The Dipole in a Uniform Magnetic Field

Torque on a Magnetic Dipole

Place a small compass needle of known magnetic moment $\vec{m}$ in a uniform magnetic field $\vec{B}$ and let it oscillate. What does the field do to it?

In a uniform field, the needle experiences no net force — but it does experience a torque:

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

In magnitude,

$$\tau = mB\sin\theta$$

where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$.

Key features to internalise:

• This is a restoring torque — it always tries to rotate $\vec{m}$ back into alignment with $\vec{B}$.
• Maximum torque $\tau_{\max} = mB$ at $\theta = 90^{\circ}$ (needle perpendicular to field).
• Zero torque at $\theta = 0^{\circ}$ and $\theta = 180^{\circ}$ — the two equilibrium orientations (we'll see one is stable, the other unstable).
• The direction of $\vec{\tau}$ is perpendicular to the plane containing $\vec{m}$ and $\vec{B}$ (right-hand rule).

Key Point: Uniform field — torque but no translation. Only a non-uniform field can exert a net force on a dipole. This single line answers a whole family of exam questions.

Magnetic Potential Energy

Because the field exerts a torque, work must be done by an external agent to rotate the dipole slowly against the field torque — so the dipole stores magnetic potential energy, exactly parallel to the electrostatic case.

For a slow rotation from $0$ to $\theta$, the external work is

$$W_{\text{ext}}=\int_0^{\theta} mB\sin\theta'\,d\theta' = mB(1-\cos\theta).$$

Thus, apart from an arbitrary additive constant,

$$U_m = -mB\cos\theta$$

or

$$U_m = -\vec{m} \cdot \vec{B}$$

Taking the constant of integration as zero fixes the zero of potential energy at $\theta = 90^{\circ}$ — the needle perpendicular to the field. (Remember from Chapter 2: the zero of PE is ours to choose.)

The energy landscape:

Orientation	$\theta$	$U_m$	Torque	Nature
$\vec{m}$ parallel to $\vec{B}$	$0^{\circ}$	$-mB$ (minimum)	0	Most stable
$\vec{m}$ perpendicular	$90^{\circ}$	0	$mB$ (maximum magnitude)	Reference level
$\vec{m}$ antiparallel	$180^{\circ}$	$+mB$ (maximum)	0	Most unstable

Work done in rotating a dipole

The external work equals the change in potential energy:

$$W = U_{\text{final}} - U_{\text{initial}} = mB(\cos\theta_1 - \cos\theta_2)$$

Two cases worth memorising: $W(0 \to 90^{\circ}) = mB$ and $W(0 \to 180^{\circ}) = 2mB$.

[NEET Important] Work done to rotate a dipole from its stable position through $180^{\circ}$ is $2mB$ — a standard Boards/NEET application.

Oscillations, and the Classic Reasoning Set

The oscillating needle

Displace the compass needle slightly from alignment and release: the restoring torque $-mB\sin\theta \approx -mB\theta$ (small $\theta$) drives angular oscillations about the field direction. This arrangement can be used to determine either $B$ or the moment $m$ of the needle.

[JEE Tip] For small oscillations, the needle behaves like a torsional pendulum with moment of inertia $\mathcal{I}$:

$$T = 2\pi\sqrt{\frac{\mathcal{I}}{mB}}$$

So $T \propto 1/\sqrt{B}$: a stronger field means faster oscillations. Measuring $T$ gives $B$ (or $m$).

The reasoning set every exam loves

1. Needle in a uniform field: torque, but no net force — it rotates in place.
2. Iron nail near a bar magnet: the nail sits in a non-uniform field; the magnet induces a magnetic moment in it, so the nail feels both a force and a torque. The force is attractive because the induced unlike pole is closer to the magnet.
3. Must every field configuration have N and S poles? No — poles exist only when the source has a net magnetic moment. A toroid has no poles at all; neither does a straight infinite conductor.
4. A magnet exerts no force or torque on itself due to its own field. (But one element of a current-carrying wire can exert a force on another element of the same wire — for a straight wire this force is zero.)

Solved Examples

Example 1: Stable and unstable orientations

A bar magnet of magnetic moment 0.32 J/T is placed in a uniform field of 0.15 T. Find the potential energy in (a) the most stable, (b) the most unstable orientation.

Solution:

1. Formula: $U_m = -mB\cos\theta$.
2. (a) Stable: $\theta = 0^{\circ}$, so $U = -mB = -0.32 \times 0.15 = -0.048$ J. Torque here is zero.
3. (b) Unstable: $\theta = 180^{\circ}$, so $U = +mB = +0.048$ J. Torque is again zero — but any nudge grows.
4. Answer: $-0.048$ J (stable, $\vec{m}$ along $\vec{B}$); $+0.048$ J (unstable, $\vec{m}$ opposite $\vec{B}$).

Example 2: Torque at an angle

A magnet of moment $0.4\,\text{A m}^2$ makes an angle of $30^{\circ}$ with a uniform field of 0.25 T. Find the torque on it.

