PhysicsCBSE Class 12 · Moving Charges and Magnetism

Torque on a Current Loop and Magnetic Dipole Class 12: τ = m × B

Updated 15 October 2026 · Fact-checked

A current loop in a uniform magnetic field feels zero net force but a torque τ = m × B, of size NIAB sin θ, where m = NIA is its magnetic dipole moment and θ the angle between its normal and B. The torque turns m towards B, so the loop acts as a magnetic dipole.

Understand Torque on a Current Loop and Magnetic Dipole

Every electric motor works because a coil carrying current twists in a magnetic field. The twist comes from the forces on the sides of the coil.

Forces on a rectangular loop. Take a rectangular loop with sides a and b carrying current I in a uniform field $B$ , free to turn about an axis through its centre. The force on a straight side of length l is $I l \times B$ . The forces on opposite sides are equal and opposite, so the net force is zero in a uniform field. But the two forces on the sides perpendicular to the rotation axis need not act along the same line; they form a couple.

Torque, plane of loop parallel to B. The two sides of length a (perpendicular to $B$ ) each feel a force $I a B$ , one up and one down, separated by b. Torque $τ = I a B \times b = I A B$ , where $A = ab$ . The other two sides are parallel to $B$ and feel no force.

General orientation. Let the normal to the loop make angle $θ$ with $B$ . The forces IaB on the two sides still act, but the perpendicular distance between their lines of action is $b sin θ$ . So $τ = I A B sin θ$ . For a coil of N turns, $τ = N I A B sin θ$ .

Magnetic dipole moment. Define $m = N I A$ , a vector of size NIA along the normal to the loop, with direction given by the right-hand thumb rule: curl the fingers along the current, and the thumb points along $m$ . Then $τ = m \times B$ , exactly like $τ = p \times E$ for an electric dipole. The SI unit of m is A m² (equivalently J/T).

$θ = 90°$ (plane of loop parallel to $B$ ): torque is largest, $τ = m B$ .
$θ = 0°$ (plane perpendicular to $B$ , $m$ along $B$ ): $τ = 0$ , stable equilibrium.
$θ = 180°$ : $τ = 0$ but unstable.

The work needed to turn the loop from $θ_{1}$ to $θ_{2}$ is $m B (cos θ_{1} - cos θ_{2})$ , and the potential energy is $U = - m \cdot B$ .

The torque does not depend on shape. The result $τ = N I A B sin θ$ holds for a plane loop of any shape, since any loop can be built from tiny rectangles. So for a fixed length of wire, the shape with the largest area (a circle) gives the largest moment.

Current loop as a magnetic dipole. On the axis of a circular loop of radius R, the Biot-Savart law gives $B = \frac{μ _{0} N I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}$ . Far away ( $x ≫ R$ ) this becomes $B = \frac{μ _{0}}{4 π} \frac{2 m}{x ^{3}}$ , the same form as the axial field of an electric dipole, $\frac{1}{4 π ε _{0}} \frac{2 p}{x ^{3}}$ . A current loop therefore behaves as a magnetic dipole: one face acts like a north pole (current anticlockwise when you look at it) and the other like a south pole (current clockwise). This picture links current loops to bar magnets in the next chapter.

Torque on a Current Loop and Magnetic Dipole: formulas and key facts

Magnetic dipole moment of a coil: $m = N I A$; Direction by the right-hand thumb rule; unit A m².
Torque on a coil: $τ = m \times B, τ = N I A B sin θ$; θ = angle between the normal (m) and B.
Net force in a uniform field: $F_{n e t} = 0$; Forces on opposite sides cancel.
Work to rotate and energy: $W = m B (cos θ_{1} - cos θ_{2}), U = - m \cdot B$; Stable at θ = 0°, unstable at 180°.
Field on the axis of a circular coil: $B = \frac{μ _{0} N I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}$; At the centre (x = 0): $\frac{μ _{0} N I}{2 R}$ .
Far field of a current loop (dipole): $B = \frac{μ _{0}}{4 π} \frac{2 m}{x ^{3}}$; For x ≫ R; same form as an electric dipole's axial field.

