Torque and Work Done in a Magnetic Field

Q: What is the magnetic dipole moment of a current loop?

M = NIA, where N is turns, I is current, and A is the area vector normal to the loop (right-hand curl rule). SI unit: A·m².

Q: What is the torque on a current loop in a magnetic field?

τ = M×B, magnitude MB sin θ = NIAB sin θ, where θ is the angle between M and B. Maximum MB when loop's plane is parallel to B; zero when M is parallel or antiparallel to B.

Q: Why is the net force on a current loop in a uniform field zero?

Net force = I(L_endpoints × B). For a closed loop, L_endpoints = 0, so net force vanishes. The loop can still feel torque.

Q: What is the potential energy of a magnetic dipole in a field?

U = -M·B = -MB cos θ. Minimum -MB when M parallel to B (stable); maximum +MB when M antiparallel (unstable).

Q: What is the work done in rotating a magnetic dipole?

External work from θ₁ to θ₂: W = MB(cos θ₁ - cos θ₂). Full flip 0° to 180° needs 2MB; 0° to 90° needs MB.

Q: Why does a current loop behave like a bar magnet?

Both produce the same 1/r³ dipole field, feel the same torque M×B, and have the same energy -M·B. All magnetism arises from tiny atomic current loops.

Q: When can a current loop feel a net force?

Only in a non-uniform field. Net force is ∇(M·B); this attracts a magnet toward the pole of another magnet (where field varies fastest).

Q: What is the frequency of small oscillations of a magnetic dipole in a field?

ω = √(MB/I_inertia), period T = 2π√(I_inertia/MB). Used in the vibration magnetometer.

Q: Is θ in the torque formula measured from the plane of the loop or from the normal?

From the normal (M). If loop's plane is parallel to B, θ = 90° and torque is maximum. If loop's plane is perpendicular to B, θ = 0° and torque is zero.

PhysicsMagnetic Effects of Current and MagnetismFor JEE aspirants

A current loop in a uniform magnetic field experiences zero net force but a non-zero torque that tries to align its magnetic moment with the field. For a loop of $N$ turns, area $A$ , carrying current $I$ , the magnetic dipole moment is $M = N I A$ , and the torque is $τ = M \times B$ , magnitude $τ = M B sin θ$ . The potential energy is $U = - M \cdot B$ , and the work done in rotating the dipole from angle $θ_{1}$ to $θ_{2}$ is $W = M B (cos θ_{1} - cos θ_{2})$ . These formulas are directly examined in JEE Main and NEET, and they are the working principle behind the moving coil galvanometer, electric motors, and the concept of magnetic dipoles in matter.

Key Formulas - Quick Reference

Magnetic dipole moment of $N$ -turn loop: $M = N I A$ , where $A$ is normal to the loop (right-hand curl from $I$ )
Torque on loop in uniform $B$ : $τ = M \times B$ , magnitude $τ = M B sin θ = N I A B sin θ$
Maximum torque ( $θ = 90°$ ): $τ_{m a x} = M B = N I A B$
Zero torque ( $θ = 0°$ or $180°$ ): loop is stable ( $θ = 0$ ) or unstable ( $θ = 180°$ )
Potential energy: $U = - M \cdot B = - M B cos θ$
Work done rotating from $θ_{1}$ to $θ_{2}$ : $W = M B (cos θ_{1} - cos θ_{2})$
Angular frequency of small oscillations about stable equilibrium: $ω = M B / I_{inertia}$
Net force on a loop in uniform $B$ : zero. In non-uniform $B$ : $F = \nabla (M \cdot B)$ .

1. Magnetic Dipole Moment

A closed current loop of area $A$ carrying current $I$ has a magnetic dipole moment:

M = I A (single turn)

M = N I A (N turns)

Here $A$ is a vector normal to the loop's plane; its direction is fixed by the right-hand curl rule: curl fingers along $I$ , thumb points along $A$ (and hence along $M$ ).