Solution:

1. Formula: $\tau = mB\sin\theta$.
2. Substitute: $\tau = 0.4 \times 0.25 \times \sin 30^{\circ} = 0.4 \times 0.25 \times 0.5$.
3. Answer: $\tau = 0.05$ N m, directed perpendicular to the plane of $\vec{m}$ and $\vec{B}$, tending to align the magnet with the field.

Example 3: Work done in rotating a magnet

A magnet of moment $2.0\,\text{A m}^2$ lies along a 0.1 T field. How much work is needed to rotate it (a) by $90^{\circ}$, (b) by $180^{\circ}$?

Solution:

1. Formula: $W = mB(\cos\theta_1 - \cos\theta_2)$ with $\theta_1 = 0^{\circ}$.
2. (a) $W = mB(1 - 0) = 2.0 \times 0.1 = 0.2$ J.
3. (b) $W = mB(1-(-1)) = 2mB = 0.4$ J.
4. Takeaway: flipping a dipole completely costs $2mB$ — twice the $90^{\circ}$ cost.

Example 4: Torque on the solenoid of Section 2

The solenoid with $m = 1.28\,\text{A m}^2$ is free to turn about a vertical axis in a horizontal field of 0.25 T. What is the torque when its axis makes $30^{\circ}$ with the field?

Solution:

1. $\tau = mB\sin\theta = 1.28 \times 0.25 \times 0.5$.
2. Answer: $\tau = 0.16$ N m. The solenoid is dynamically identical to a bar magnet of the same moment.

Example 5: Energy cost of a 30-degree turn

For the same solenoid ($m = 1.28\,\text{A m}^2$, $B = 0.25$ T), find the increase in potential energy when it turns from alignment to $30^{\circ}$.

Solution:

1. $\Delta U = -mB\cos 30^{\circ} - (-mB\cos 0^{\circ}) = mB(1 - \cos 30^{\circ})$.
2. $= 1.28 \times 0.25 \times (1 - 0.866) = 0.32 \times 0.134$.
3. Answer: $\Delta U \approx 0.043$ J — this is also the work done against the field torque.

Example 6: Needle vs nail

A magnetised needle in a uniform field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why?

Solution:

1. In a uniform field the forces on the two poles of the needle are equal and opposite — net force zero, torque generally non-zero.
2. The nail near a bar magnet sits in a non-uniform field. The magnet induces a moment in the nail.
3. The induced unlike pole (say S) is closer to the magnet's N than the induced N, so attraction wins: net force plus torque.

Example 7: Does a toroid have poles?

Must every magnetic configuration have a north and a south pole?

Solution:

1. Poles correspond to a net magnetic moment of the source.
2. A toroid confines its field inside; it has no net moment and no poles. The same holds for a straight infinite conductor.
3. Answer: No — N and S poles exist only when the source has a net non-zero magnetic moment (e.g. a bar magnet or solenoid).

Example 8: Oscillation period (JEE pattern)

A compass needle has magnetic moment $6.7 \times 10^{-2}\,\text{A m}^2$ and moment of inertia $7.5 \times 10^{-6}\,\text{kg m}^2$. It oscillates in a horizontal field of 0.01 T. Find its period.

Solution:

1. Formula: $T = 2\pi\sqrt{\mathcal{I}/(mB)}$.
2. $mB = 6.7 \times 10^{-2} \times 0.01 = 6.7 \times 10^{-4}$ N m.
3. $\mathcal{I}/(mB) = 7.5 \times 10^{-6} / 6.7 \times 10^{-4} = 1.12 \times 10^{-2}$ s$^2$; square root $= 0.106$ s.
4. Answer: $T = 2\pi \times 0.106 \approx 0.67$ s.

Example 9: Where is torque half its maximum?

At what angle between $\vec{m}$ and $\vec{B}$ is the torque on a dipole half of its maximum value?

Solution:

1. $\tau = mB\sin\theta$; maximum $mB$ at $90^{\circ}$.
2. Half-max: $\sin\theta = 1/2 \Rightarrow \theta = 30^{\circ}$ or $150^{\circ}$.
3. Note: at $30^{\circ}$ the energy is $-mB\cos 30^{\circ} = -0.866\,mB$, not half of $-mB$ — torque and energy vary differently with angle.

Example 10: Reading the energy landscape

A dipole is released from rest at $\theta = 90^{\circ}$ in a uniform field. Describe its subsequent motion and the energy conversion.

Solution:

1. At $90^{\circ}$, $U = 0$ and torque is maximum $mB$ — the dipole starts rotating towards $\theta = 0^{\circ}$.
2. PE converts to rotational KE: at $\theta = 0^{\circ}$, $U = -mB$, so KE $= mB$.
3. With no damping it overshoots and oscillates between the two positions perpendicular to the field; with damping it settles into the stable alignment $\theta = 0^{\circ}$.