Magnetic Dipole Moment

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Chris Vuong - Spring 2026

Main Idea

A magnetic dipole moment describes the strength and orientation of a current loop as a magnetic source. For a loop carrying current, the magnetic dipole moment is a vector that points perpendicular to the plane of the loop, with direction determined by the right-hand rule.

The magnitude of the magnetic dipole moment is

[math]\displaystyle{ \mu = IA }[/math]

where [math]\displaystyle{ I }[/math] is the current and [math]\displaystyle{ A }[/math] is the area enclosed by the loop. For a coil with [math]\displaystyle{ N }[/math] turns, the magnitude becomes

[math]\displaystyle{ \mu = NIA }[/math]

The magnetic dipole moment is useful because it helps predict how a loop behaves in an external magnetic field. In particular, it is used to calculate torque and magnetic potential energy. Current loops, bar magnets, and even atoms can often be modeled as magnetic dipoles.

Mathematical Model

The magnetic dipole moment is represented by the vector [math]\displaystyle{ \boldsymbol{\mu} }[/math]. Its magnitude is equal to the current in the loop multiplied by the area enclosed by the loop. For a circular loop, the area is [math]\displaystyle{ \pi R^2 }[/math]. The direction of [math]\displaystyle{ \boldsymbol{\mu} }[/math] is given by the right-hand rule. The SI units of magnetic dipole moment are ampere-meters squared.

[math]\displaystyle{ |\boldsymbol{\mu}| = \mu = IA = I\pi R^2 }[/math]

For a coil with [math]\displaystyle{ N }[/math] turns, the magnetic dipole moment becomes

[math]\displaystyle{ |\boldsymbol{\mu}| = \mu = NIA }[/math]

• The magnitude of the magnetic field at the center of a current loop simplifies as (see Magnetic Field of a Loop):

[math]\displaystyle{ \begin{align} B & = \frac{\mu_0}{4 \pi} \frac{2 \pi I R^2}{(z^2 + R^2)^{3/2}} \text{, where } z = 0 \\ & = \frac{\mu_0}{4 \pi} \frac{2 \pi I R^2}{R^3} \text{, and the } R^2 \text{ cancels} \\ & = \frac{\mu_0}{4 \pi} \frac{2 \pi I}{R} \\ & = \frac{\mu_0}{4 \pi} \frac{2 \pi \frac{\mu}{\pi R^2}}{R} \text{, where } I = \frac{\mu}{\pi R^2} \\ B & = \frac{\mu_0}{4 \pi} \frac{2 \mu}{R^3} \end{align} }[/math]

• The magnitude of the magnetic moment [math]\displaystyle{ \mu }[/math] of an electron in circular orbit is given by

[math]\displaystyle{ |\boldsymbol{\mu}| = \mu = IA = -\frac{e}{2m_e}L }[/math]

where [math]\displaystyle{ L }[/math] is the magnitude of the angular momentum of the electron.

• The torque experienced by a current loop in an external magnetic field is defined by the cross product of the magnetic dipole moment and the external magnetic field:

[math]\displaystyle{ \boldsymbol{\tau} = \boldsymbol{\mu} \times \boldsymbol{B} }[/math]

• The magnetic potential energy of the loop is defined as the negative dot product of the magnetic dipole moment and the external magnetic field:

[math]\displaystyle{ U(\theta) = -\boldsymbol{\mu}\cdot\boldsymbol{B} = -\mu B \cos(\theta) }[/math]

Here, [math]\displaystyle{ \theta }[/math] is the angle between the magnetic moment and the external magnetic field.

• For example, if [math]\displaystyle{ \theta = 0^\circ }[/math], then

[math]\displaystyle{ U(0^\circ) = -\mu B \cos(0^\circ) = -\mu B }[/math]

• For [math]\displaystyle{ \theta = 180^\circ }[/math], we get

[math]\displaystyle{ U(180^\circ) = -\mu B \cos(180^\circ) = \mu B }[/math]

This shows that when the magnetic moment is aligned with the magnetic field, the loop is at its lowest magnetic potential energy. When the two directions are anti-parallel, the loop is at its highest magnetic potential energy.

