In short

A small flat current loop behaves, as seen from far away, like a magnetic dipole. It is completely described by a single vector — the magnetic dipole moment:

\boxed{\;\vec{m} \;=\; NI\vec{A}\;}

where I is the loop current, N is the number of turns, \vec{A} is the area vector (magnitude = enclosed area, direction = right-hand thumb when the fingers curl along I). SI unit: A·m².

Torque. In a uniform external field \vec{B}, the loop feels no net force (the forces on opposite sides cancel), but it does feel a torque that tries to align \vec{m} with \vec{B}:

\boxed{\;\vec{\tau} \;=\; \vec{m}\times\vec{B}\;}, \qquad |\vec{\tau}| = mB\sin\theta.

Potential energy. Because the torque is conservative, there is a potential energy associated with the orientation:

\boxed{\;U \;=\; -\vec{m}\cdot\vec{B}\;} \;=\; -mB\cos\theta.

The energy is lowest (U = -mB) when \vec{m} is aligned with \vec{B} — the stable equilibrium. It is highest (U = +mB) when \vec{m} is anti-aligned — unstable. Halfway between (\theta = 90^\circ) U = 0.

Why it matters. Every bar magnet, every compass needle, every electron spin, every proton aligning in an MRI scanner, and every ferromagnetic domain is described by exactly this vector \vec{m} and this pair of formulas. Once you know \vec{m}, you know the torque, the energy, and even (at large distances) the field the object produces itself — it looks, far away, exactly like an electric dipole field but with \vec{m} in place of \vec{p}.

A compass needle in the Thar desert near Jaisalmer spins in its brass housing, settles down, and points north. It has done this for a thousand years, with no external source of energy except the Earth's own magnetic field — a gentle, invisible field roughly 50 microteslas strong, fifty times weaker than a fridge magnet. The needle settles because there is a torque on it that keeps shoving it toward alignment with the field, and once it is aligned, the torque drops to zero. A tiny oscillation remains, damped by friction, and the needle is still.

Inside an AIIMS scanner, something similar — but on a different scale — is happening to four quintillion hydrogen nuclei per cubic millimetre of a patient's brain. Each proton has its own magnetic moment, sixteen orders of magnitude smaller than the compass needle's. In the scanner's 3 T field, each proton is nudged toward alignment with the field. Most stay aligned; a tiny fraction — about ten parts per million at body temperature — are aligned rather than anti-aligned. That ten-parts-per-million excess is what produces the entire MRI image. The whole medical imaging business rests on the quantitative theory of the magnetic dipole moment.

From a current loop to the dipole moment

Start with the simplest system: a single rectangular loop of wire, sides a and b, carrying current I, placed in a uniform external magnetic field \vec{B}. Put the loop flat on a table for now; \vec{B} will come in horizontally. You want to find out what the external field does to the loop.

The force on each of the four sides is given by the current-conductor law from chapter 185: \vec{F} = I\vec{L}\times\vec{B}, where \vec{L} points along the current direction with magnitude equal to the side's length.

Top and bottom cancel; left and right cancel. The net force on the loop is zero. This is the first key fact: in a uniform external field, a current loop feels no translational force — it cannot be pushed as a whole; it can only be rotated.

The torque — where the action is

Even though the net force is zero, the lines of action of the four forces are different, so they create a torque. Here is the clean derivation.

Set up the geometry. Place the loop in the xy-plane with its centre at the origin, sides a along \hat{x} and b along \hat{y}. Let the external field be \vec{B} = B\hat{x} — pointing along the x-axis, in the plane of the loop. The current flows counter-clockwise when viewed from the +z side (right-hand rule: fingers along current, thumb along +\hat{z}).