SI unit of $M$ : $A \cdot m^{2}$ (or joule/tesla).
Direction: anticlockwise current gives $M$ out of the plane; clockwise gives $M$ into the plane.
For a revolving charge $q$ in a circle of radius $R$ with speed $v$ : equivalent current $i = q v / (2 π R)$ , so $M = i A = \frac{1}{2} q v R$ .

2. Torque on a Current Loop

Consider a rectangular loop of length $L$ and width $b$ , in uniform $B$ , with the normal $A$ at angle $θ$ to $B$ . The two sides of length $L$ (perpendicular to $B$ ) feel equal and opposite forces $F = B I L$ separated by $b sin θ$ , forming a couple:

$τ = F \cdot (b sin θ) = (B I L) (b sin θ) = B I A sin θ$

For $N$ turns: $τ = N B I A sin θ$ . In vector form:

τ = M \times B τ = M B sin θ

Figure 1: The two sides perpendicular to

B

feel equal and opposite forces

F = B I L

. Their lines of action are separated by

b sin θ

, so the couple has moment

τ = (B I L) (b sin θ) = B I A sin θ = M B sin θ

Figure 2: The same situation viewed along the rotation axis. Current is out of the page (

⊙

) on one side and into it (

\otimes

) on the other, so the two forces

B I L

point oppositely. Only the perpendicular separation

b sin θ

contributes to the moment, which is why

τ \propto sin θ

2.1 Equilibrium positions

$θ = 0°$ ( $M ∥ B$ ): $τ = 0$ , $U = - M B$ (minimum) - stable.
$θ = 180°$ ( $M ∥ - B$ ): $τ = 0$ , $U = + M B$ (maximum) - unstable.
$θ = 90°$ : $τ = M B$ (maximum torque).

Figure 3: Both

θ = 0°

and

θ = 180°

give zero torque, but only

0°

is stable. Note that at

θ = 90°

the plane of the loop contains

B

- this is the maximum-torque case, not the zero-torque case.

This result - torque is

M \times B

- is general: it holds for any planar current loop, not just rectangular. The area vector

A

carries all the geometric information.

3. Potential Energy and Work Done

The work done by the magnetic torque as the loop rotates from angle $θ_{1}$ to $θ_{2}$ (with respect to $B$ ):

$W_{mag} = \int_{θ_{2}}^{θ_{1}} τ d θ = \int_{θ_{2}}^{θ_{1}} M B sin θ d θ = M B (cos θ_{2} - cos θ_{1})$

(Note the sign: the torque tries to decrease $θ$ , i.e., align $M$ with $B$ .) The corresponding potential energy is:

U (θ) = - M \cdot B = - M B cos θ

and the work done by an external agent in rotating the dipole from $θ_{1}$ to $θ_{2}$ (quasi-statically):

W_{ext} = U (θ_{2}) - U (θ_{1}) = M B (cos θ_{1} - cos θ_{2})

Figure 4:

τ = M B sin θ

peaks at

θ = 90°

, while

U = - M B cos θ

is a minimum at

θ = 0°

(stable) and a maximum at

180°

(unstable). The shaded area under the torque curve is exactly the external work

W = M B (1 - cos θ)

Rotate from $0°$ to $θ$ : $W_{ext} = M B (1 - cos θ)$ .
Rotate from $0°$ to $180°$ : $W_{ext} = 2 M B$ (maximum energy stored).
Rotate from $0°$ to $90°$ : $W_{ext} = M B$ .

3.1 Small-oscillation frequency

For small angular displacements $θ$ from stable equilibrium, $sin θ \approx θ$ , so:

$I_{inertia} \ddot{θ} = - M B θ ⟹ ω = \frac{M B}{I _{inertia}}; T = 2 π \frac{I _{inertia}}{M B}$

This is the basis of the vibration magnetometer and the equivalent oscillation of a bar magnet in a horizontal field.