Computational Model

The following GlowScript model gives a simple visualization of the magnetic field produced by a current loop. The arrows represent sample magnetic field vectors along the axis of the loop.

<glowscript> from vpython import * R = 0.1 I = 2 mu = I*pi*R**2

ring = ring(pos=vector(0,0,0), axis=vector(0,0,1), radius=R)

1. Visualize sample magnetic field vectors

for z in range(-5,6):

   B = vector(0,0, (mu)/(R**2+z**2)**(3/2))
   arrow(pos=vector(0,0,z*0.05), axis=B*0.1, color=color.blue)

</glowscript>

Examples

Simple

A [math]\displaystyle{ 1,200 }[/math] turn circular coil of radius [math]\displaystyle{ 10 \text{ cm} }[/math] carries [math]\displaystyle{ 2 \text{ Amps} }[/math].

a) What is the magnitude of the magnetic dipole moment of the coil. What direction does it point?

We can use our Mathematical Model to answer this:

[math]\displaystyle{ \mu = N I A = N I \cdot \pi R^2 }[/math]

In this problem, the following values have been specified for us:

[math]\displaystyle{ N = 1,200 \text{ turns or loops} }[/math]

[math]\displaystyle{ R = 0.1 \text{ m} }[/math]

[math]\displaystyle{ I = 2 \text{ Amps} }[/math]

Therefore, [math]\displaystyle{ \mu }[/math] is:

[math]\displaystyle{ \mu = 1,200 \times 2 \times \pi \times (0.1)^2 = 75.4 \text{ A} \cdot \text{m}^2 }[/math]

Middling

A current loop with a radius of [math]\displaystyle{ 2 \text{ m} }[/math] and a current of [math]\displaystyle{ 5 \text{ Amps} }[/math] is in an external magnetic field. This magnetic field is given by:

[math]\displaystyle{ \mathbf{B_{ext}} = (x, y, z) = (1, 1, 1) = \sqrt{3} \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) }[/math]

A unit vector normal to the area of the current loop (and has a direction consistent with the right-hand rule) is given by:

[math]\displaystyle{ \mathbf{\hat n} = (x, y, z) = \left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) }[/math]

a) What is the magnetic potential energy of the current loop?

The magnetic potential energy of a current loop is defined as:

[math]\displaystyle{ U(\theta) = - \boldsymbol{\mu} \bullet \mathbf{B_{ext}} = - [\mu B_{ext}] \ \text{cos}(\theta) \ \mathbf{(1)} }[/math]

Since [math]\displaystyle{ \mathbf{\hat n} }[/math] is consistent with the direction of [math]\displaystyle{ \boldsymbol{\mu} }[/math] and has a magnitude of [math]\displaystyle{ 1 }[/math], we can define [math]\displaystyle{ \boldsymbol{\mu} }[/math] as the its magnitude times its unit direction:

[math]\displaystyle{ \boldsymbol{\mu} = \mu \mathbf{\hat n} \ \mathbf{(2)} }[/math]

We know [math]\displaystyle{ \mu }[/math] is:

[math]\displaystyle{ \mu = I A = I \cdot \pi R^2 \ \mathbf{(3)} }[/math]

Putting 2 and 3 together shows:

[math]\displaystyle{ \boldsymbol{\mu} = \left( I \cdot \pi R^2 \right) \mathbf{\hat n} }[/math]

Plugging this into 1 gives:

[math]\displaystyle{ U(\theta) = - \left( I \cdot \pi R^2 \right) \mathbf{\hat n} \bullet \mathbf{B_{ext}} }[/math]

The dot product is calculated as:

