# Magnetic Dipole in a External Magnetic Field

## Important Formula

## What do the formula symbols mean?

## Magnetic dipole moment

`$$ \class{red}{\boldsymbol{\mu}} $$`Unit

`$$ \mathrm{A} \cdot \mathrm{m}^2 $$`

A torque acts on a magnetic dipole in an external magnetic field, resulting in the orientation of the dipole along the magnetic field direction.

## Electric current

`$$ \class{red}{\boldsymbol I} $$`Unit

`$$ \mathrm{A} = \frac{ \mathrm C }{ \mathrm s } $$`

## Area

`$$ \boldsymbol{A} $$`Unit

`$$ \mathrm{m}^2 $$`

## Table of contents

You probably already know **magnetic dipoles** in the form of a *bar magnet* with a north pole and a south pole. The magnetic field lines of a bar magnet run from the north to the south pole. A compass needle placed in the magnetic field of the bar magnet aligns itself so that the north pole of the compass needle aligns with the south pole. If the bar magnet is divided into two halves, the two will form a north and south pole. Also, if the generated bar magnets are hypotetically further divided into two halves, they always possess both poles.

## Magnetic dipole moment of a circular current

In classical electrodynamics, this tiny bar magnet is actually a tiny *current-carrying loop*. A circling electron, for example, thus generates a circular current, which can be regarded as a current-carrying loop. A moving charge generates a magnetic field. A collection of such tiny circular currents then align themselves so that the mutual interaction of the current loops, on a macroscopic level, creates the magnetic field of a bar magnet. Of course, it can also be a magnet of any other shape.

You can find out where the north or south pole of the current-carrying loop is by using the right hand rule. For example, consider a circular loop with an **electric current** \(I\) flowing through it. If the loop is enclosed with the right hand, the outstretched thumb points in the direction of the south pole. Inside the loop, the magnetic field lines run to the south pole.

The circular current encloses a surface represented by a vector \(\boldsymbol{A}\) which is orthogonal to the enclosed surface. Its magnitude gives the area \(A\) of the enclosed surface and its direction is determined by the right-hand rule.

The product of the current \(I\) with the surface orthogonal vector \(\boldsymbol{A}\) is called **magnetic dipole moment** \(\boldsymbol{A}\):

The unit of the dipole moment is: \([\mu] = \mathrm{A}\,\mathrm{m}^2 \). The magnetic dipole moment points in the same direction as the surface orthogonal vector \(\boldsymbol{A}\). The dipole moment is a characteristic quantity of the magnetic dipole and it occurs in important other quantities as you will see in a moment. It also occurs in the description of para-, dia- and ferromagnetism.

## Dipole in an external magnetic field

If now the magnetic dipole is brought into an external (homogeneous) **magnetic field** \( \class{violet}{\boldsymbol{B}} \), then the dipole will interact with the external magnetic field. The dipole rotates. Why the dipole rotates is due to the Lorentz force (magnetic force).

Consider a rectangular loop in a homogeneous magnetic field. Let the loop be suspended in such a way that it can only rotate along one axis \(\text{C}\) (see illustration).The **current** \(I\) flows in a piece of conductor of **length** \(a\) parallel to this axis of rotation. The Lorentz force \(\boldsymbol{F}\) acts on this current-carrying wire, the direction of which can be determined using the right-hand rule. In the second parallel piece of wire, the current flows in the opposite direction \(-I\), which is why the Lorentz force acts in the opposite direction here.

Due to the opposing forces acting on both wires, the rectangular loop performs a rotation. As soon as the magnetic dipole moment \(\boldsymbol{\mu}\) points in the same direction as the magnetic field \(\class{violet}{\boldsymbol{B}}\), the loop stops rotating, because the Lorentz force now pulls the wires outwards.

If the magnetic dipole moment \(\boldsymbol{\mu}\) points in the opposite direction as the magnetic field \(\class{violet}{\boldsymbol{B}}\) (antiparallel), then the two Lorentz forces on the conductor pieces act inward (into the loop). In perfect equilibrium, the loop would not rotate. But a tiny perturbation is enough to make the loop rotate so that \(\boldsymbol{\mu}\) and \(\class{violet}{\boldsymbol{B}}\) align in parallel.

## Torque on the dipole

The rotation of the dipole generates a **torque** \(\boldsymbol{M}\). The torque \(\boldsymbol{M}\) is the cross product between the magnetic dipole moment \(\boldsymbol{\mu}\) and the external magnetic field \(\class{violet}{\boldsymbol{B}}\). Thus \(\boldsymbol{M}\) is always orthogonal to \(\boldsymbol{\mu}\) and \(\class{violet}{\boldsymbol{B}}\) due to the property of the cross product.

See: Derivation of the torque. The magnitude \(M\) of the torque is thus:

Here \(\varphi\) is the angle enclosed by \(\mu\) and \(\class{violet}{B}\). If the angle \(\varphi = 0 \) (dipole moment and magnetic field are parallel), the torque disappears. Also at \(\varphi = 180^{\circ} \) (dipole moment and magnetic field are antiparallel) the torque disappears. However, as described earlier, this is not a stable equilibrium of the loop in the magnetic field. At\(\varphi = 90^{\circ} \) however, \(\sin(90^{\circ}) \) is equal to 1 and thus the torque becomes the maximum value: \(\mu \, \class{violet}{B} \).