Rectangular current loop in a uniform external magnetic field A rectangular current loop in the xy-plane with side a along x and side b along y. The external field B points along +x. The forces on the top and bottom sides (parallel to B) are zero. The forces on the left and right sides (perpendicular to B) are equal and opposite but on different lines of action, producing a torque that tries to rotate the loop to align its normal with B. I → I ← I ↑ I ↓ x (along B) y B F_left = IbB (+z) F_right = IbB (−z) m (out of page)
Top view of a rectangular current loop in a uniform field $\vec{B}$ along $+\hat{x}$. The forces on the top and bottom sides (parallel to $\vec{B}$) are zero. The forces on the left and right sides point out of and into the page respectively — producing a torque that rotates the loop so its normal $\hat{n}$ aligns with $\vec{B}$.

Step 1. Force on the top side. The current in the top goes along +\hat{x}; the field is also along +\hat{x}. The cross product \hat{x}\times\hat{x} = 0.

\vec{F}_\text{top} \;=\; Ia\hat{x}\times B\hat{x} \;=\; 0.

Similarly \vec{F}_\text{bot} = 0. Why: the current in the top and bottom sides is parallel (or anti-parallel) to \vec{B}, so those sides feel no force at all.

Step 2. Force on the right side. Current along +\hat{y}; field along +\hat{x}.

\vec{F}_\text{right} \;=\; Ib\hat{y}\times B\hat{x} \;=\; IbB(\hat{y}\times\hat{x}) \;=\; -IbB\hat{z}.

Why: use the cyclic rule \hat{x}\times\hat{y} = \hat{z}, so \hat{y}\times\hat{x} = -\hat{z}. The force on the right side is -IbB\hat{z} — into the page (in our top-view convention).

Step 3. Force on the left side. Current along -\hat{y}; field along +\hat{x}.

\vec{F}_\text{left} \;=\; -Ib\hat{y}\times B\hat{x} \;=\; +IbB\hat{z}.

Why: flipping the sign of the current flips the force. Left side points out of the page.

Step 4. The two forces are equal in magnitude (IbB), opposite in direction, and separated by a lever arm of a — the spacing between the left and right sides. That is the definition of a couple: zero net force, nonzero net torque.

Step 5. Compute the torque about the centre of the loop. The right side is at position \vec{r}_\text{right} = (a/2)\hat{x} with force \vec{F}_\text{right} = -IbB\hat{z}. The left side is at \vec{r}_\text{left} = -(a/2)\hat{x} with force \vec{F}_\text{left} = +IbB\hat{z}.

\vec{\tau} \;=\; \vec{r}_\text{right}\times\vec{F}_\text{right} + \vec{r}_\text{left}\times\vec{F}_\text{left}
\;=\; (a/2)\hat{x}\times(-IbB\hat{z}) + (-a/2)\hat{x}\times(IbB\hat{z})
\;=\; -(a/2)IbB(\hat{x}\times\hat{z}) - (a/2)IbB(\hat{x}\times\hat{z})
\;=\; -IabB(\hat{x}\times\hat{z}) \;=\; IabB\hat{y}.

Why: \hat{x}\times\hat{z} = -\hat{y} (going against the cyclic order x\to y\to z). The two contributions add, not cancel — this is the signature of a torque couple.

Step 6. Notice that ab = A, the area of the loop. So

\vec{\tau} \;=\; IAB\hat{y}.

The direction \hat{y} is perpendicular to \vec{B} and to the loop's normal \hat{n} = \hat{z}. This torque rotates the loop about the y-axis — turning it so that \hat{n} swings toward \vec{B}.

Defining the dipole moment and writing the torque as \vec{m}\times\vec{B}

Define

\vec{m} \;\equiv\; I\vec{A}, \qquad \vec{A} \;=\; A\hat{n}

where \hat{n} is the normal to the loop (right-hand rule: fingers curl along I, thumb points along \hat{n}). For our setup, \vec{m} = IA\hat{z}.

Rewrite the torque from step 6:

\vec{\tau} \;=\; IAB\hat{y} \;=\; IA\hat{z}\times B\hat{x} \;\cdot\; (\text{because }\hat{z}\times\hat{x} = \hat{y}).

That is:

\vec{\tau} \;=\; (IA\hat{z})\times(B\hat{x}) \;=\; \vec{m}\times\vec{B}.

