Magnetic Dipole: Field, Torque and Energy

Derivation

Magnetic Dipole: Field, Torque and Energy

D-054 Home PU-102 Threads force · energy · fields Depends on Magnetic Vector Potential and Gauge Freedom, Multipole Expansion of the Potential

Statement

For a steady, spatially localized current distribution \(\vec{J}(\vec{r}')\), the leading non-vanishing term of the vector potential is the magnetic dipole term, characterized by a single vector \(\vec{m}=\frac{1}{2}\int \vec{r}'\times\vec{J}\,d^3r'\) (for a planar loop, \(\vec{m}=I\vec{a}=Ia\,\hat{n}\)). We derive the far vector potential \(\vec{A}=\frac{\mu_0}{4\pi}\frac{\vec{m}\times\hat{r}}{r^2}\), the field \(\vec{B}=\frac{\mu_0}{4\pi}\frac{3(\vec{m}\cdot\hat{r})\hat{r}-\vec{m}}{r^3}\), the torque \(\vec{N}=\vec{m}\times\vec{B}\) on the moment in a uniform field, and the orientation energy \(U=-\vec{m}\cdot\vec{B}\).

Why it matters

The magnetic moment is the single vector that survives when a complicated current loop is viewed from far away: atoms, nuclei, bar magnets, MRI spins, planetary dynamos and antenna coils are all described, to leading order, by \(\vec{m}\) alone. It is the magnetic counterpart of the electric dipole moment, and the torque and energy expressions are what make a compass point north, an NMR spin precess, and a motor turn.

Because the magnetic monopole term vanishes identically for steady currents, the dipole is the leading multipole — there is no \(1/r\) tail as there is in electrostatics. That single fact controls the entire structure of magnetostatic fields far from their sources.

Assumptions

Steady currents, \(\nabla\cdot\vec{J}=0\).Without this the current-conservation identities used to kill the monopole term and to symmetrize the first moment fail; a time-varying source radiates and the static multipole expansion is no longer valid. Localized source, field point outside it (\(r\gg\) source size).The Taylor expansion of \(1/|\vec{r}-\vec{r}'|\) in powers of \(r'/r\) diverges inside or near the distribution, so the dipole form is only the far-field limit. Uniform external \(\vec{B}\) for the torque and energy.If \(\vec{B}\) varies over the loop there is also a net force \(\vec{F}=\nabla(\vec{m}\cdot\vec{B})\) and the torque acquires higher-moment corrections; \(\vec{N}=\vec{m}\times\vec{B}\) holds only to leading order about the loop centre. Rigid moment of fixed magnitude.If \(|\vec{m}|\) depends on orientation or field (induced/diamagnetic response), the energy is no longer simply \(-\vec{m}\cdot\vec{B}\) and a factor-of-two subtlety appears for field-induced moments.

Derivation

\[ \vec{A}(\vec{r})=\frac{\mu_0}{4\pi}\int \frac{\vec{J}(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d^3r' \]

Magnetostatic vector potential in Coulomb gauge, taken as a prior result. A

\[ \frac{1}{|\vec{r}-\vec{r}'|}=\frac{1}{r}+\frac{\hat{r}\cdot\vec{r}'}{r^2}+\mathcal{O}\!\left(\frac{r'^2}{r^3}\right) \]

Multipole (Taylor) expansion for \(r\gg r'\); the coefficients are Legendre polynomials in \(\cos\gamma=\hat{r}\cdot\hat{r}'\). A

\[ \int \vec{J}\,d^3r'=0\ \Rightarrow\ \text{monopole term vanishes} \]

For steady currents \(\int J_i\,d^3r'=\int \nabla\cdot(x_i\vec{J})\,d^3r'=\oint x_i\,\vec{J}\cdot d\vec{a}=0\), using \(\nabla\cdot\vec{J}=0\) and \(\vec{J}=0\) on a bounding surface. There is no magnetic monopole. C

\[ \vec{A}(\vec{r})=\frac{\mu_0}{4\pi r^2}\int (\hat{r}\cdot\vec{r}')\,\vec{J}(\vec{r}')\,d^3r' \]