Solved Example 1

A circular coil of

12

turns, radius

0.10

m, carries

5

A. It is placed in a uniform

B = 0.60

T. Find (i) the magnetic moment, (ii) the maximum torque, (iii) the torque when the moment makes

60°

with

B

Solution:

Area $A = π R^{2} = π (0.10)^{2} = 0.0314 m^{2}$ .

(i) $M = N I A = 12 \times 5 \times 0.0314 = 1.885 A\cdotm^{2}$ .

(ii) $τ_{m a x} = M B = 1.885 \times 0.60 = 1.13 N\cdotm$ .

(iii) $τ = M B sin 60° = 1.13 \times (3 /2) \approx 0.98 N\cdotm$ .

Solved Example 2

For a given length

L

of wire carrying current

I

, how many circular turns give the maximum magnetic moment, and what is its value?

Solution:

If $N$ turns are made, each has circumference $2 π R = L / N$ , so $R = L / (2 π N)$ and $A = π R^{2} = L^{2} / (4 π N^{2})$ .

$M = N I A = N I \cdot \frac{L ^{2}}{4 π N ^{2}} = \frac{I L ^{2}}{4 π N}$

$M$ decreases with $N$ , so maximum is at $N = 1$ : $M_{m a x} = \frac{I L ^{2}}{4 π}$ . A single loop beats any multi-turn coil for a fixed length of wire.

Solved Example 3

A bar magnet of moment

M = 1.0 \times 1 0^{- 4}

J/T sits in a horizontal field

B = 4 \times 1 0^{- 5}

T, free to rotate. Find the work done in rotating it slowly from parallel-to-field to

60°

from the field.

Solution:

$W_{ext} = M B (cos 0° - cos 60°) = M B (1 - 0.5) = 0.5 M B$

$W = 0.5 \times (1.0 \times 1 0^{- 4}) \times (4 \times 1 0^{- 5}) = 2 \times 1 0^{- 9} J = 2 nJ$ .

4. Force on a Dipole in a Non-Uniform Field

In a uniform field, net force on any closed loop is zero. In a non-uniform field, the two ends of the dipole sit at different field strengths, producing a net force:

$F_{x} = M \frac{\partial B}{\partial x}$ (along the direction of variation, when $M ∥ B$ )

More generally: $F = \nabla (M \cdot B)$ . This is why a bar magnet is attracted more strongly to the poles of another magnet than to its middle (where field is weakest).

Figure 5: In a non-uniform field the two poles sit at different field strengths, so the forces no longer cancel. The net force

F = \nabla (M \cdot B)

drags an aligned dipole towards the region of stronger field. In a uniform field the two forces are equal and the net force is exactly zero.

Common Mistakes to Avoid

Watch out

Confusing the angle in the torque formula: $θ$ is the angle between $M$ (normal to the loop) and $B$ , not between the plane of the loop and $B$ . If the loop's plane is parallel to $B$ , then $M ⊥ B$ and torque is maximum ( $θ = 90°$ ), not zero.
Forgetting the $N$ factor for a multi-turn coil.
Wrong sign of potential energy: $U = - M \cdot B$ is minimum (most negative) when $M ∥ B$ ; this is stable equilibrium.
Applying $F = M \times B$ in a uniform field: it is the torque, not the force. Net force in a uniform field is zero.
Confusing work done by torque vs work done by external agent: they have opposite signs. $W_{ext} = U (θ_{2}) - U (θ_{1})$ ; $W_{mag} = - W_{ext}$ .
Assuming small-oscillation formula applies for large angles: the SHM $ω = M B / I$ is valid only for small $θ$ . Large-angle motion is periodic but not simple harmonic.

Frequently Asked Questions

Q1. What is the magnetic dipole moment of a current loop?

For a loop of $N$ turns carrying current $I$ enclosing area $A$ , the magnetic dipole moment is $M = N I A$ , where $A$ is the area vector perpendicular to the loop with direction fixed by the right-hand curl rule from the current direction. SI unit: $A \cdot m^{2}$ .