Why: we pulled the expression IAB\hat{y} apart and recognised it as a cross product using \hat{z}\times\hat{x} = \hat{y}. This repackages the torque in a form that makes the physics jump out — the torque is the dipole moment crossed with the external field.

For a loop with N turns carrying the same current, each turn contributes its own moment IA\hat{n}, all adding in the same direction:

\boxed{\;\vec{m} \;=\; NI\vec{A}\;}

And the torque

\boxed{\;\vec{\tau} \;=\; \vec{m}\times\vec{B}\;}, \qquad |\vec{\tau}| = mB\sin\theta

where \theta is the angle between \vec{m} and \vec{B}. Although we derived this for a rectangle, the result holds for any planar loop of any shape — only the enclosed area A matters.

A quick sanity check: at \theta = 0 (moment aligned with field), the torque is zero. At \theta = 90^\circ (moment perpendicular to field), the torque is maximum. At \theta = 180^\circ (anti-aligned), the torque is again zero — but unstably so, as you will see in a moment.

Potential energy — U = -\vec{m}\cdot\vec{B}

The torque \vec{\tau} = \vec{m}\times\vec{B} is conservative: the work done by the torque depends only on the initial and final orientations, not on the path. So there is a potential energy U(\theta).

Step 1. The work done by the torque when the loop rotates by a small angle d\theta (in the direction of decreasing \theta, since the torque pulls the moment toward the field) is

dW \;=\; \tau\,d\theta' \;=\; -mB\sin\theta\,d\theta.

Why: \tau = mB\sin\theta is the magnitude of the torque at angle \theta; the negative sign is because rotating in the direction that increases \theta does negative work (the torque pulls the other way). When you rotate from alignment (\theta = 0) toward \theta, you do positive work against the torque.

Step 2. The change in potential energy is minus the work done by the torque:

dU \;=\; -dW \;=\; +mB\sin\theta\,d\theta.

Step 3. Integrate from some reference orientation \theta_0 to the current \theta:

U(\theta) - U(\theta_0) \;=\; \int_{\theta_0}^{\theta} mB\sin\theta'\,d\theta' \;=\; mB[\cos\theta_0 - \cos\theta].

Step 4. Choose the standard convention: set U = 0 at \theta_0 = 90^\circ (where \cos\theta_0 = 0). Then

U(\theta) \;=\; -mB\cos\theta.

Step 5. Recognise that mB\cos\theta = \vec{m}\cdot\vec{B} by the definition of the dot product. So

\boxed{\;U \;=\; -\vec{m}\cdot\vec{B}\;}

Why: the dot product bundles the magnitudes and the angle into a single scalar. The minus sign encodes the physical fact that alignment is the low-energy configuration — aligned moments "want to be there."

Reading the energy landscape

The function U(\theta) = -mB\cos\theta has a simple shape on [0, 180^\circ].

This energy landscape is what compass needles minimise, what proton spins in an MRI scanner settle toward (partially, against thermal agitation), and what makes ferromagnetic domains align with an applied field.

Interactive — explore the torque and energy as you rotate the dipole

Interactive: torque and energy of a magnetic dipole versus angle Two curves showing the torque (proportional to sin theta) and the potential energy (proportional to minus cosine theta) of a magnetic dipole in a uniform field, as the angle between the dipole and the field is varied. A draggable marker moves along the theta axis and reads out both quantities. angle θ (degrees) between m and B normalised value +1 0 −1 0 90 180 τ/mB = sin θ U/(−mB) = cos θ, so U = −mB cos θ drag the red point along θ
Drag the red point along the $\theta$-axis to change the angle between the dipole and the field. The red curve is the normalised torque $\tau/(mB) = \sin\theta$. The dark curve is the normalised potential energy $U/(mB) = -\cos\theta$. Note: torque is maximum at $\theta = 90°$, where energy is $0$; torque is zero at both $\theta = 0$ and $\theta = 180°$, which are the equilibria (stable and unstable respectively).

Notice the interplay: where the torque is largest (\theta = 90^\circ), the energy is steepest (the slope dU/d\theta = mB\sin\theta is maximum). Where the torque is zero, the energy is flat (an extremum — either a minimum or maximum). The two curves tell the same story from two angles.