Keep the first surviving (dipole) term; \(\hat{r}\) and \(r\) are constants of the integration. A

\[ \int \left(x_i' J_j + x_j' J_i\right)d^3r'=0 \]

Lemma from \(\nabla\cdot\vec{J}=0\): \(\int \nabla\cdot(x_i x_j\vec{J})\,d^3r'=0\) gives \(\int (x_iJ_j+x_jJ_i)\,d^3r'=0\), so the symmetric part of the first current moment vanishes and only the antisymmetric part survives. C

\[ \int (\hat{r}\cdot\vec{r}')\,\vec{J}\,d^3r'=-\frac{1}{2}\,\hat{r}\times\int (\vec{r}'\times\vec{J})\,d^3r' \]

Contract the antisymmetric moment: using \([\hat{r}\times(\vec{r}'\times\vec{J})]_i=x_i'(\hat{r}\cdot\vec{J})-J_i(\hat{r}\cdot\vec{r}')\) together with the Step-5 identity, the mixed term reorganizes into this cross product. C

\[ \vec{m}\equiv\frac{1}{2}\int \vec{r}'\times\vec{J}(\vec{r}')\,d^3r'\ \Longrightarrow\ \vec{A}(\vec{r})=\frac{\mu_0}{4\pi}\frac{\vec{m}\times\hat{r}}{r^2} \]

Define the magnetic dipole moment; substituting Step 6 into Step 4 gives the dipole vector potential. For a filamentary loop \(\vec{J}\,d^3r'\to I\,d\vec{\ell}\), so \(\vec{m}=\frac{I}{2}\oint \vec{r}'\times d\vec{\ell}=I\vec{a}\). B

\[ \vec{B}=\nabla\times\vec{A}=\frac{\mu_0}{4\pi}\frac{3(\vec{m}\cdot\hat{r})\hat{r}-\vec{m}}{r^3} \]

Curl of \((\vec{m}\times\vec{r})/r^3\) using \(\nabla(1/r^3)\) and \(\nabla\cdot(\vec{r}/r^3)=4\pi\delta^3(\vec{r})\); the delta term is dropped for the far field \(r\neq0\). This is the dipole field. C

\[ \vec{N}=\int \vec{r}'\times(\vec{J}\times\vec{B})\,d^3r'=\vec{m}\times\vec{B} \]

Torque about the centre for uniform \(\vec{B}\). Expanding \(\vec{r}'\times(\vec{J}\times\vec{B})=\vec{J}(\vec{r}'\cdot\vec{B})-\vec{B}(\vec{r}'\cdot\vec{J})\) and applying the Step-5 symmetry identity to each Cartesian component collapses the integral to \(\left(\frac12\int\vec{r}'\times\vec{J}\right)\times\vec{B}=\vec{m}\times\vec{B}\). C

\[ U=-\int \vec{N}\,d\theta=-\int_{\pi/2}^{\theta} mB\sin\theta'\,d\theta'=-mB\cos\theta=-\vec{m}\cdot\vec{B} \]

Work done against the torque in rotating the rigid moment from \(\theta=\pi/2\) to \(\theta\); the reference is fixed by \(U(\pi/2)=0\). B

Result

\[ \vec{m}=\frac{1}{2}\int \vec{r}'\times\vec{J}\,d^3r'=I\vec{a} \]

\[ \vec{B}(\vec{r})=\frac{\mu_0}{4\pi}\frac{3(\vec{m}\cdot\hat{r})\hat{r}-\vec{m}}{r^3} \]

\[ \vec{N}=\vec{m}\times\vec{B},\qquad U=-\vec{m}\cdot\vec{B} \]

Reading. A current loop looks, from far away, like a point dipole of moment \(\vec{m}\) whose field falls as \(1/r^3\) and has the same angular shape as an electric dipole's. Placed in an external field, the dipole feels a torque \(\vec{m}\times\vec{B}\) that tries to align it with \(\vec{B}\), and its orientation energy is lowest (\(U=-mB\)) when aligned, highest (\(+mB\)) when anti-aligned.