How to solve Torque on a Current Loop and Magnetic Dipole questions

Use this method for torque, magnetic moment and energy questions on coils.

1Find the area A in m² (rectangle ab, circle πr²) and the magnetic moment m = NIA.
2Find θ, the angle between the normal to the coil and B. If the question gives the angle between the plane of the coil and B, θ = 90° minus that angle.
3Torque: τ = mB sin θ. Largest when the plane of the coil is parallel to B.
4For work or energy, use $m B (cos θ_{1} - cos θ_{2})$ or $U = - m B cos θ$ .
5For 'same wire, different shape or number of turns' questions, express A in terms of the wire length first.
6Give the direction of m with the right-hand thumb rule and say which way the coil turns (towards m along B).

Torque on a Current Loop and Magnetic Dipole in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Torque in an electric motor

Suppose the rectangular coil of a toy motor has 200 turns, each 4 cm by 5 cm, carries 1.5 A and sits in a uniform field of 0.4 T.

(a) Find the magnetic moment of the coil. (b) Find the largest torque and the coil position at which it acts. (c) Why does the motor need a split-ring commutator to keep turning?

Try it first, then show the solutionSolution

$m = N I A = 200 \times 1.5 \times (0.04 \times 0.05) = 0.6$ A m².
Largest torque when the plane of the coil is parallel to B (θ = 90°): $τ = m B = 0.6 \times 0.4 = 0.24$ N m.
When the coil passes θ = 0°, the torque would reverse and pull it back. Reversing the current every half turn keeps the torque in the same sense, so the coil keeps rotating.

Answer: (a) 0.6 A m²; (b) 0.24 N m, with the plane of the coil parallel to B; (c) to reverse the current each half turn so the torque does not reverse.

Memory hook: A motor is a current loop chasing alignment with B, never allowed to arrive.

Stable and unstable positions of a dipole

Imagine a small coil of magnetic moment 0.05 A m² hanging from a thread in a uniform field of 0.02 T, free to turn.

(a) In which position does it settle? (b) What is the largest torque on it? (c) How much work turns it from the settled position until its plane is parallel to the field?

Try it first, then show the solutionSolution

It settles with $m$ along $B$ (plane of coil perpendicular to the field), where U = −mB is least.
$τ_{ma x} = m B = 0.05 \times 0.02 = 1 0^{- 3}$ N m.
Plane parallel to B means θ = 90°: $W = m B (cos 0° - cos 90°) = m B = 1 0^{- 3}$ J.

Answer: (a) with m along B; (b) $1 0^{- 3}$ N m; (c) $1 0^{- 3}$ J.

Memory hook: A free coil is a compass: its moment points along B.

Tricks to remember Torque on a Current Loop and Magnetic Dipole

Short memory aids. Say each one aloud once, then use it on a question the same day.

Plane parallel, torque peak
Torque is maximum when the plane of the coil is parallel to B (normal at 90°), zero when the plane faces B.
Use it when: Avoiding the plane-versus-normal angle trap.
m = NIA: 'Nice Indian Answer'
Turns × current × area.
Use it when: Magnetic moment of any coil.
Electric and magnetic twins
τ = p × E and τ = m × B; U = −p · E and U = −m · B.
Use it when: Carrying results across from the electric dipole.
Anticlockwise north
Looking at a face, anticlockwise current means a north pole; clockwise means south.
Use it when: Polarity of a current loop.

Torque on a Current Loop and Magnetic Dipole: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Torque on a rectangular coil · 2 marks · Very short answerBoard level

A rectangular coil of 100 turns, 10 cm × 5 cm, carries 2 A in a uniform field of 0.2 T. The normal to the coil makes 30° with the field. Find (a) the magnetic moment, (b) the torque, and (c) the largest possible torque.