Worked examples

Example 1: A compass needle in Jaisalmer

A compass needle can be modelled as a magnetic dipole with moment m = 3.0\times 10^{-2} A·m² (a realistic value for a small steel needle). It sits in the Earth's magnetic field, whose horizontal component at Jaisalmer (latitude ≈ 26.9° N, magnetic latitude ≈ 19° N) is about B_H = 3.6\times 10^{-5} T. Find (a) the maximum torque on the needle when it is held perpendicular to the field, and (b) the work required to slowly flip the needle from pointing north to pointing south.

Compass needle in the horizontal component of Earth's field A compass needle pointing north, aligned with the horizontal Earth field. A second position shows the needle turned to an angle theta, with the torque trying to rotate it back toward alignment. aligned (θ=0) m B (north) tilted (θ=90°) m B (north) θ
Left: the needle aligned with $\vec{B}$ — zero torque, minimum energy. Right: the needle tilted by $\theta = 90°$ — maximum torque pushing it back toward alignment.

Step 1. (a) Maximum torque.

\tau_\text{max} \;=\; mB\sin 90^\circ \;=\; mB.

Why: the torque magnitude mB\sin\theta is largest at \theta = 90^\circ where \sin = 1.

Step 2. Plug in numbers.

\tau_\text{max} \;=\; (3.0\times 10^{-2})(3.6\times 10^{-5}) \;=\; 1.08\times 10^{-6}\ \text{N·m}.

Why: 3 \times 3.6 = 10.8, and 10^{-2}\times 10^{-5} = 10^{-7}, so 10.8\times 10^{-7} = 1.08\times 10^{-6}.

Step 3. (b) Work to flip from \theta = 0 (north) to \theta = 180^\circ (south).

W \;=\; \Delta U \;=\; U(180^\circ) - U(0) \;=\; -mB\cos 180^\circ - (-mB\cos 0) \;=\; +mB - (-mB) \;=\; 2mB.

Why: the needle goes from the minimum energy -mB (aligned with field) to the maximum +mB (anti-aligned). The difference is 2mB.

Step 4. Plug in.

W \;=\; 2\,(3.0\times 10^{-2})(3.6\times 10^{-5}) \;=\; 2.16\times 10^{-6}\ \text{J}.

Result. (a) Maximum torque is \approx 1.1\ \mu\text{N·m}. (b) Flipping the needle requires \approx 2.2\ \mu\text{J} of work. Tiny numbers — which is why it takes almost no effort to disturb a compass, and why the needle settles back to alignment under even gentle restoring torques. The microscopic work required for the flip equals the energy stored in an anti-aligned magnetic configuration.

Example 2: A proton in an AIIMS 3-tesla MRI scanner

An MRI scanner at AIIMS Delhi uses a superconducting magnet producing B = 3.0 T inside the bore. The magnetic moment of a single proton is m_p = 1.41\times 10^{-26} A·m² (this is the magnitude of the proton's intrinsic spin magnetic moment, fixed by nature). Find the energy difference between a proton aligned with the field and one anti-aligned. Express the answer in joules and in frequency units \Delta E / h (called the Larmor frequency).

Step 1. Energy of alignment vs anti-alignment.

U_\text{aligned} \;=\; -m_p B, \qquad U_\text{anti} \;=\; +m_p B.

The gap between the two levels is

\Delta E \;=\; U_\text{anti} - U_\text{aligned} \;=\; 2m_p B.

Why: in quantum mechanics the proton spin has only two allowed projections along \vec{B} — "up" (aligned) and "down" (anti-aligned). Their energies are \mp m_p B, and the gap between them is 2m_p B.

Step 2. Plug in numbers.

\Delta E \;=\; 2\,(1.41\times 10^{-26})(3.0) \;=\; 8.46\times 10^{-26}\ \text{J}.

Why: 2\times 1.41\times 3 = 8.46, exponents stay at 10^{-26}.

