Magnetic field and dipoles

A current loop is a magnetic dipole. A flat loop of area $A$ carrying current $I$ produces, far away, a field pattern indistinguishable from a tiny bar magnet's. It is characterised by a single vector, the magnetic dipole moment:

where: $I$ is the loop current, $A$ is the loop's enclosed area, $N$ is the number of turns, and $\vec A$ (hence $\vec m$) points along the loop's normal, fixed by the right-hand rule (curl fingers along $I$, thumb gives $\vec m$). Field lines emerge from one face of the loop — its effective north pole — loop around, and re-enter the other face, its effective south pole.

An orbiting electron is a microscopic current loop. In the simple Bohr picture, an electron of charge $e$ circling at radius $r$ with speed $v$ completes one revolution in period $T = 2\pi r/v$, so it constitutes an average current $I = e/T = ev/(2\pi r)$. Treating this as a tiny current loop of area $\pi r^2$ gives its orbital magnetic dipole moment:

where: $e$ is the electron's charge, $v$ its orbital speed, and $r$ its orbital radius. This is the atomic-scale seed of all magnetism in matter — every electron orbiting a nucleus is a current loop with its own dipole moment, and stacking billions of these (mostly cancelling, but not quite, in magnetic materials) is what makes bulk matter magnetic at all.

A bar magnet behaves like an equivalent solenoid. Slice a solenoid into a stack of current loops along its axis; each loop is a tiny dipole, and the sum of their fields — both outside and, qualitatively, inside — matches a bar magnet's field almost exactly. Field lines emerge from the north pole, arc through the surrounding space, re-enter at the south pole, and continue inside the magnet back to the north pole, closing into loops. This is why a solenoid carrying a suitable current can stand in for a bar magnet in demonstrations: same field pattern, same dipole behaviour, different microscopic source. The prescribed treatment of the bar magnet's own field along special lines is qualitative only — no formula is required:

• On the axial line (extended along the magnet's own axis, beyond a pole), the field points along the axis and weakens rapidly with distance, strongest close to the pole.
• On the equatorial line (the perpendicular bisector of the magnet), the field is weaker than the axial field at the same distance, and points antiparallel to $\vec m$ — a fact students often get backwards.

Quantitative treatment of Earth's magnetism (declination, dip, horizontal component) is outside this syllabus and is not needed here.

A bar magnet of dipole moment $m = 0.8\ \text{A}\,\cdot\,\text{m}^2$ sits in a uniform magnetic field of $B = 0.25\ \text{T}$, tilted at $60°$ to the field. Find the torque on it, and then find the orbital magnetic dipole moment of an electron orbiting at radius $r = 5\times10^{-11}\ \text{m}$ with speed $v = 2\times10^6\ \text{m/s}$ (take $e = 1.6\times10^{-19}\ \text{C}$).

Step 1. For the bar magnet, use $\tau = mB\sin\theta$ with $m = 0.8$, $B = 0.25$, $\theta = 60°$ so $\sin60° = 0.866$.

Step 2. Multiply: $0.8\times0.25 = 0.2$, then $0.2\times0.866 = 0.1732$, so $\tau \approx 0.173\ \text{N}\,\cdot\,\text{m}$, directed to rotate $\vec m$ toward $\vec B$.

Step 3. For the electron, use $m = evr/2$ with $e = 1.6\times10^{-19}$, $v = 2\times10^6$, $r = 5\times10^{-11}$.

Step 4. Multiply the numerator: $1.6\times2 = 3.2$, then $3.2\times5 = 16$, giving $16\times10^{-19+6-11} = 16\times10^{-24}$; dividing by 2 gives $m = 8\times10^{-24}\ \text{A}\,\cdot\,\text{m}^2$.

Final answers: the torque on the bar magnet is about $\mathbf{0.173\ N\cdot m}$, and the electron's orbital dipole moment is $\mathbf{8\times10^{-24}\ A\cdot m^2}$ — a scale comparable to the "Bohr magneton" that sets atomic magnetism.

Competitive stretch

If the same bar magnet were rotated from $\theta = 60°$ to $\theta = 90°$ (perpendicular to $B$), the torque would rise to its maximum $\tau_{max} = mB = 0.8\times0.25 = 0.2\ \text{N}\,\cdot\,\text{m}$ — only slightly more than the $0.173\ \text{N}\,\cdot\,\text{m}$ at $60°$, since $\sin60°$ is already close to $1$.

• Treating the axial/equatorial bar-magnet field as something to derive from a formula — CISCE keeps this qualitative; know the pattern (axial stronger, equatorial weaker and antiparallel to $\vec m$), not an equation.
• Mixing up the direction convention for $\vec m$: right-hand rule for a current loop, but south-to-north pole for a bar magnet — both are used in this topic, on different objects.
• Forgetting the equatorial field points opposite to $\vec m$, not along it like the axial field does.
• Dragging in quantitative Earth's-magnetism content (dip, declination, horizontal component) — that is a separate, out-of-scope topic here.

Ask the tutor: Want a walkthrough comparing how the right-hand rule for a current loop and the south-to-north convention for a bar magnet both point to the same underlying dipole moment?