[math]\displaystyle{ \begin{align} \mathbf{\hat n} \bullet \mathbf{B_{ext}} & = \left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) \bullet \sqrt{3} \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \\ & = \sqrt{3} \left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) \bullet \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \\ & = \sqrt{3} \Biggr[ \left(-\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} \right) + \left(-\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} \right) + \left(-\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} \right) \Biggr] \\ & = \sqrt{3} \Biggr[ -\frac{1}{3} -\frac{1}{3} -\frac{1}{3} \Biggr] \\ & = \sqrt{3} \Bigr[- 1 \Bigr] \\ \mathbf{\hat n} \bullet \mathbf{B_{ext}} & = -\sqrt{3} \\ \end{align} }[/math]

We can now find the magnetic potential energy of the loop:

[math]\displaystyle{ U(\theta) = \sqrt{3} \times 5 \times \pi \times (2)^2 = 20 \sqrt{3} \pi \approx 108.83 \text{ J} }[/math]

Keep in mind we could have found the angle between [math]\displaystyle{ \mathbf{\hat n} }[/math] and [math]\displaystyle{ \mathbf{B_{ext}} }[/math] and used

[math]\displaystyle{ U(\theta) = - [\mu B_{ext}] \ \text{cos}(\theta) }[/math]

to find the magnetic potential energy. The angle found would be [math]\displaystyle{ \pm 180&deg; }[/math], corresponding to an anti-parallel relation between the magnetic field and the magnetic moment, and therefore a max potential energy.

Difficult

A thin uniform ring of radius [math]\displaystyle{ R }[/math] and mass [math]\displaystyle{ M }[/math], carrying a uniformly distributed charge [math]\displaystyle{ +Q }[/math], rotates clockwise (as seen from above) about its axis with a constant angular speed [math]\displaystyle{ \omega }[/math]. It lies in the xy-plane.

a) Find the ratio of the magnitude of the ring's magnetic dipole moment [math]\displaystyle{ (\mu) }[/math] to its angular momentum [math]\displaystyle{ (L) }[/math].

The current in the ring can be calculated using the period of the rotation [math]\displaystyle{ (T) }[/math] and the charge on the ring as:

[math]\displaystyle{ I = \frac{dQ}{dt} = \frac{\Delta Q}{\Delta t} = \frac{Q}{T} = \frac{Q \omega}{2 \pi} }[/math]

The magnetic moment is:

[math]\displaystyle{ \mu = I A = \frac{Q \omega}{2 \pi} \cdot \pi R^2 = \frac{1}{2} QωR^2 }[/math]

The angular momentum is calculated as the loops rotational inertia (moment of inertia) times its angular speed. The rotational inertia of a ring is given by [math]\displaystyle{ m R^2 }[/math]:

[math]\displaystyle{ L = I \omega = m R^2 \omega }[/math]

Therefore, the ratio is:

[math]\displaystyle{ \frac{\mu}{L} = \frac{\frac{1}{2} Q \omega R^2}{m R^2 \omega} = \frac{Q}{2m} }[/math]

b) What is the torque experienced by the ring if an external magnetic field described by [math]\displaystyle{ \mathbf{B} = (2, 4, -5) }[/math] is applied to the ring?

We know the torque experienced by a current carrying loop is described by:

[math]\displaystyle{ \boldsymbol{\tau} = \boldsymbol{\mu} \times \mathbf{B} \ \mathbf{(1)} }[/math]

The magnitude of [math]\displaystyle{ \boldsymbol{\mu} }[/math] was found to be [math]\displaystyle{ \frac{1}{2} Q \omega R^2 }[/math] in the previous part. To find the direction of [math]\displaystyle{ \boldsymbol{\mu} }[/math], we consider the direction of the current and where the ring lies. Since the ring is lying in the xy-plane and rotating clockwise, using the right-hand rule shows [math]\displaystyle{ \boldsymbol{\mu} }[/math] must point in the [math]\displaystyle{ -z }[/math] direction. A unit vector in this direction can be described as [math]\displaystyle{ \mathbf{\hat n} = (0, 0, -1) }[/math].