Step 3. Convert to frequency using E = hf, where h = 6.626\times 10^{-34} J·s.

f \;=\; \frac{\Delta E}{h} \;=\; \frac{8.46\times 10^{-26}}{6.626\times 10^{-34}}\ \text{Hz}.

Dividing: 8.46 / 6.626 = 1.277; exponents: -26 - (-34) = +8.

f \;\approx\; 1.28\times 10^{8}\ \text{Hz} \;=\; 128\ \text{MHz}.

Why: a proton in a 3 T field absorbs/emits photons at about 128 MHz — the Larmor frequency at 3 T. This is the radio-wave frequency used by the MRI scanner's RF coil to flip proton spins and read out the image. The frequency is a fingerprint of the field strength: 1.5 T machines use 64 MHz, 3 T machines use 128 MHz, 7 T research machines use 300 MHz.

Step 4. Sanity-check with the rule-of-thumb.

The gyromagnetic ratio for a proton is \gamma_p/2\pi = 42.58 MHz/T. For 3 T: f = 3 \times 42.58 = 127.7 MHz. Matches within rounding.

Result. The energy gap is \Delta E \approx 8.5\times 10^{-26} J, corresponding to the Larmor frequency f_L \approx 128 MHz at 3 T. Every resonance pulse inside an AIIMS 3-tesla scanner has to be tuned to exactly this frequency. The whole enterprise of magnetic resonance imaging rests on the dipole formula U = -\vec{m}\cdot\vec{B} applied to four quintillion protons per cubic millimetre of tissue.

The bar magnet — a dipole from atoms all the way up

You know a bar magnet from the physics lab — a rectangular block of magnetised steel with a north end and a south end. The surprising claim of the dipole theory is that a bar magnet, seen from sufficiently far away, is indistinguishable from a small current loop of the appropriate area and current. You cannot tell, from the external field, whether you are looking at a loop or a magnet.

Why? Because the bar magnet's magnetisation is the sum of microscopic current loops — each iron atom has unpaired electron spins whose intrinsic magnetic moments add up. All the interior loops cancel (adjacent atoms' moments point the same way, so their currents cancel on shared boundaries). What survives is an effective current running around the outside surface of the magnet — exactly what a macroscopic single-turn solenoid would look like. Seen from outside, this surface current is the source of the bar magnet's field.

The equivalence. A bar magnet of length L and cross-section A, with pole strength p (defined so that the magnetic moment m = pL — a legacy convention), is equivalent to a current loop of the same area A with current I and N turns satisfying

m_\text{loop} \;=\; NIA \;=\; pL \;=\; m_\text{bar}.

At large distances, both produce the same dipole field:

\vec{B}_\text{axial}(r) \;=\; \frac{\mu_0}{4\pi}\,\frac{2\vec{m}}{r^3} \qquad\text{(on the axis, far away)}
\vec{B}_\text{equatorial}(r) \;=\; -\frac{\mu_0}{4\pi}\,\frac{\vec{m}}{r^3} \qquad\text{(on the perpendicular bisector plane)}.

These are the exact analogues of the electric dipole field with \vec{m}\leftrightarrow\vec{p} and \mu_0/(4\pi)\leftrightarrow 1/(4\pi\varepsilon_0). The 1/r^3 fall-off is the hallmark of every dipole field — electric or magnetic — at large distances. Closer in (within a few magnet-lengths), the field deviates from the ideal dipole form and you need the full Biot-Savart integral, but that is a finer correction.

Common confusions

If you came here to understand the dipole moment, apply \vec{\tau} = \vec{m}\times\vec{B} and U = -\vec{m}\cdot\vec{B} in standard problems, you have what you need. What follows is for readers who want the general formula, the connection to the field a dipole produces, and the quantum story that makes bar magnets exist in the first place.

The general dipole moment — \vec{m} = \tfrac{1}{2}\oint\vec{r}\times I\,d\vec{l}

For an arbitrary current loop — planar or not, polygonal or smooth — the dipole moment is defined by

\vec{m} \;=\; \tfrac{1}{2}\oint_\text{loop} \vec{r}\times I\,d\vec{l}

where \vec{r} is the position vector of the current element I\,d\vec{l} with respect to any chosen origin inside the loop. This looks abstract but reduces cleanly to NIA for a planar loop.