## Potential energy of the dipole

The **potential energy** \(W_{\mu}\) of the magnetic dipole is the *scalar product* between the dipole moment \(\boldsymbol{\mu}\) and the external magnetic field \(\class{violet}{\boldsymbol{B}}\).

See: Deriving the potential energy of a magnetic dipole.

If \(\boldsymbol{\mu}\) and \(\class{violet}{\boldsymbol{B}}\) are aligned *parallel* to each other, then the potential energy is minimal (the most negative value): \(-\mu \, \class{violet}{B}\). This is exactly why a minus occurs in the equation, to minimize the potential energy in stable equilibrium. The dipole tries to reach this state of minimum energy by rotation in the homogeneous magnetic field.

In contrast, the dipole has the maximum potential energy when the dipole moment and the magnetic field are aligned *antiparallel* to each other. Then the scalar product is negative, which altogether with the other minus gives a positive and maximum energy of the dipole: \(\mu \, \class{violet}{B}\).

The potential energy of the dipole is zero (but not minimal!) if \(\boldsymbol{\mu}\) and \(\class{violet}{\boldsymbol{B}}\) are *orthogonal* to each other.

A magnetic dipole deflected by the angle \(\varphi\) (between \(\boldsymbol{\mu}\) and \(\class{violet}{\boldsymbol{B}}\)) has the following energy according to Eq. 4

:

Without friction, the energy supplied to the dipole is conserved. For the energy to be conserved, the dipole must oscillate (as in a deflected pendulum) about its equilibrium position (\(\boldsymbol{\mu}\) and \(\class{violet}{\boldsymbol{B}}\) in parallel). In the case of friction, the dipole loses its energy and remains with its dipole moment parallel to the magnetic field.

## Force on the magnetic dipole

The **force** \(\boldsymbol{F}\) on a magnetic dipole is the negative gradient of the potential energy:

Here, \(\nabla\) is the nabla operator that contains the derivatives with respect to the spatial variables.

Assuming that \(\mu\) is independent of position, the force on the dipole can also be written as follows:

Note that the gradient of a vector field: \(\nabla \class{violet}{\boldsymbol{B}}\) (in the three-dimensional case) is a 3x3 matrix. This matrix yields the gradient for each component \(\class{violet}{B}_{\text x}\), \(\class{violet}{B}_{\text y}\), \(\class{violet}{B}_{\text z}\) of \(\class{violet}{\boldsymbol{B}}\). Then vector \(\boldsymbol{\mu}\) is multiplied by the matrix, which again gives a vector. This vector is a force on the magnetic dipole.

A homogeneous magnetic field is constant at any position, therefore the gradient of \( \class{violet}{\boldsymbol{B}} \)(i.e. spatial derivatives) vanish: \( \nabla \class{violet}{\boldsymbol{B}} = 0 \). Thus also the force on the dipole disappears.

In an *inhomogeneous* magnetic field, on the other hand, \(\class{violet}{\boldsymbol{B}}\) is position-dependent. The gradient \(\nabla \class{violet}{\boldsymbol{B}}\) is NOT zero and so is NOT the force. Since the gradient points in the direction of the steepest slope of the magnetic field, the force also points in the direction where the magnetic field increases the most.

## Atomic magnetic dipole

In a hydrogenium atom (H atom), in the Bohr atom model, the electron of charge \(q = -e \), mass \(m = m_{\text e}\) performs a circular motion. Due to this circular motion, the electron experiences both a **torque** (and thus also an **angular momentum**: \( \boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p} \) ). It has also a magnetic dipole moment. The angular momentum \(\boldsymbol{L}\) and the magnetic dipole moment \(\boldsymbol{\mu}\) are related as follows:

The proportionality factor \(\frac{q}{2m}\) is called **gyromagnetic ratio**. It determines the ratio of the magnetic dipole moment to the angular momentum. Small mass, large charge means: large dipole moment, small angular momentum. Large mass, small charge, on the other hand, means: small dipole moment, large angular momentum.

The equation 8

describes a circling atomic particle purely classically. Quantum mechanically (in the Bohr atom model) the angular momentum \(L\) is quantized. \(L\) occurs as a multiple of the reduced Planck's constant \(\hbar\): \(L = l \, \hbar \), with \( l \) as an integer quantum number. In the ground state, the electron in the H atom has the angular momentum \(L = \hbar\). Putting this into the magnitude of 8

gives the following equation:

Equation 9

is the smallest possible magnetic dipole moment, which is called **Bohr magneton** \(\mu_{\text B}\).