Try it first, then show the solutionSolution

(a) $m = N I A = 100 \times 2 \times 0.1 \times 0.05 = 1$ A m². (½)
(b) $τ = m B sin 30° = 1 \times 0.2 \times 0.5 = 0.1$ N m. (1)
(c) $τ_{ma x} = m B = 0.2$ N m (plane parallel to B). (½)

Answer: (a) 1 A m²; (b) 0.1 N m; (c) 0.2 N m.

Example 2: Derivation: torque on a current loop · 3 marks · Short answerBoard level

Derive an expression for the torque on a rectangular current loop of area A carrying current I in a uniform magnetic field B, when the normal to the loop makes an angle θ with the field. Express it in vector form.

Try it first, then show the solutionSolution

Loop sides a and b; the sides of length b lie in the plane containing B and the normal, and the sides of length a are perpendicular to B. Forces on the b-sides are equal, opposite and along the same line: they cancel. (1)
Each a-side feels $F = I a B$ , in opposite directions, so the net force is zero but they form a couple. The perpendicular distance between their lines of action is $b sin θ$ . (1)
$τ = I a B \times b sin θ = I A B sin θ$ ; for N turns $τ = N I A B sin θ$ , i.e. $τ = m \times B$ with $m = N I A$ . (1)

Answer: $τ = N I A B sin θ$ ; $τ = m \times B$ .

Example 3: Square coil or circular coil? · 3 marks · Short answerHarder board level

The same length of wire is bent once into a single square loop and once into a single circular loop. Both carry the same current in the same field. Find the ratio of the largest torques on them.

Try it first, then show the solutionSolution

Let the wire length be L. Square: side $\frac{L}{4}$ , area $\frac{L ^{2}}{16}$ . (1)
Circle: $2 π r = L$ , $r = \frac{L}{2 π}$ , area $π r^{2} = \frac{L ^{2}}{4 π}$ . (1)
$τ_{ma x} = I A B$ , so $\frac{τ _{c i r c l e}}{τ _{s q u a r e}} = \frac{L ^{2} /4 π}{L ^{2} /16} = \frac{4}{π} \approx 1.27$ . (1)

Answer: Circle : square = 4 : π ≈ 1.27 : 1; the circular loop feels more torque.

Example 4: Work in turning a coil · 2 marks · Very short answerHarder board level

A coil of magnetic moment 1 A m² is in a uniform field of 0.2 T. Find (a) the work needed to turn it from the stable position to the unstable position, and (b) the torque on it when its plane is parallel to the field.

Try it first, then show the solutionSolution

(a) Stable θ = 0°, unstable θ = 180°: $W = m B (cos 0° - cos 180°) = 2 m B = 0.4$ J. (1)
(b) Plane parallel means normal perpendicular: $τ = m B sin 90° = 0.2$ N m. (1)

Answer: (a) 0.4 J; (b) 0.2 N m.

Example 5: One turn or many? · 5 marks · Long answerHOTS / competency

A wire of fixed length L is wound into a flat circular coil of n turns and carries current I in a uniform field B.
(a) Show that the magnetic moment is $\frac{I L ^{2}}{4 π n}$ . (2)
(b) Hence find the number of turns that gives the largest torque. (1)
(c) Evaluate the moment for L = 0.628 m and I = 1 A with n = 1 and n = 2. (2)

Try it first, then show the solutionSolution

(a) Each turn has circumference $\frac{L}{n}$ , so $r = \frac{L}{2 π n}$ . (1)
$m = n I π r^{2} = n I π \frac{L ^{2}}{4 π ^{2} n ^{2}} = \frac{I L ^{2}}{4 π n}$ . (1)
(b) m is inversely proportional to n, so n = 1 (a single large turn) gives the largest moment and torque. (1)
(c) n = 1: $r = 0.1$ m, $m = π \times 0.01 \approx 0.0314$ A m². (1)
n = 2: $r = 0.05$ m, $m = 2 \times π \times 0.0025 \approx 0.0157$ A m², half as much. (1)

Answer: (a) $m = \frac{I L ^{2}}{4 π n}$ ; (b) n = 1; (c) about 0.0314 A m² and 0.0157 A m².