Units check. \([\vec{m}]=\text{A}\cdot\text{m}^2\). Field: \(\frac{\mu_0}{4\pi}\frac{m}{r^3}=\frac{(\text{T}\cdot\text{m/A})(\text{A}\cdot\text{m}^2)}{\text{m}^3}=\text{T}\). Torque: \(mB=(\text{A}\cdot\text{m}^2)(\text{T})=\text{A}\cdot\text{m}^2\cdot\frac{\text{kg}}{\text{A}\cdot\text{s}^2}=\text{N}\cdot\text{m}\). Energy: the same product \(\to\text{J}\). All consistent.

Limiting cases

On axis (\(\hat{r}\parallel\vec{m}\)): \(\vec{B}=\frac{\mu_0}{4\pi}\frac{2\vec{m}}{r^3}\), pointing along \(\vec{m}\) — the strongest direction.
Equatorial plane (\(\hat{r}\perp\vec{m}\)): \(\vec{B}=-\frac{\mu_0}{4\pi}\frac{\vec{m}}{r^3}\), i.e. anti-parallel to \(\vec{m}\) and half the axial magnitude.
Aligned dipole (\(\theta\to0\)): \(\vec{N}\to0\), \(U\to-mB\) (stable equilibrium).
Perpendicular (\(\theta=90^\circ\)): torque is maximal, \(N=mB\), while \(U=0\).
Small oscillations about alignment: \(U\approx-mB+\tfrac12 mB\,\theta^2\), a harmonic well, so a compass needle librates like a torsional pendulum.

Breaks when

Time-varying currents. If \(\partial\rho/\partial t\neq0\) then \(\nabla\cdot\vec{J}\neq0\), the identities of Steps 3, 5 and 9 collapse, the source radiates, and the static \(1/r^3\) dipole field is replaced by retarded radiation fields (\(1/r\) at large distance).
Near or inside the source. For \(r\) comparable to the loop size the multipole series does not converge; the true field includes quadrupole and higher terms, and close in one must use Biot–Savart directly. A contact term \(\frac{2\mu_0}{3}\vec{m}\,\delta^3(\vec{r})\) is also needed to reproduce the volume-integrated field.
Strongly non-uniform external field. \(\vec{N}=\vec{m}\times\vec{B}\) and \(U=-\vec{m}\cdot\vec{B}\) assume \(\vec{B}\) constant over the source; a gradient adds a translational force \(\vec{F}=\nabla(\vec{m}\cdot\vec{B})\) and reference-point ambiguity in the torque.
Field-induced or saturating moments. If \(\vec{m}\) is itself a response to \(\vec{B}\) (diamagnets, superconductors, ferromagnet saturation), \(U=-\vec{m}\cdot\vec{B}\) is wrong — induced moments carry an extra \(\tfrac12\) and the constitutive relation must be tracked.

Failure modes

Dropping the \(\tfrac12\). Writing \(\vec{m}=\int \vec{r}'\times\vec{J}\) instead of \(\tfrac12\int\) — the factor comes from symmetrizing the first moment (Step 5), and forgetting it doubles every predicted field and torque.
Using the electric-dipole field constant. The angular structure \(3(\vec{m}\cdot\hat{r})\hat{r}-\vec{m}\) is identical to the electric case, but students sometimes insert \(1/(4\pi\varepsilon_0)\) instead of \(\mu_0/4\pi\).
Sign of the equatorial field. Assuming \(\vec{B}\) points along \(\vec{m}\) everywhere; in the equatorial plane it is anti-parallel.
Confusing torque angle and energy angle. \(N\propto\sin\theta\) but \(U\propto-\cos\theta\); using \(\sin\theta\) in the energy (or \(\cos\theta\) in the torque) is a frequent slip.
Applying \(U=-\vec{m}\cdot\vec{B}\) to the total energy. It is only the orientation energy in an external field; it omits the self-energy and the work done by the source keeping \(I\) constant.
Forgetting \(\vec{N}=\vec{m}\times\vec{B}\) requires uniform \(\vec{B}\). Applying it in a gradient and ignoring the net force.

