop does not translate. But the two perpendicular forces form a couple — and a couple produces a torque.

Deriving the Torque — $\vec{\tau} = \vec{m}\times\vec{B}$

The two equal-and-opposite forces $F = NI\,b\,B$ act at the centres of the two perpendicular sides, separated by a perpendicular distance $a\sin\alpha$ (where $\alpha$ is the angle between the area vector $\hat{n}$ and $\vec{B}$).

Torque magnitude

$$\tau = F\cdot(\text{lever arm}) = (NI\,b\,B)\cdot(a\sin\alpha)$$

Since $ab$ is the area $A$ of the loop:

$$\boxed{\;\tau = NIAB\sin\alpha\;}$$

This formula holds for any planar loop (not just rectangular) — the $A$ in the formula is just the area enclosed.

The magnetic dipole moment $\vec{m}$

Define the magnetic dipole moment of the loop as

$$\vec{m} = NI\vec{A}$$

where $\vec{A} = A\hat{n}$ is the area vector (perpendicular to the loop, sense given by the right-hand thumb rule: curl fingers along the current, thumb points along $\hat{n}$).

In terms of $\vec{m}$, the torque on the loop is

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

with magnitude $\tau = mB\sin\alpha$.

Units of $\vec{m}$

From $\vec{m} = NI\vec{A}$: units are A·m$^2$. Equivalent units (from $\tau = mB$ when $\sin\alpha = 1$): J/T.

Direction of the torque

The torque vector $\vec{\tau}$ is perpendicular to both $\vec{m}$ and $\vec{B}$. It always tries to rotate $\vec{m}$ toward alignment with $\vec{B}$ — i.e., toward $\alpha = 0$. This is exactly analogous to the way an electric field aligns an electric dipole.

Potential Energy and Work Done in Rotating the Loop

When a torque acts on a system, the work done in rotating against it is stored as potential energy. Defining the zero of potential energy at $\alpha = 90°$ (loop's area vector perpendicular to $\vec{B}$):

$$\boxed{\;U(\alpha) = -\vec{m}\cdot\vec{B} = -mB\cos\alpha\;}$$

Stable and unstable equilibrium

• $\alpha = 0$ (loop's $\vec{m}$ parallel to $\vec{B}$): $U = -mB$, minimum energy → stable equilibrium. Torque is zero, and small disturbances drive the loop back.
• $\alpha = 180°$ (loop's $\vec{m}$ antiparallel to $\vec{B}$): $U = +mB$, maximum energy → unstable equilibrium. Torque is zero, but the slightest disturbance sends the loop flipping over.

Work done in rotating from $\alpha_1$ to $\alpha_2$

By the work-energy theorem,

$$W_{ext} = U(\alpha_2) - U(\alpha_1) = mB(\cos\alpha_1 - \cos\alpha_2)$$

Some quick special cases:

Rotation	Work done by external agent
$0° \to 90°$	$mB(1 - 0) = mB$
$0° \to 180°$	$mB(1 - (-1)) = 2mB$
$90° \to 180°$	$mB(0 - (-1)) = mB$

The work done to flip the loop completely, from aligned to anti-aligned, is 2mB — the largest possible single-flip energy cost.

[JEE Tip] Many problems ask for “work done to rotate from a position of stable equilibrium”. This always means starting at $\alpha = 0$, where $U = -mB$. Answer: $W = mB(1 - \cos\alpha_2)$.

Loop = Bar Magnet; Atomic Dipoles; Bohr Magneton

The torque law $\vec{\tau} = \vec{m}\times\vec{B}$ and the energy $U = -\vec{m}\cdot\vec{B}$ are identical in form to the electric-dipole laws $\vec{\tau} = \vec{p}\times\vec{E}$, $U = -\vec{p}\cdot\vec{E}$ from Chapter 1.

Current loop = tiny bar magnet

A current loop carrying current $I$ has a magnetic dipole moment $\vec{m} = IA\hat{n}$. Externally, it looks magnetically identical to a tiny bar magnet of the same dipole moment — same field pattern, same torque in an external $\vec{B}$, same energy. This is the loop–bar-magnet equivalence, central to understanding the magnetism of bulk matter.