Derivation for a planar loop. Let the loop lie in the xy-plane. Then \vec{r} = x\hat{i} + y\hat{j} and d\vec{l} = dx\,\hat{i} + dy\,\hat{j}, so

\vec{r}\times d\vec{l} \;=\; (x\,dy - y\,dx)\hat{k}.

The enclosed area (by the shoelace theorem) is A = \tfrac{1}{2}\oint(x\,dy - y\,dx). Therefore

\tfrac{1}{2}\oint\vec{r}\times d\vec{l} \;=\; \tfrac{1}{2}\oint(x\,dy - y\,dx)\hat{k} \;=\; A\hat{k} \;=\; \vec{A}.

Multiplying by I:

\vec{m} \;=\; I\vec{A}.

Why: the shoelace theorem expresses the area of any planar polygon (and by a limit argument, any planar region) as an integral around its boundary. The current magnifies this geometric quantity into a physical dipole moment.

For N turns, each contributes the same \vec{A}, so \vec{m} = NI\vec{A} as before.

Force on a dipole in a non-uniform field — \vec{F} = \nabla(\vec{m}\cdot\vec{B})

A fixed dipole in a uniform field feels no net force (the derivation at the start of this article). In a non-uniform field, however, the two sides of the loop sit in slightly different \vec{B}, and the cancellation breaks down.

Heuristic: expand \vec{B} to first order around the loop's centre: \vec{B}(\vec{r}) = \vec{B}(0) + (\vec{r}\cdot\nabla)\vec{B} + \ldots. The constant term \vec{B}(0) gives zero net force as before. The gradient term, integrated over the loop, gives

\vec{F} \;=\; \nabla(\vec{m}\cdot\vec{B}).

Why: the potential energy U = -\vec{m}\cdot\vec{B} depends on position through \vec{B}(\vec{r}). The force is minus the gradient of the potential energy: \vec{F} = -\nabla U = \nabla(\vec{m}\cdot\vec{B}).

This formula is what lets a refrigerator magnet stick to the fridge. The steel surface magnetises locally, creating many tiny dipoles that then feel a force \vec{F} = \nabla(\vec{m}\cdot\vec{B}) in the magnet's non-uniform field, pointing toward regions of higher \vec{B}. It is also what drives the Stern-Gerlach experiment, where a beam of silver atoms (each with net magnetic moment from its single unpaired valence electron) passes through a non-uniform field and splits into two discrete beams — the first direct evidence for quantised spin.

The magnetic field of a dipole — analogy with the electric dipole

The source side of the dipole — not just how it responds to fields, but how it produces its own — is captured by the dipole field formula. At position \vec{r} from a point dipole \vec{m}, with \hat{r} = \vec{r}/r:

\vec{B}(\vec{r}) \;=\; \frac{\mu_0}{4\pi}\,\frac{3(\vec{m}\cdot\hat{r})\hat{r} - \vec{m}}{r^3}.

Compare to the electric dipole field:

\vec{E}(\vec{r}) \;=\; \frac{1}{4\pi\varepsilon_0}\,\frac{3(\vec{p}\cdot\hat{r})\hat{r} - \vec{p}}{r^3}.

Identical form, different constants. Replace \vec{p}\to\vec{m} and 1/(4\pi\varepsilon_0)\to\mu_0/(4\pi) and you go from one to the other. This is not a coincidence — it is a reflection of the fact that far from any localised source, the first non-trivial term in a multipole expansion is the dipole term, and its angular dependence is universal.

Special cases. On the axis (\vec{r}\parallel\vec{m}): \vec{m}\cdot\hat{r} = m, so the formula reduces to \vec{B} = (\mu_0/4\pi)(2\vec{m}/r^3). On the equatorial plane (\vec{r}\perp\vec{m}): \vec{m}\cdot\hat{r} = 0, so \vec{B} = -(\mu_0/4\pi)(\vec{m}/r^3) — pointing opposite to \vec{m} and half the axial magnitude.