Discussion

The deep reason the magnetic multipole expansion starts at the dipole is charge conservation: \(\nabla\cdot\vec{J}=0\) forbids a net "magnetic charge," so the \(1/r\) monopole potential that dominates electrostatics simply does not exist here. Everything magnetic that we see from a distance — the Earth's field, a fridge magnet, an electron — is dipolar to leading order, and the vector \(\vec{m}\) is the compact carrier of that information.

The torque and energy expressions unify a remarkable range of physics. A compass needle is a permanent moment seeking the \(-\vec{m}\cdot\vec{B}\) minimum; an electric motor exploits \(\vec{N}=\vec{m}\times\vec{B}\) with a commutator to keep the torque one-signed; NMR and MRI rest on the fact that a moment not aligned with \(\vec{B}\) feels a torque perpendicular to both \(\vec{m}\) and \(\vec{B}\), which for a moment tied to angular momentum (\(\vec{m}=\gamma\vec{L}\)) produces steady Larmor precession rather than alignment.

The gyromagnetic link \(\vec{m}=\gamma\vec{L}\) is where classical and quantum descriptions meet. Classically a rotating charge gives \(\gamma=q/2M\), so orbital moments come in units of the Bohr magneton \(\mu_B=e\hbar/2m_e\). Spin, however, carries an anomalous factor \(g\approx2\) (from the Dirac equation, corrected by QED to \(g=2.00232\ldots\)), so the same torque law \(\vec{N}=\vec{m}\times\vec{B}\) governs both, but with a magnitude no classical loop model can reproduce. The precession frequency \(\omega=\gamma B\) is what a spectrometer actually measures.

Common misconceptions. The dipole field formula is a far-field result; it is not the field at the centre of a real loop (that is \(\mu_0 I/2R\), finite, not divergent). Also, \(\vec{m}\times\vec{B}\) tends to align a static moment with \(\vec{B}\), but a moment carrying angular momentum precesses instead of aligning — alignment requires dissipation. Finally, "the loop moves to strong field" is only true for a moment already aligned; an anti-aligned moment is pushed toward weak field.

Worked examples

Far axial field of a small current loop.

A single circular loop of radius \(R=5.0\ \text{cm}\) carries \(I=2.0\ \text{A}\). Find its moment and the field on the axis at \(z=0.50\ \text{m}\). A

\[ m=Ia=I\pi R^2=(2.0)\pi(0.050)^2=1.571\times10^{-2}\ \text{A}\,\text{m}^2 \]

Planar loop, \(a=\pi R^2\). A

\[ B_{\text{axis}}=\frac{\mu_0}{4\pi}\frac{2m}{z^3}=(10^{-7})\frac{2(1.571\times10^{-2})}{(0.50)^3} \]

Axial case \(\hat{r}\parallel\vec{m}\), and \(z=0.50\ \text{m}\gg R\) so the dipole limit is valid. B

\[ B_{\text{axis}}=(10^{-7})\frac{3.142\times10^{-2}}{0.125}=2.51\times10^{-8}\ \text{T}\approx25\ \text{nT} \]

Reading. At ten loop-radii the field is only tens of nanotesla — about the Earth's field divided by a thousand — and directed along \(\vec{m}\).

Units check. \((\text{T}\,\text{m/A})(\text{A}\,\text{m}^2)/\text{m}^3=\text{T}\). Consistent.

Torque and energy of the same loop in a uniform field.

The loop (\(m=1.571\times10^{-2}\ \text{A}\,\text{m}^2\)) sits in a uniform \(B=0.30\ \text{T}\) with \(\vec{m}\) at \(\theta=30^\circ\) to \(\vec{B}\). Find the torque magnitude and the orientation energy. A

\[ N=mB\sin\theta=(1.571\times10^{-2})(0.30)(\sin30^\circ)=(1.571\times10^{-2})(0.30)(0.500) \]

Magnitude of \(\vec{m}\times\vec{B}\). A

\[ U=-mB\cos\theta=-(1.571\times10^{-2})(0.30)(\cos30^\circ)=-(1.571\times10^{-2})(0.30)(0.866) \]

Orientation energy relative to the \(\theta=90^\circ\) reference. A

\[ N=2.36\times10^{-3}\ \text{N}\,\text{m},\qquad U=-4.08\times10^{-3}\ \text{J} \]

Reading. The torque drives the loop toward alignment; the energy is negative because \(\theta<90^\circ\) is on the favourable side. Flipping it fully to \(\theta=180^\circ\) would cost \(U(180^\circ)-U(30^\circ)=\left[+4.71-(-4.08)\right]\times10^{-3}=8.79\times10^{-3}\ \text{J}\).