## Exercises with Solutions

Use this formula eBook if you have problems with physics problems.### Exercise #1: Dipole Moment of a Rotating Hollow Cylinder

A hollow cylinder of length \(L\), uniformly charged with electric charge density \(\rho\), rotates with angular velocity \(\omega\) about its longitudinal axis. The cylinder is not infinitely thin; its inner wall has radius \(r_{\text i}\) and outer wall has radius \(r_{\text e}\).

- What is the
**magnetic dipole moment**of the hollow cylinder? - What would be the
**magnetic moment**of the hollow cylinder if it were infinitely thin and carried a surface charge density \(\sigma\)?

#### Solution to Exercise #1.1

The starting point is the definition of the magnetic dipole moment for a current loop carrying current \(I\) and enclosing an area \(A\):
`1
\[ \mu = A \, I \]
`

The rotating uniformly charged cylinder has many current loops, each enclosing different currents and different areas. The current loop with radius \(r_1\) generates a different dipole moment than a current loop with radius \(r_2\). The area and current, and therefore the magnetic moment, depend on position:
`2
\[ \mu(r) = A(r) \, I(r) \]
`

The aim of this exercise is to find the total magnetic dipole moment and not just the dipole moment at radius \(r\). Therefore, all current loops of the cylinder must be summed up using an integral. Before doing that, let's first consider an infinitesimal dipole moment \(\text{d}\mu\) generated by an infinitesimal current element \(\text{d}I\). And the area \(A(r)\) is the area of a circle (\(\pi \, r^2\)), whose radius must vary naturally:
`3
\[ \text{d}\mu = \pi \, r^2 \, \text{d}I \]
`

The current element can be expressed in terms of the definition of electric current with infinitesimal charge \(\text{d}Q\) per period \(T\).
`4
\[ \text{d}\mu = \pi \, r^2 \, \frac{\text{d}Q}{T} \]
`

Now, the given angular velocity \(\omega\) can be brought into play, as the period \(T\) is inversely related to the frequency \(f\) (\( T = 1/f\)) and the frequency is related to the angular velocity solely through the factor \(2\pi\) (\(\omega = 2\pi \, f \)). Thus, the period can be expressed with the known angular velocity:
`5
\[ \text{d}\mu = \pi \, r^2 \, \frac{\omega}{2\pi} \, \text{d}Q \]
`

The charge is not known, so it also needs to be replaced. Here comes the given charge density \(\rho\) into play. The total charge on the cylinder is the charge density multiplied by the volume of the cylinder (\(Q=\rho \, V\)). However, since an infinitesimal charge \( \text{d}Q \) is considered here, the volume enclosed by this charge is also infinitesimal:
`6
\[ \text{d}\mu = \frac{\omega}{2} \, r^2 \, \rho \, \text{d}v \]
`

In the cylinder coordinates suitable for the problem, \(\text{d}v = r \, \text{d}\varphi \, \text{d}r \, \text{d}z \):
`7
\[ \text{d}\mu = \frac{\omega \, \rho}{2} \, r^2 \, r \, \text{d}\varphi \, \text{d}r \, \text{d}z \]
`

The integral over the z-coordinate would result in the length \(L\) of the cylinder, and the integral over the \(\varphi\)-coordinate around the circle would result in \(2\pi\):
`8
\[ \text{d}\mu = \frac{\omega \, \rho}{2} \, r^3 \, 2\pi \, L \, \text{d}r \]
`

Now, it's just a matter of integrating over the radius \(r\) from the inner wall to the outer wall of the cylinder:
`9
\[ \int \text{d}\mu = \pi \omega \, \rho \, L \, \int^{r_{\text e}}_{r_{\text i}} r^3 \, \text{d}r \]
`

The integral yields:
`10
\[ \mu = \pi \omega \, \rho \, L \, \left[ \frac{1}{4} \, r^4 \right]^{r_{\text e}}_{r_{\text i}} \]
`

Substituting the integration limits yields the total magnetic dipole moment of the hollow cylinder:
`11
\[ \mu = \frac{\pi}{4} \, \omega \, \rho \, L \, \left( r_{\text e}^4 - r_{\text i}^4 \right) \]
`

#### Solution to Exercise #1.2

The starting point is, as in Exercise #1.1, the definition of the magnetic dipole moment of a current loop:
`12
\[ \mu = A \, I \]
`

As in equation 4

, here the area \(A = \pi \, r^2 \) and the current \( I = Q/T\) are replaced. The difference now is that due to an infinitely thin hollow cylinder, there are now only current loops with radius \(r\):
`13
\[ \mu = \pi r^2 \, \frac{Q}{T} \]
`

Analogous to equation 5

, replace the period \(T\):
`14
\[ \mu = \pi r^2 \, \frac{\omega}{2\pi} \, Q \]
`

Now, the charge \(Q\) is expressed in terms of the given surface charge density (charge per area) and the area of the cylinder (circumference times height):
`15
\[ \mu = \frac{\omega}{2} \, r^2 \, \lambda \, 2\pi \, r \, L \]
`

This results in the magnetic dipole moment of a rotating, infinitely thin, charged hollow cylinder:
`16
\[ \mu = \pi \, \omega \, \lambda \, L \, r^3 \]
`