Magnetic moment of an orbiting electron

A simple quantum-classical hybrid calculation: an electron in a circular orbit of radius r with speed v forms a current loop of current I = e/T = ev/(2\pi r) and area A = \pi r^2. Its dipole moment is

m \;=\; IA \;=\; \frac{ev}{2\pi r}\cdot\pi r^2 \;=\; \frac{evr}{2}.

The electron's angular momentum about the orbit centre is L = m_e v r. So

m \;=\; \frac{e}{2m_e}\,L.

Why: divide and multiply by m_e, identify m_e vr as the angular momentum L. The factor e/(2m_e) is called the classical gyromagnetic ratio for an electron — the ratio of magnetic moment to angular momentum for an orbital current loop.

In Bohr's semi-classical atom, L is quantised in units of \hbar = h/(2\pi), so the natural unit of magnetic moment is

\mu_B \;\equiv\; \frac{e\hbar}{2m_e} \;\approx\; 9.274\times 10^{-24}\ \text{A·m}^2.

This is the Bohr magneton — the quantum of magnetic moment. Every atomic and electronic magnetic moment is measured in units of \mu_B. The electron's intrinsic spin magnetic moment turns out to be approximately 1\,\mu_B (actually \sim 1.00116\,\mu_B once quantum-electrodynamics corrections are included — a precision check that has agreed with theory to 10 decimal places).

From atomic moments to bar magnets — the material story

Every iron atom has four unpaired electron spins, contributing a net magnetic moment of about 4\,\mu_B \approx 3.7\times 10^{-23} A·m². In a bar of iron, these atomic moments spontaneously align within microscopic regions called magnetic domains, each about a micrometre across. An unmagnetised bar has domains pointing in random directions, so the bulk moment is zero. An external field causes domains aligned with the field to grow at the expense of others; the bulk becomes magnetised. A strong enough field saturates the material — all domains aligned, bulk moment at its maximum.

Estimate the bar-magnet moment. A rectangular ferrite magnet of volume V = 1\ \text{cm}^3 = 10^{-6}\ \text{m}^3 contains about N = V/V_\text{atomic} \approx 10^{29} atoms. If all their moments align perfectly (saturation), the bulk dipole moment would be m = N\times 4\mu_B \approx 10^{29}\times 3.7\times 10^{-23} \approx 4\ \text{A·m}^2. Real ferrite magnets reach about 1 A·m² per cm³, which matches this estimate to within the fraction of domains that actually align.

This is the microscopic origin of macroscopic magnetism: the magnetic dipole moments of unpaired electron spins, added up over ten to the twenty-ninth atoms, produce the bulk moment that lifts a paper clip or holds a magnet to the fridge.

Why MRI works — the dipole-energy story, quantitatively

Inside an AIIMS 3 T scanner, each proton has energy U = \mp m_p B depending on whether its spin is aligned or anti-aligned with \vec{B}. The gap 2m_p B = 8.5\times 10^{-26} J (from Example 2). The thermal energy at body temperature (T = 310 K) is k_B T = 1.38\times 10^{-23}\times 310 \approx 4.3\times 10^{-21} J — about 50\,000 times larger than the spin-splitting energy.

So the two spin states are almost equally populated. The Boltzmann factor is

\frac{N_\text{aligned}}{N_\text{anti}} \;=\; \exp\!\left(\frac{2m_pB}{k_BT}\right) \;\approx\; 1 + \frac{2m_pB}{k_BT} \;=\; 1 + 2.0\times 10^{-5}.

A 20 parts-per-million excess of aligned spins. In a cubic millimetre of tissue (\sim 10^{18} water molecules, \sim 10^{19} protons), that is about 2\times 10^{14} unpaired spins — enough to produce a detectable net magnetisation that the scanner's RF coils can flip, precess, and read out. Every MRI image is built from the small imbalance of dipole populations in the dipole-energy formula.

This is what it looks like when a single equation — U = -\vec{m}\cdot\vec{B}, written for one proton — cascades up through a mole of spins and a body of tissue to produce a modality that has transformed medicine.

Where this leads next