Dipole in a uniform magnetic field

This topic sits inside Unit V, Magnetism and Matter, immediately after the bar-magnet-as-equivalent-solenoid picture and the axial/equatorial field expressions for a dipole. CBSE marks the torque result as qualitative only — the formula $\vec\tau = \vec m\times\vec B$ "may be quoted for reference," and no derivation from first-principles pole forces is examinable, even though a short physical justification (as given below) helps most students remember it. The associated energy expression $U = -\vec m\cdot\vec B$ and the stable/unstable equilibrium orientations are standard board-exam material built directly on top of the torque formula, so this article treats torque, energy, and equilibrium as one connected package rather than three separate facts to memorise.

Model the bar magnet as two point magnetic poles: a north pole of strength $+q_m$ and a south pole of strength $-q_m$, separated by a small vector distance $2\vec l$ pointing from south to north, so that the dipole moment is $\vec m = q_m(2\vec l)$, with SI unit $\text{A}\,\cdot\,\text{m}^2$. Place this pair in a field $\vec B$ that is uniform, meaning it has the same magnitude and direction everywhere the magnet sits — a reasonable approximation between wide magnet poles or over the small footprint of a lab bench. The north pole feels a force $q_m\vec B$ along $\vec B$; the south pole feels a force $-q_m\vec B$, equal in size and exactly opposite in direction. Add them and the net force is exactly zero, for any orientation of the magnet, because the field does not vary from one pole to the other. But the two forces do not act along the same line — they act at the two ends of the magnet, a distance $2l\sin\theta$ apart when measured perpendicular to the force direction, where $\theta$ is the angle between $\vec m$ and $\vec B$. A pair of equal, opposite, offset forces is a couple, and a couple produces a pure turning effect with a well-defined torque about any point, independent of where you choose to take moments. That turning effect is

where: $\vec\tau$ is the torque vector (unit $\text{N}\,\cdot\,\text{m}$), acting along an axis perpendicular to both $\vec m$ and $\vec B$ by the right-hand rule; $m = |\vec m|$ is the dipole moment magnitude; $B = |\vec B|$ is the uniform field strength (tesla); and $\theta$ is the angle between the dipole axis and the field, measured $0°$ to $180°$. The physical reading of $\sin\theta$ is that the torque vanishes when the dipole is already lined up with the field ($\theta = 0°$, nothing left to turn it) or fully reversed ($\theta = 180°$, the forces are still equal-and-opposite but their offset — the effective lever arm — has shrunk to zero along the perpendicular), and peaks when the dipole sits perpendicular to the field ($\theta = 90°$, maximum lever arm). The energy stored in this configuration follows from the work an external agent must do, against this torque, to rotate the dipole from some reference angle to $\theta$. By convention the reference is $\theta = 90°$ (where $U$ is defined to be zero), giving

where: $U$ is the potential energy of the dipole–field system (joules); the dot product turns the vector alignment into a single number that is most negative when $\vec m$ and $\vec B$ point the same way. Physically, $U$ is lowest (most negative, $-mB$) at $\theta = 0°$ — the field has "done work" pulling the dipole into alignment, so this is the energetically favoured, stable state — and highest ($+mB$) at $\theta = 180°$, the fully reversed, unstable state that a magnet only occupies if held there by an external agent.

Board level - torque on a bar magnet at a given angle

A small bar magnet of dipole moment $m = 0.40\ \text{A}\,\cdot\,\text{m}^2$ is placed in a uniform magnetic field of $B = 0.25\ \text{T}$, making an angle of $30°$ with the field direction. Find the torque on the magnet.

Step 1. Write down the torque formula for a dipole in a uniform field: $\tau = mB\sin\theta$.

Step 2. Substitute the given values, $m = 0.40\ \text{A}\,\cdot\,\text{m}^2$, $B = 0.25\ \text{T}$, $\theta = 30°$: $\tau = 0.40 \times 0.25 \times \sin30°$.

Step 3. Evaluate using $\sin30° = 0.5$: $\tau = 0.40 \times 0.25 \times 0.5 = 0.05\ \text{N}\,\cdot\,\text{m}$. The magnet experiences a torque of $\tau = 0.05\ \text{N}\,\cdot\,\text{m}$, directed so as to rotate $\vec m$ toward alignment with $\vec B$ — this is the restoring torque a compass needle feels when displaced $30°$ from magnetic north.

