greatitself.
Continue
advanced29 min studyLesson 242 in Physics

Magnetic dipole fields and torque

Separate the distant dipole field, the singular ideal point source, and the torque of a small moment in an external field.

Before you begin
On this page

The question for today

What you’ll understand

  • Compute a loop moment and its distant axial or equatorial field.
  • Calculate torque and rotational work with a declared external field.
  • Explain why a point-dipole contact term is not an ordinary finite field value.

A small current loop and a permanent magnet can produce the same leading magnetic field far from their sources. The magnetic moment packages this shared distant behaviour into one vector. It also determines the leading torque in a slowly varying external field. Keeping the source field separate from the external field prevents a common mistake: inserting a dipole's singular self-field into the formula for the torque acting on that dipole.

A moment summarizes a localized current

For a planar loop the moment is current times oriented area, with the normal fixed by the right-hand rule. Identical closely spaced turns multiply the moment by their number. The unit is ampere square metre, equivalently joule per tesla. The general localized steady-current definition uses one half of the volume integral of position crossed with current density; the factor one half is essential and disappears only after reducing the expression to a single planar loop.

The leading distant field has an inverse-cube radial dependence and an angular pattern set by the moment. On the positive axis it points with the moment and has twice the magnitude found at an equatorial point at the same distance, where it points against the moment. Higher multipoles encode source details that the dipole approximation discards. Being small relative to the source-observer separation, rather than merely being a physically small object, controls this expansion.

m=NIa,Bdip(r)=μ04πr3[3(m⋅r^)r^−m](r>0)\mathbf m=NI\mathbf a,\qquad \mathbf B_{\mathrm{dip}}(\mathbf r)=\frac{\mu_0}{4\pi r^3}\left[3(\mathbf m\cdot\hat{\mathbf r})\hat{\mathbf r}-\mathbf m\right]\quad(r>0)
N is turn count, a is oriented loop area in m², I is current in A, and m has units A m². The field formula is the leading vacuum field at distances much larger than the source size; as an ideal point-dipole expression it applies away from its origin.↔ Scroll wide equations horizontally.

Torque and energy use the external field

In an approximately uniform imposed field, forces on opposite sides of a current loop cancel in translation but form a couple. The moment rotates toward alignment. Torque vanishes for parallel and antiparallel orientations, yet only the parallel orientation minimizes the interaction energy of a fixed permanent moment. A small angular disturbance distinguishes stable alignment from the unstable antiparallel configuration. Zero torque at an instant is therefore insufficient evidence of rotational stability.

The interaction energy concerns a fixed moment in a prescribed external field; maintaining currents can require work by their supplies. A slow rotation with negligible kinetic-energy change requires external work equal to the interaction-energy change, which can be negative when the agent extracts energy. A spatial gradient may additionally give a net force. These qualifications matter because a uniform field can produce substantial torque while producing no leading translational force on the dipole.

τ=m×Bext,U=−m⋅Bext,Wext=ΔU\boldsymbol\tau=\mathbf m\times\mathbf B_{\mathrm{ext}},\qquad U=-\mathbf m\cdot\mathbf B_{\mathrm{ext}},\qquad W_{\mathrm{ext}}=\Delta U
Torque is in N m and interaction energy is in J. These dipole expressions assume the external field varies little across the source, m is fixed, and the rotation is quasistatic when Wext=ΔU is used. The external field excludes the dipole's self-field.↔ Scroll wide equations horizontally.

The origin is a distributional question

The inverse-cube expression cannot be assigned a finite value at the origin. For an ideal point dipole obtained from a shrinking current loop at fixed moment, its distributional field includes the contact contribution (2μ0/3)m δ3(r)(2\mu_0/3)\mathbf m\,\delta^3(\mathbf r), together with the spherical principal-value interpretation of the singular exterior tensor. The delta is a distribution acting under integration, not an extra uniform field near the magnet or a number that can be read at its centre.

A real finite loop has a regular interior field determined by its actual geometry. Replacing it by the exterior dipole approximation inside the loop is invalid, even if the approximation works beautifully at a distant sensor. Contact terms matter in distributions and some microscopic interaction calculations; ordinary torque examples with separated sources use the nonsingular external field. Also, electron spin magnetism is intrinsically quantum: representing its distant field by a dipole does not establish that an electron is a classical rotating charged ball.

Work it through

Turning a small coil toward a field

A 20-turn coil carries 0.40 A, each turn enclosing 0.0030 m². Its moment makes 60° with a uniform 0.050 T external field. Find its moment, initial torque magnitude, and quasistatic external work to align it.

  1. Multiply turn count, current, and area to obtain m=20(0.40)(0.0030)=0.024 A m2m=20(0.40)(0.0030)=0.024\,\mathrm{A\,m^2}. The oriented normal, rather than the wire tangent, gives the direction of this moment.

  2. Use the perpendicular component for torque: τ=(0.024)(0.050)sin⁡60∘=1.04×10−3 N m\tau=(0.024)(0.050)\sin60^\circ=1.04\times10^{-3}\,\mathrm{N\,m}. The torque direction is given by the cross product and tends to reduce the angle.

  3. Initially Ui=−(0.024)(0.050)cos⁡60∘=−0.00060 JU_i=-(0.024)(0.050)\cos60^\circ=-0.00060\,\mathrm J. Finally Uf=−0.00120 JU_f=-0.00120\,\mathrm J. Their difference is −0.00060 J, the work by a slowly controlling external agent.

The result

The moment is 0.024 A m², the initial torque magnitude is 1.04 millinewton metres, and the external work is −0.60 millijoules: the agent extracts energy during alignment.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

Parallel projection and dipole energy

Reset and set both controls to Angle 60° and Magnitude 2. Treat x as the imposed-field direction and predict the parallel projection 1. For a fresh comparison, Reset and set both controls to Angle 0° and Magnitude 2; predict projection 2. Energy is proportional to the negative of that projection. This instrument displays vector components, not magnetic torque or a field map.

Parallel projection and dipole energyPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAlignment increases the parallel projection and lowers fixed-dipole interaction energy. The lab does not simulate rotation, self-fields, or the singular dipole contact term.

Pause & reason

Check your understanding

At the origin of the ideal current-loop point-dipole model, how should the magnetic field be interpreted?

Your turn

Try it yourself

A localized dipole has moment 0.10 A m² along +z. Find its vacuum field at distance 0.20 m on the positive axis and at the same distance on the equatorial plane, assuming the source is much smaller than 0.20 m.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

On axis the bracket in the dipole formula becomes twice the moment; in the equatorial plane its scalar product with the radial unit vector vanishes.

Compare with the solution

The cube of the distance is 0.203=0.0080 m30.20^3=0.0080\,\mathrm{m^3}. On axis, Bz=(2×10−7)(0.10)/0.0080=2.5×10−6 TB_z=(2\times10^{-7})(0.10)/0.0080=2.5\times10^{-6}\,\mathrm T. At the equator, Bz=−(10−7)(0.10)/0.0080=−1.25×10−6 TB_z=-(10^{-7})(0.10)/0.0080=-1.25\times10^{-6}\,\mathrm T. Thus the fields are +2.5 and −1.25 microteslas along z. Their ratio of magnitudes is two, and the reversal at the equator is an angular effect rather than a negative source strength.

Keep the idea

Take it with you

  • A magnetic moment determines the leading distant field of a localized steady source.
  • Dipole torque and interaction energy use an externally imposed field.
  • An ideal contact distribution cannot replace the interior field of a finite magnet.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.