Thus, [math]\displaystyle{ \boldsymbol{\mu} }[/math] can be described by:

[math]\displaystyle{ \boldsymbol{\mu} = \mu \mathbf{\hat n} }[/math]

Plugging this into 1 shows:

[math]\displaystyle{ \begin{align} \boldsymbol{\tau} & = \mu \bigr[ \mathbf{\hat n} \times \mathbf{B} \bigr] \\ & = \mu \bigr[(0, 0, -1) \times (2, 4, -5) \bigr] \\ & = \mu \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & -1 \\ 2 & 4 & -5 \end{vmatrix} \\ & = \mu \biggr[ \ \mathbf{i} \begin{vmatrix} 0 & -1 \\ 4 & -5 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 0 & -1 \\ 2 & -5 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 0 & 0 \\ 2 & 4 \end{vmatrix} \biggr] \\ & = \mu \biggr[\mathbf{i} \bigr[(0) - (-4) \bigr] - \mathbf{j} \bigr[(0) - (-2) \bigr] + \mathbf{k} \bigr[(0) - (0) \bigr] \biggr] \\ & = \mu \bigr[4 \mathbf{i} - 2 \mathbf{j} + 0 \mathbf{k} \bigr] \\ & = \frac{1}{2} Q \omega R^2 \cdot (4, -2, 0) \\ & = Q \omega R^2 \cdot (2, -1, 0) \\ \boldsymbol{\tau} & = \sqrt{5} \cdot Q \omega R^2 \left(\frac{2}{\sqrt{5}}, -\frac{1}{\sqrt{5}}, 0 \right) \end{align} }[/math]

Connectedness

Magnetic dipole moment connects to many topics in E&M:

• Used in torque and potential energy problems involving loops and magnets
• Basis of ferromagnetism and MRI imaging
• Fundamental idea behind solenoids, coils, motors and inductors
• Relates to Ampere’s Law and Maxwell’s Equations
• Connects to quantum magnetic moments (spin, Bohr magneton)

History

The study of magnetism dates far back in time, but it was not until 1825 that Andre Ampere showed that magnetism is due to perpetually flowing current through loops of wire. He then went on to derive Ampere's Law, which connected magnetic fields to electric currents. This equation was then further adapted to simplify the on axis magnetic field generated by a loop of current to use the magnetic dipole moment.

At first, people thought magnets, both natural and man-made, are made up of countless magnetic dipoles, each consisting of positive and negative magnetic charges separated by a small distance. This consequently defined the concept of magnetic dipole moment, a vector, pointing from the negative magnetic charge to the positive magnetic charge. When scientists discovered that magnetic charges, or "magnetic monopoles" do not exist, the magnetic properties of matter were believed to be generated by the alignment of molecular currents. Since then, magnetic dipole moment has been redefined using current [math]\displaystyle{ I }[/math], allowing the idea of magnetic dipole to remain.

Real-World Applications

• MRI machines rely on magnetic dipole alignment in hydrogen atoms
• Credit card strips encode data using magnetic dipoles
• Electric motors generate torque from μ × B
• Earth's magnetic field acts as a planetary dipole moment

See also

Here are some more resources with extra information on magnetic dipole moments.

Further reading

• Lecture PDF on Magnetic Dipole Moment
• Wikipedia Magnetic dipole
• MRI Questions Magnetic Moment
• http://web.mit.edu/8.02t/www/mitxmaterials/Presentations/Presentation_W07D2.pdf MIT PDF Magnetic Moment]

External links

• HyperPhysics Magnetic Moment
• Bozeman Science Magnetic Dipole Moment

References

• Lecture PDF on Magnetic Dipole Moment
• HyperPhysics: Magnetic Moment
• MRI Questions: Magnetic Dipole Moment