Competitive level - work to flip the dipole, then the equivalent restoring behaviour near equilibrium

The same bar magnet ($m = 0.40\ \text{A}\,\cdot\,\text{m}^2$, $B = 0.25\ \text{T}$) is rotated quasi-statically by an external agent from $\theta = 0°$ (aligned) to $\theta = 180°$ (fully reversed). Find the work done by the external agent, and then find the torque the magnet would feel if displaced by a small angle $\theta = 5°$ from the stable orientation, to see how strongly it is "held" near $\theta = 0°$.

Step 1. Use $U = -mB\cos\theta$ to find the potential energy at $\theta = 0°$: $U_1 = -0.40\times0.25\times\cos0° = -0.40\times0.25\times1 = -0.10\ \text{J}$.

Step 2. Find the potential energy at $\theta = 180°$: $U_2 = -0.40\times0.25\times\cos180° = -0.40\times0.25\times(-1) = +0.10\ \text{J}$.

Step 3. Since the rotation is quasi-static (negligible kinetic energy throughout), the work done by the external agent equals the change in potential energy: $W = U_2 - U_1 = 0.10-(-0.10) = 0.20\ \text{J}$.

Step 4. Now check the restoring torque at a small displacement $\theta = 5°$ from the stable orientation, using $\tau = mB\sin\theta$ with $\sin5° \approx 0.0872$: $\tau = 0.40\times0.25\times0.0872 \approx 8.7\times10^{-3}\ \text{N}\,\cdot\,\text{m}$. So $W = 0.20\ \text{J}$ is required to flip the dipole from stable to unstable equilibrium, and near the stable orientation the restoring torque grows roughly in proportion to the small displacement angle ($\tau \approx mB\theta$ for small $\theta$ in radians), which is exactly the linear-restoring-torque condition behind why a compass needle, once nudged, oscillates back and forth rather than flipping over — the small-angle torque here, $8.7\times10^{-3}\ \text{N}\,\cdot\,\text{m}$, is tiny next to the $0.05\ \text{N}\,\cdot\,\text{m}$ found at $30°$, showing the torque grows fastest as the dipole approaches $90°$, not near $0°$. Timing note: the board-level question should take under a minute — one substitution, one sine value. The competitive-level question packs two ideas (energy-as-work, and the small-angle limit of $\sin\theta$) into roughly three minutes; the key time-saver is recognising instantly that $\cos0°=1$ and $\cos180°=-1$ without a calculator, and that $\sin\theta \approx \theta$ (radians) for small angles saves re-deriving anything from scratch.

Examiners repeatedly probe the same handful of confusions in this topic, because the torque and energy formulas look deceptively similar and the physical intuition about forces is easy to get backwards. Thinking a uniform field exerts a net force on a dipole is the single most common error — it does not; the net force is exactly zero for any orientation in a uniform field, and only a torque results. A genuine net force on a dipole requires a non-uniform field (the kind that lets one magnet attract or repel another), which lies beyond this qualitative CBSE topic. Confusing which equilibrium is stable is another frequent slip: $\theta = 0°$ (aligned with the field) is the stable, minimum-energy orientation, and $\theta = 180°$ (reversed) is the unstable, maximum-energy orientation — not the other way round, and it is worth remembering that "stable" always means the state a small nudge returns to, never the state a small nudge pushes away from. Assuming torque is largest when the dipole is aligned with the field is a natural but wrong instinct — torque is actually zero at $\theta = 0°$ (nothing left to turn) and maximum at $\theta = 90°$ (perpendicular), because torque scales with $\sin\theta$, not with how "magnet-like" the alignment looks. Mixing up the torque and energy formulas costs easy marks on almost every paper: $\tau = mB\sin\theta$ (no negative sign, sine) is a very different function of $\theta$ from $U = -mB\cos\theta$ (negative sign, cosine) — writing $\tau = -mB\cos\theta$ or $U = mB\sin\theta$ is a common transcription error under exam pressure and worth a deliberate double-check before submitting.

• Explain, using the two-pole picture, why a magnetic dipole in a uniform field feels zero net force but a nonzero torque.
• State and apply $\tau = mB\sin\theta$ to compute the torque on a dipole at a given angle to the field.
• Identify the stable ($\theta = 0°$) and unstable ($\theta = 180°$) equilibrium orientations, and explain which is which in terms of energy.
• State and apply $U(\theta) = -mB\cos\theta$, including computing the work done in reorienting a dipole between two angles.
• Recognise the exact mathematical parallel between this result and the electric dipole in a uniform field, $U = -\vec p\cdot\vec E$.
• Avoid the common sign and trig-function mix-ups between the torque ($\sin$, no minus sign) and energy ($\cos$, minus sign) formulas.

Ask the tutor: Want a few more mixed torque-and-energy numericals, or a walkthrough of how this exact structure reappears for the electric dipole in Electrostatic Potential and Capacitance?