Example 6: Coil as a magnetic dipole · 4 marks · Case-basedHarder board level

Read the passage and answer.

A circular coil of 50 turns and radius 5 cm carries a current of 2 A.
(a) Find its magnetic moment. (1)
(b) Using the dipole formula, find the field on its axis 50 cm from the centre. (1)
(c) The exact formula gives about $1.24 \times 1 0^{- 6}$ T there. Why do the two values differ slightly? (1)
(d) Find the largest torque on the coil in a uniform field of 0.1 T. (1)

Try it first, then show the solutionSolution

(a) $m = N I π R^{2} = 50 \times 2 \times π \times 0.0025 \approx 0.785$ A m². (1)
(b) $B = \frac{μ _{0}}{4 π} \frac{2 m}{x ^{3}} = 1 0^{- 7} \times \frac{2 \times 0.785}{0.125} \approx 1.26 \times 1 0^{- 6}$ T. (1)
(c) The dipole form assumes $x ≫ R$ ; here x = 10R, so it differs from the exact value by about 1.5%. (1)
(d) $τ_{ma x} = m B = 0.785 \times 0.1 \approx 0.0785$ N m. (1)

Answer: (a) about 0.785 A m²; (b) about $1.26 \times 1 0^{- 6}$ T; (c) the dipole formula is exact only for x ≫ R; (d) about 0.0785 N m.

Where students lose marks in Torque on a Current Loop and Magnetic Dipole

The slips examiners see most often, and what to write instead.

Mistake: Taking θ as the angle between the plane of the coil and B.
The formula uses the angle with the normal.
Wrong working
Plane at 30° to B: τ = NIAB sin 30°
Right working
Plane at 30° to B means normal at 60°: τ = NIAB sin 60°
Do this: Convert to the angle between m (normal) and B every time.
Usually costs: 1 mark.
Mistake: Saying the loop moves along the field.
Mixing torque with force.
Do this: In a uniform field the net force is zero; the loop only rotates.
Usually costs: 1 mark in reasoning or assertion-reason questions.
Mistake: Leaving out the diagram in the torque derivation.
The arm b sin θ is hard to follow without a top view.
Do this: Draw a top view showing the two forces, the normal, B and θ.
Usually costs: 1 mark of a 3-mark derivation.
Mistake: Forgetting the number of turns N.
Using IA instead of NIA.
Wrong working
m = 2 × 0.005 = 0.01 A m² (100 turns)
Right working
m = 100 × 2 × 0.005 = 1 A m²
Do this: Write m = NIA in full each time.
Usually costs: The final answer mark.
Mistake: Using cm² for area.
Forgetting 1 cm² = 10⁻⁴ m².
Do this: Convert the sides to metres before multiplying.
Usually costs: The final answer mark.

Exam tips

Show both forces on the coil, explain why they cancel as a net force, and then find the couple.
State the vector result $τ = m \times B$ and the meaning and direction of m.
Compare with the electric dipole when asked why a current loop is called a magnetic dipole.
Recent board papers have asked this derivation or its use for 3 to 5 marks (for example 2025 55/1/1 Q32 and 2026 55/5/1 Q32(a)); the square-versus-circular coil comparison appeared in 2023 (55/5/1 Q18).
The magnetic moment of a revolving electron is not in the 2026-27 syllabus; do not prepare it.