Q2. What is the torque on a current loop in a magnetic field?

$τ = M \times B$ , magnitude $τ = M B sin θ = N I A B sin θ$ , where $θ$ is the angle between $M$ and $B$ . Maximum torque $M B$ occurs when the loop's plane is parallel to $B$ ( $θ = 90°$ ); zero torque when $M ∥ B$ or $M ∥ - B$ .

Q3. Why is the net force on a current loop in a uniform field zero?

For any wire in a uniform field, the net force is $I (L_{endpoints} \times B)$ . A closed loop has coincident start and end, so $L_{endpoints} = 0$ and net force vanishes. The loop can still feel a torque.

Q4. What is the potential energy of a magnetic dipole in a field?

$U = - M \cdot B = - M B cos θ$ . Minimum ( $U = - M B$ ) when $M ∥ B$ (stable equilibrium); maximum ( $U = + M B$ ) when $M ∥ - B$ (unstable). Zero energy is chosen at $θ = 90°$ .

Q5. What is the work done in rotating a magnetic dipole in a uniform field?

Work done by an external agent in rotating from $θ_{1}$ to $θ_{2}$ : $W_{ext} = M B (cos θ_{1} - cos θ_{2})$ . Rotating from stable equilibrium ( $0°$ ) to $θ$ requires $W = M B (1 - cos θ)$ ; a full flip ( $0°$ to $180°$ ) needs $W = 2 M B$ .

Q6. Why does a current loop behave like a bar magnet?

Both produce the same magnetic field pattern at distances large compared to their size (axial field $\propto M / r^{3}$ ), both feel the same torque $M \times B$ in an external field, and both have the same potential energy $- M \cdot B$ . In fact, at the atomic level, all magnetism arises from electron orbital and spin motion - tiny current loops.

Q7. When can a current loop feel a net force?

Only in a non-uniform magnetic field. Different parts of the loop then sit in different field strengths, so the forces don't cancel. The net force is $F = \nabla (M \cdot B)$ ; this is why a paper clip is attracted toward the pole of a magnet, not toward its middle.

Q8. What is the frequency of small oscillations of a magnetic dipole in a field?

For small angular deviations $θ$ from stable equilibrium, the restoring torque $τ = - M B θ$ produces SHM with angular frequency $ω = M B / I_{inertia}$ and period $T = 2 π I_{inertia} / (M B)$ . This principle is used in the vibration magnetometer.

Q9. Is $θ$ in the torque formula measured from the plane of the loop or from the normal?

From the normal ( $M$ ), not the plane. If the loop's plane is parallel to $B$ (so $M ⊥ B$ ), then $θ = 90°$ and torque is maximum. If the loop's plane is perpendicular to $B$ ( $M ∥ B$ ), then $θ = 0°$ and torque is zero. Confusion here is a very common exam error.

Previous year questions on Torque and Work Done in a Magnetic Field

7 questions from past papers, each with a step-by-step solution.

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Practice questions →Take a mock test

1. Magnetic Dipole Moment

2. Torque on a Current Loop

2.1 Equilibrium positions

3. Potential Energy and Work Done

3.1 Small-oscillation frequency

4. Force on a Dipole in a Non-Uniform Field

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the magnetic dipole moment of a current loop?

Q2. What is the torque on a current loop in a magnetic field?

Q3. Why is the net force on a current loop in a uniform field zero?

Q4. What is the potential energy of a magnetic dipole in a field?

Q5. What is the work done in rotating a magnetic dipole in a uniform field?

Q6. Why does a current loop behave like a bar magnet?

Q7. When can a current loop feel a net force?

Q8. What is the frequency of small oscillations of a magnetic dipole in a field?

Q9. Is θ in the torque formula measured from the plane of the loop or from the normal?

Previous year questions on Torque and Work Done in a Magnetic Field

Ready to master Magnetic Effects of Current and Magnetism?

Q9. Is $θ$ in the torque formula measured from the plane of the loop or from the normal?