Magnetic Dipoles: Torque, Energy and Far Field
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You must be comfortable with the Biot–Savart law, the definition of the vector potential \(\vec{A}\), and basic vector calculus identities, particularly the curl of a cross product.
Why this matters
The magnetic dipole is the fundamental building block for understanding how magnets interact with fields and each other.
In electrostatics, the electric dipole is the first correction to the monopole field. In magnetostatics, because there are no magnetic monopoles, the dipole is the leading term of the field expansion for any localized current distribution. Understanding the dipole moment, its torque in an external field, and its potential energy allows us to model everything from compass needles to atomic magnetic moments using a single, unified vector quantity \(\vec{m}\).
The Magnetic Dipole Moment
Imagine a small loop of wire carrying a steady current \(I\). If you stand far away from this loop, the details of the wire's shape become irrelevant; the field looks exactly like that of a tiny bar magnet. We define a vector quantity, the magnetic dipole moment \(\vec{m}\), to capture this behavior. For a planar loop, \(\vec{m}\) points perpendicular to the plane of the loop, following the right-hand rule with respect to the current direction. Its magnitude is the product of the current and the area enclosed, \(m = I a\). This vector \(\vec{m}\) is the source of the magnetic field in the far region, analogous to how an electric dipole moment \(\vec{p}\) sources the electric field. The key insight is that \(\vec{m}\) is a property of the source distribution itself, independent of where you measure the field, provided you are far enough away.
The Far-Field Vector Potential
Starting from the general expression for the vector potential of a steady current distribution, we perform a multipole expansion. Since the magnetic monopole term vanishes identically (due to \(\nabla \cdot \vec{J} = 0\)), the first non-zero term is the dipole term. The derivation (see D-054) shows that the vector potential at a point \(\vec{r}\) far from the origin is given by:
This expression is remarkably simple. It depends only on the dipole moment \(\vec{m}\), the distance \(r\), and the direction \(\hat{r}\). The cross product ensures that \(\vec{A}\) is always perpendicular to both \(\vec{m}\) and the radial direction \(\hat{r}\). This form is the magnetic analogue of the electric dipole potential, but with a crucial difference: the electric dipole potential is a scalar, while the magnetic dipole potential is a vector.
The Magnetic Dipole Field
To find the magnetic field \(\vec{B}\), we take the curl of the vector potential: \(\vec{B} = \nabla \times \vec{A}\). Applying the vector calculus identity for the curl of a cross product and using the fact that \(\vec{m}\) is constant, we obtain the standard dipole field expression:
This field has a characteristic shape. Along the axis of the dipole (where \(\hat{r}\) is parallel to \(\vec{m}\)), the field points in the same direction as \(\vec{m}\) and has a magnitude of \(\frac{\mu_0}{2\pi} \frac{m}{r^3}\). In the equatorial plane (where \(\hat{r}\) is perpendicular to \(\vec{m}\)), the field points opposite to \(\vec{m}\) and has a magnitude of \(\frac{\mu_0}{4\pi} \frac{m}{r^3}\). The field falls off as \(1/r^3\), which is faster than the \(1/r^2\) fall-off of a point charge field, reflecting the fact that the dipole is a higher-order multipole.
Torque and Potential Energy
When a magnetic dipole is placed in an external magnetic field \(\vec{B}_{\text{ext}}\), it experiences a torque that tends to align \(\vec{m}\) with the field. The torque is given by the cross product:
This torque is zero when \(\vec{m}\) is parallel or antiparallel to \(\vec{B}_{\text{ext}}\), and maximum when they are perpendicular. The system also has a potential energy associated with its orientation. By integrating the work done against the torque, we find the energy to be:
The negative sign indicates that the energy is minimized when \(\vec{m}\) is aligned with \(\vec{B}_{\text{ext}}\). This is the magnetic analogue of the electric dipole energy \(U = -\vec{p} \cdot \vec{E}\). It explains why a compass needle aligns with the Earth's magnetic field: the system seeks the lowest energy state.
The derivation
Magnetic Dipole: Field, Torque and Energy
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 \(U=-\vec{m}\cdot\vec{B}\).
A common misconception is that the magnetic dipole field is identical to the field of a pair of magnetic monopoles. While the far-field expressions look similar, the near-field behavior is fundamentally different. A true magnetic dipole (like a current loop) has a field that is continuous and divergence-free everywhere, whereas a hypothetical monopole pair would have singularities at the monopole locations. The dipole field derived here is valid only in the region where \(r\) is much larger than the size of the current distribution; inside or close to the loop the field does not follow the \(1/r^3\) dipole law (the full Biot–Savart integral is needed there). C
Worked example
Problem. A circular loop of radius \(R = 0.10\ \mathrm{m}\) carries a current \(I = 5.0\ \mathrm{A}\). It is placed in a uniform external magnetic field \(\vec{B}_{\text{ext}} = 0.20\ \mathrm{T}\) directed along the \(x\)-axis. The plane of the loop is initially in the \(xy\)-plane, so its magnetic moment \(\vec{m}\) is directed along the \(z\)-axis. Calculate the magnitude of the torque on the loop and the potential energy of the dipole in this orientation.
- Determine the magnetic dipole moment \(\vec{m}\). For a circular loop, \(m = I A = I \pi R^2\). Substituting the values: \(m = 5.0 \times \pi \times (0.10)^2 = 0.05\pi\ \mathrm{A \cdot m^2}\). The direction is \(\hat{z}\), so \(\vec{m} = 0.05\pi\ \hat{z}\ \mathrm{A \cdot m^2}\).
- Calculate the torque \(\vec{N} = \vec{m} \times \vec{B}_{\text{ext}}\). Since \(\vec{m}\) is along \(\hat{z}\) and \(\vec{B}_{\text{ext}}\) is along \(\hat{x}\), the angle between them is \(90^\circ\). The magnitude is \(N = m B_{\text{ext}} \sin(90^\circ) = m B_{\text{ext}}\). Substituting: \(N = (0.05\pi) \times 0.20 = 0.01\pi\ \mathrm{N \cdot m}\).
- Calculate the potential energy \(U = -\vec{m} \cdot \vec{B}_{\text{ext}}\). Since \(\vec{m}\) is perpendicular to \(\vec{B}_{\text{ext}}\), the dot product is zero. Thus, \(U = 0\ \mathrm{J}\).
Units check. Torque: \(\mathrm{A \cdot m^2} \times \mathrm{T} = \mathrm{A \cdot m^2} \times \mathrm{N/(A \cdot m)} = \mathrm{N \cdot m}\). Energy: \(\mathrm{A \cdot m^2} \times \mathrm{T} = \mathrm{N \cdot m} = \mathrm{J}\). Both are consistent.
Three quick tests
- What is the direction of the magnetic dipole moment for a current loop?
Answer
Perpendicular to the plane of the loop, determined by the right-hand rule (curl fingers in direction of current, thumb points in direction of \(\vec{m}\)). - How does the magnetic dipole field fall off with distance?
Answer
As \(1/r^3\). - What is the potential energy of a magnetic dipole aligned with an external field?
Answer
\(U = -mB\), which is the minimum possible energy.
Problem set for PU-102 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.