Practice questions on Torque on a Current Loop and Magnetic Dipole

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A current loop is placed in a uniform magnetic field. The net force on it is
(a) NIAB (b) zero (c) NIAB sin θ (d) depends on the shape
Hint
Opposite sides.
Answer
(b) zero.
Q2 · 1 mark · Very short answerBoard level
Write the SI unit of magnetic dipole moment and state how its direction is found for a current loop.
Hint
m = NIA.
Answer
A m² (or J/T). Its direction is along the normal to the loop, given by the right-hand thumb rule: fingers curled along the current, thumb along m.
Q3 · 2 marks · Very short answerHarder board level
A coil is placed with its plane perpendicular to a uniform magnetic field. What is the torque on it, and is it in equilibrium? Is the equilibrium stable?
Hint
Plane perpendicular means the normal is along B.
Answer
θ = 0° (or 180°), so τ = 0: it is in equilibrium. If m points along B it is stable; if m points opposite to B it is unstable.
Q4 · 2 marks · Very short answerHarder board level
A circular coil of 50 turns and radius 4 cm carries 3 A in a field of 0.25 T, its normal at 60° to the field. Find the torque.
Hint
m = NIπr².
Answer
$m = 50 \times 3 \times π \times 0.0016 \approx 0.754$ A m²; $τ = m B sin 60° \approx 0.754 \times 0.25 \times 0.866 \approx 0.163$ N m.
Q5 · 1 mark · Very short answerHarder board level
A small loop of magnetic moment 0.4 A m² is placed at the origin. Find the magnetic field on its axis at 20 cm.
Hint
Dipole formula.
Answer
$B = \frac{μ _{0}}{4 π} \frac{2 m}{x ^{3}} = 1 0^{- 7} \times \frac{0.8}{0.008} = 1 0^{- 5}$ T.
Q6 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): The torque on a current-carrying coil in a uniform field is greatest when the plane of the coil is parallel to the field.
Reason (R): τ = mB sin θ, and θ, the angle between m and B, is then 90°.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
m is normal to the plane.
Answer
(a).
Q7 · 3 marks · Short answerHOTS / competency
A circular coil of N turns is rewound, using the same wire, into a coil of 2N turns. How do (a) its radius, (b) its magnetic moment and (c) the largest torque on it in the same field (same current) change?
Hint
Wire length fixed: N × 2πr is constant.
Answer
(a) $2 N \cdot 2 π r^{'} = N \cdot 2 π r$ , so $r^{'} = \frac{r}{2}$ . (b) $m^{'} = 2 N \cdot I \cdot π \frac{r ^{2}}{4} = \frac{m}{2}$ . (c) The largest torque mB also halves.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Torque on a coil of N turns?Show answer
τ = NIAB sin θ.
2.Magnetic dipole moment of a coil?Show answer
m = NIA, along the normal by the right-hand thumb rule.
3.When is the torque on a coil maximum?Show answer
When the plane of the coil is parallel to B.
4.Net force on a current loop in a uniform field?Show answer
Zero.
5.Potential energy of a magnetic dipole?Show answer
$U = - m \cdot B$ .
6.Far-field on the axis of a current loop?Show answer
$\frac{μ _{0}}{4 π} \frac{2 m}{x ^{3}}$ .

Torque on a Current Loop and Magnetic Dipole: frequently asked questions

Why is a current loop called a magnetic dipole?

It has a dipole moment m = NIA, feels a torque m × B in a uniform field, and far away its field has the same form as the field of an electric dipole, with m in place of p. One face behaves like a north pole and the other like a south pole.

Does the torque depend on the shape of the loop?

No. For a plane loop the torque is NIAB sin θ, whatever the shape; only the area matters. For a fixed wire length, though, a circle encloses the most area and so gives the most torque.

Why is there no net force on a current loop in a uniform field?

The forces on opposite parts of the loop are equal and opposite, because the current flows in opposite directions there. They cancel as forces but can still form a couple.

What is the difference between stable and unstable equilibrium of a coil?

With m along B (θ = 0°) a small twist creates a torque that brings the coil back: stable. With m opposite to B (θ = 180°) a small twist makes it swing away: unstable.

Which rule gives the direction of the magnetic moment?

The right-hand thumb rule: curl the fingers of the right hand along the current in the loop; the thumb points along m.