Orbital magnetic moment of an electron

Treat an electron in a Bohr orbit (radius $r$, speed $v$) as a current loop: current $I = e/T = ev/(2\pi r)$, area $A = \pi r^2$:

$$m_{orbital} = IA = \dfrac{ev}{2\pi r}\cdot\pi r^2 = \dfrac{evr}{2} = \dfrac{e}{2m_e}\cdot(m_e v r) = \dfrac{e}{2m_e}L$$

where $L = m_e v r$ is the orbital angular momentum. So

$$\vec{m}_{orbital} = -\frac{e}{2m_e}\vec{L}$$

(negative because electron charge is $-e$).

The Bohr magneton

For the ground-state hydrogen electron, $L = \hbar$ (Bohr's quantisation). The smallest “natural” unit of magnetic moment is therefore

$$\boxed{\;\mu_B = \frac{e\hbar}{2m_e} = 9.27\times10^{-24}\ \text{J/T}\;}$$

— called the Bohr magneton. It sets the scale of atomic magnetism: typical atomic moments are a few $\mu_B$, and this is what makes substances paramagnetic or ferromagnetic.

Memory Capsule

A compact recap before moving to Section 11.

Torque on a current loop

$$\vec{\tau} = \vec{m}\times\vec{B},\quad \tau = NIAB\sin\alpha$$

Quantity	Meaning
$\vec{m} = NI\vec{A}$	Magnetic dipole moment of the loop
$\hat{n}$	Area-vector direction (right-hand rule on current)
$\alpha$	Angle between $\vec{m}$ and $\vec{B}$
Units of $m$	A·m$^2$ (equivalently J/T)

Forces and the couple

• Net force on a loop in a uniform $\vec{B}$ is zero.
• The two sides not parallel to $\vec{B}$ form a couple that produces torque.

Potential energy

$$U = -\vec{m}\cdot\vec{B} = -mB\cos\alpha$$

• $\alpha = 0$: $U = -mB$ (minimum) — stable equilibrium.
• $\alpha = 90°$: $U = 0$.
• $\alpha = 180°$: $U = +mB$ (maximum) — unstable equilibrium.

Work to rotate from $\alpha_1$ to $\alpha_2$

$$W = mB(\cos\alpha_1 - \cos\alpha_2)$$

Loop ↔ bar magnet

A current loop is magnetically equivalent to a bar magnet of moment $m = NIA$. External magnetic effects of both are indistinguishable.

Atomic origin

• Orbital moment: $\vec{m}_{orb} = -(e/2m_e)\vec{L}$.
• Bohr magneton: $\mu_B = e\hbar/(2m_e) = 9.27\times10^{-24}$ J/T — the natural unit of atomic magnetism.

One-line takeaway

A current loop is a magnetic dipole — it feels a torque $\vec{m}\times\vec{B}$ in an external field and stores energy $-\vec{m}\cdot\vec{B}$, exactly mirroring the electric-dipole laws of Chapter 1.

Solved Examples

Example 1: Torque on a single rectangular loop

A rectangular coil of 100 turns and area $4\times 10^{-3}$ m$^2$ carries a current of 2 A. It is placed in a uniform magnetic field of 0.5 T such that the plane of the coil is parallel to the field. Find the torque on the coil.

Solution. When the plane of the coil is parallel to $\vec{B}$, the area vector $\hat{n}$ is perpendicular to $\vec{B}$, so $\alpha = 90°$, $\sin\alpha = 1$.

Formula:

$$\tau = NIAB\sin\alpha$$

Substitute $N = 100$, $I = 2$ A, $A = 4\times 10^{-3}$ m$^2$, $B = 0.5$ T, $\sin\alpha = 1$:

$$\tau = (100)(2)(4\times 10^{-3})(0.5)(1) = 0.4\ N·m$$

Answer: $\tau = 0.4$ N·m.

Example 2: Magnetic dipole moment of a loop

A circular coil of radius 5 cm has 50 turns and carries 4 A. Compute its magnetic dipole moment.

Solution.

$$m = NIA = NI\pi r^2$$

Substitute $N = 50$, $I = 4$ A, $r = 0.05$ m:

$$m = (50)(4)\pi(0.05)^2 = 200\pi\times 2.5\times 10^{-3} = 0.5\pi \approx 1.57\ A·m^2$$

Answer: $m \approx 1.57$ A·m$^2$.

Example 3: Work done to rotate a loop

A magnetic dipole of moment $0.5$ A·m$^2$ is placed in a magnetic field of $0.2$ T. Find the work required to rotate it from a position of stable equilibrium to a position perpendicular to the field.

