📂Electrodynamics

Multipole Expansion of Vector Potentials and Magnetic Dipole Moments

Multipole Expansion of Vector Potentials¹

The multipole expansion of vector potentials refers to the expression of vector potentials as a power series approximation to $\dfrac{1}{r^{n}}$ when currents are concentrated, sufficiently far from the source. Initially, the vector potential due to a current loop is as follows.

$$\mathbf{A}(\mathbf{r})=\dfrac{\mu_{0} I}{4\pi}\oint \dfrac{1}{\cR}d\mathbf{l}^{\prime}$$

Under the conditions shown in the figure, the following equation holds.

$$\dfrac{1}{\cR} =\dfrac{1}{\sqrt{r^2+(r^{\prime})^2-2rr^{\prime}\cos\alpha}} = \dfrac{1}{r}\sum \limits_{n=0}^{\infty} \left( \dfrac{r^{\prime}}{r} \right)^{n} P_{n}(\cos \alpha )$$

Substituting this into the vector potential equation yields the following.

$$\mathbf{A}(\mathbf{r})=\dfrac{\mu_{0} I}{4\pi} \sum \limits_{n=0}^{\infty} \dfrac{1}{r^{n+1}} \oint (r^{\prime})^n P_{n}(\cos\alpha)d\mathbf{l}^{\prime}$$

Expanding and organizing $\sum$ gives the following.

$$\mathbf{A}(\mathbf{r})=\dfrac{\mu_{0} I}{4\pi} \left[ \dfrac{1}{r}\oint d\mathbf{l}^{\prime} + \dfrac{1}{r^2}\oint r^{\prime} \cos \alpha d\mathbf{l}^{\prime}+\dfrac{1}{r^3}\oint (r^{\prime})^2 \left( \dfrac{3}{2}\cos ^2 \alpha -\dfrac{1}{2}\right) d\mathbf{l}^{\prime} \cdots \right]$$

In this series, the first term is called the (magnetic) monopole, the second term the dipole, and the $n$rd term the $2^{n-1}$-pole. Since the displacement integrated along a closed curve is always $0$, the magnetic monopole term is always $0$.

$$\oint d \mathbf{l}^{\prime}=0$$

This equation explains that magnetic monopoles do not exist in nature. Therefore, a single pole cannot create a magnetic field, and at least a dipole (a pair of opposite charges at minimum) is necessary to generate a magnetic field. Unlike charges, where a point charge $\pm q$ can alone generate an electric field, cutting a magnet in half does not separate it into N and S poles but creates two new magnets. There are no single poles that emit magnetic fields. Magnetic monopoles are also called magnetic charges.

Magnetic Dipole Moment

As magnetic monopoles do not exist, the most important term in the multipole expansion of vector potentials is the dipole term. Since $\cos \alpha = \hat{\mathbf{r}} \cdot \hat{\mathbf{r}^{\prime}}$,

$$\mathbf{A}_{\text{dip}}(\mathbf{r}) = \dfrac{\mu_{0} I}{4 \pi r^2}\oint r^{\prime} \cos \alpha d\mathbf{l}^{\prime} = \dfrac{\mu_{0} I}{4\pi r^2}\oint (\hat{\mathbf{r}}\cdot \mathbf{r}^{\prime})d\mathbf{l}^{\prime}$$

The integral expression can be rewritten as follows, but the proof is omitted.

$$\oint(\hat{\mathbf{r}}\cdot \mathbf{r}^{\prime})d\mathbf{l}^{\prime}=-\hat{\mathbf{r}}\times \int d \mathbf{a}^{\prime}$$

Then,

$$\mathbf{A}_{\text{dip}}(\mathbf{r}) = \dfrac{\mu_{0}}{4\pi r^2} (-\hat{\mathbf{r}})\times \int I d\mathbf{a}^{\prime}$$

The result of this integral is called the magnetic dipole moment and is denoted by $\mathbf{m}$.

$$\mathbf{A}_{\text{dip}}(\mathbf{r})=\dfrac{\mu_{0}}{4 \pi} \dfrac{\mathbf{m} \times \hat{\mathbf{r}}}{r^2}$$

$$\mathbf{m}=I\int d\mathbf{a}^{\prime}=I\mathbf{a}^{\prime}$$

$\mathbf{a}^{\prime}$ represents the vector area of the loop. If the loop is flat, the magnitude of $\mathbf{a}$ is the area enclosed by the loop, and its direction is determined by the right-hand rule.

Example

Calculate the magnetic dipole moment for the ㄴ-shaped loop shown below. All sides have length $w$, and a current of $I$ flows through it.

The ㄴ-shaped loop is equivalent to two flat square loops joined together. Adding the magnetic dipole moments from each,

• The dipole moment of loop 1 is $Iw^2 \hat{\mathbf{y}}$
• The dipole moment of loop 2 is $Iw^2 \hat{\mathbf{z}}$ Therefore, $\mathbf{m}=Iw^2 \hat{\mathbf{y}}+Iw^2 \hat{\mathbf{z}}$