Units check. \((\text{A}\,\text{m}^2)(\text{T})=\text{N}\,\text{m}\) for torque and \(\text{J}\) for energy. Consistent.

Problems

A flat coil of \(N=50\) turns and area \(a=1.2\times10^{-3}\ \text{m}^2\) carries \(I=0.40\ \text{A}\). Find its magnetic moment.
Solution
\(m=NIa=(50)(0.40)(1.2\times10^{-3})=2.4\times10^{-2}\ \text{A}\,\text{m}^2\). The turns add coherently, so an \(N\)-turn coil has \(N\) times the single-loop moment.
For a point dipole \(\vec{m}\), find the ratio of the field magnitude on the axis to that in the equatorial plane at the same distance \(r\), and state the direction in each case.
Solution
Axial: \(B_{\text{ax}}=\frac{\mu_0}{4\pi}\frac{2m}{r^3}\), along \(+\vec{m}\). Equatorial: \(B_{\text{eq}}=\frac{\mu_0}{4\pi}\frac{m}{r^3}\), along \(-\vec{m}\). Ratio \(|B_{\text{ax}}|/|B_{\text{eq}}|=2\). The axial field is twice as strong and points opposite to the equatorial field.
The loop of Worked Example 1 (\(m=1.571\times10^{-2}\ \text{A}\,\text{m}^2\)) is placed in \(B=0.50\ \text{T}\). Find (a) the maximum torque and (b) the work needed to rotate it from fully aligned to fully anti-aligned.
Solution
(a) \(N_{\max}=mB=(1.571\times10^{-2})(0.50)=7.85\times10^{-3}\ \text{N}\,\text{m}\), occurring at \(\theta=90^\circ\). (b) \(W=U(180^\circ)-U(0^\circ)=(+mB)-(-mB)=2mB=2(7.85\times10^{-3})=1.57\times10^{-2}\ \text{J}\).
A compass needle of magnetic moment \(m=0.80\ \text{A}\,\text{m}^2\) and moment of inertia \(I_r=3.0\times10^{-6}\ \text{kg}\,\text{m}^2\) oscillates in a horizontal field \(B=2.0\times10^{-5}\ \text{T}\). Find the period of small oscillations.
Solution
Equation of motion \(I_r\ddot\theta=-mB\sin\theta\approx-mB\,\theta\), so \(\omega=\sqrt{mB/I_r}=\sqrt{(0.80)(2.0\times10^{-5})/(3.0\times10^{-6})}=\sqrt{5.33}=2.31\ \text{rad/s}\). Period \(T=2\pi/\omega=2.72\ \text{s}\). This is exactly the torsional-pendulum analogue with \(mB\) playing the role of a torsion constant.
Show that a classical particle of charge \(q\) and mass \(M\) in a circular orbit has \(\vec{m}=\frac{q}{2M}\vec{L}\), and evaluate the moment for an electron with orbital angular momentum \(L=\hbar\).
Solution
Orbital current \(I=q/T=qv/(2\pi R)\); moment \(m=I\pi R^2=qvR/2\). Angular momentum \(L=MvR\), so \(vR=L/M\) and \(m=\frac{q}{2M}L\); as vectors \(\vec{m}=\frac{q}{2M}\vec{L}\). For an electron (\(q=-e\)) with \(L=\hbar\): \(|m|=\frac{e\hbar}{2m_e}=\mu_B=\frac{(1.602\times10^{-19})(1.055\times10^{-34})}{2(9.109\times10^{-31})}=9.27\times10^{-24}\ \text{A}\,\text{m}^2\), the Bohr magneton. The moment is anti-parallel to \(\vec{L}\) because the charge is negative.