Solution. Start at $\alpha_1 = 0$ (stable), end at $\alpha_2 = 90°$.

Formula:

$$W = mB(\cos\alpha_1 - \cos\alpha_2) = mB(\cos 0 - \cos 90°) = mB(1 - 0)$$

$$W = (0.5)(0.2) = 0.1\ \text{J}$$

Answer: $W = 0.1$ J.

Example 4: Work to flip a dipole through 180°

For the dipole of Example 3, find the work required to rotate it from the stable to the unstable equilibrium position.

Solution. Start at $\alpha_1 = 0$, end at $\alpha_2 = 180°$.

$$W = mB(\cos 0 - \cos 180°) = mB(1 - (-1)) = 2mB$$

$$W = 2(0.5)(0.2) = 0.2\ \text{J}$$

Answer: $W = 2mB = 0.2$ J — twice the work of the 90° rotation, as expected.

Example 5: Maximum torque

A square coil of side 10 cm has 200 turns and carries 0.5 A. What is the maximum torque it can experience in a 0.4 T field?

Solution. Maximum torque occurs at $\sin\alpha = 1$, i.e., when the loop's plane is parallel to $\vec{B}$.

$$\tau_{max} = NIAB = NI(\text{side}^2)B$$

Substitute $N = 200$, $I = 0.5$ A, $A = (0.10)^2 = 10^{-2}$ m$^2$, $B = 0.4$ T:

$$\tau_{max} = (200)(0.5)(10^{-2})(0.4) = 0.4\ N·m$$

Answer: $\tau_{max} = 0.4$ N·m.

Example 6: Potential energy difference

A coil of magnetic moment $m = 2$ A·m$^2$ is placed in a field $B = 0.3$ T. Find the potential energy of the coil when (a) $\alpha = 0$, (b) $\alpha = 60°$, (c) $\alpha = 180°$.

Solution. Use $U = -mB\cos\alpha$ with $mB = 0.6$ J.

• (a) $\alpha = 0$: $U = -0.6\cos 0 = -0.6$ J
• (b) $\alpha = 60°$: $U = -0.6\cos 60° = -0.6\times 0.5 = -0.3$ J
• (c) $\alpha = 180°$: $U = -0.6\cos 180° = -0.6\times (-1) = +0.6$ J

Answer: (a) $-0.6$ J, (b) $-0.3$ J, (c) $+0.6$ J.

Example 7: Orbital magnetic moment of a hydrogen electron

In the Bohr model of hydrogen, the electron orbits with $v = 2.2\times 10^6$ m/s at radius $r = 5.3\times 10^{-11}$ m. Compute its orbital magnetic moment and compare with the Bohr magneton.

Solution. From the derivation:

$$m_{orb} = \dfrac{evr}{2}$$

Substitute $e = 1.6\times 10^{-19}$ C, $v = 2.2\times 10^6$ m/s, $r = 5.3\times 10^{-11}$ m:

$$m_{orb} = \dfrac{(1.6\times 10^{-19})(2.2\times 10^6)(5.3\times 10^{-11})}{2}$$

$$m_{orb} = \dfrac{1.866\times 10^{-23}}{2} \approx 9.33\times 10^{-24}\ \text{J/T}$$

This agrees with the Bohr magneton, $\mu_B = 9.27\times 10^{-24}$ J/T (small difference due to rounding in $v$ and $r$). Answer: $m_{orb} \approx \mu_B \approx 9.27\times 10^{-24}$ J/T.

Example 8: Equilibrium frequency of small oscillations

A magnetic dipole of moment $m$ and moment of inertia $I_0$ (about the rotation axis through its centre) is slightly displaced from its stable equilibrium in a uniform field $B$. Show that it executes simple harmonic motion and find the period.

Solution. At angle $\alpha$ from the field (small),

$$\tau = -mB\sin\alpha \approx -mB\alpha\ \text{(restoring)}$$

Newton's second law for rotation: $I_0\ddot{\alpha} = -mB\alpha$

$$\ddot{\alpha} = -\dfrac{mB}{I_0}\alpha$$

This is SHM with angular frequency $\omega = \sqrt{mB/I_0}$ and period

$$T = 2\pi\sqrt{\dfrac{I_0}{mB}}$$

Answer: Period $T = 2\pi\sqrt{I_0/(mB)}$. This formula is widely used in magnetometers — measure $T$, and (if $I_0$ is known) you can determine $B$.