Question

I worked out the multipole expansion for the vector potential of a line current because that's the most common type, and in some respects the easiest to handle. For a volume current $\mathit{J}$:

(a) Write down the multipole expansion, analogous to Eq. 5.80.

(b) Write down the monopole potential, and prove that it vanishes.

(c) Using Eqs. 1.107 and 5.86, show that the dipole moment can be written

$\mathit{m}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\int }\left(r\times J\right)\mathit{d}\mathit{\tau }$

Short answer

(a) The multipole expansion is$\frac{{\mu }_0}{4\mathrm{\pi }}\sum _{n-0}^{\infty }{r}^{\frac{1}{n+1}}{\int }_v{\left(r^{\text{'}}\right)}^n{P}_n\left(\cos \alpha \right)\stackrel{⏜}{J}\left({\overrightarrow{r}}^{\text{'}}\right)d\tau$ .

(b) The monopole expansion vanishes.

(c) The dipole moment is$\frac{1}{2}\int \left(\overrightarrow{r}\times \overrightarrow{j}\right)d\tau$ .

Step-by-step solution

Significance of the multipole expansion

The multipole expansion is described as the mathematical series that mainly depends on the angle of an object. This type of expansion can be truncated as they provide a better approximation on the original function.

(a) Determination of the multipole expansion

The equation 5.80. can be expressed as:

$A\left(r\right)=\frac{{\mu }_{0}I}{4\mathrm{\pi }}\sum _{n-0}^{\infty }{r}^{\frac{1}{n+1}}\oint {\left(r^{\text{'}}\right)}^n{P}_n\left(\cos \alpha \right)d{I}^{\text{'}}$ …(i)

Here, $A\left(r\right)$is the current loop’s vector potential,${\mu }_0$is the permeability,$I$is the current,$\alpha$is the angle between$r^{\text{'}}$and r, r is the distance of the line current to the point inside the magnetic field,$r^{\text{'}}$is the first derivative of r, $d{I}^{\text{'}}$is the derivative of the increase in the length and $P_n$is the probability distribution till the$n$th term.

For a volume current ,$\overrightarrow{J}$from the above equation, the equation of the current will be expressed as:

$IdI\to \overrightarrow{J}d\tau$

Substitute for in the above equation.

role="math" localid="1657623058227" $A\left(r\right)=\frac{{\mu }_0}{4\mathrm{\pi }}\sum _{n-0}^{\infty }{r}^{\frac{1}{n+1}}{\left(r^{\text{'}}\right)}^n{P}_n\left(\cos \alpha \right)\overrightarrow{J}\left({\overrightarrow{r}}^{\text{'}}\right)d\tau$

Thus, the multipole expansion is $\frac{{\mu }_0}{4\mathrm{\pi }}\sum _{n-0}^{\infty }{r}^{\frac{1}{n+1}}{\left(r^{\text{'}}\right)}^n{P}_n\left(\cos \alpha \right)\overrightarrow{J}\left({\overrightarrow{r}}^{\text{'}}\right)d\tau$.

(b) Determination of the monopole expansion

The equation of the monopole moment is expressed as:

$$\begin{gathered} {\overrightarrow{A}}_0=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r}{\int }_v\overrightarrow{J}d\tau \\ =\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r}\frac{d\overrightarrow{p}}{dt} \end{gathered}$$

Here,$\overrightarrow{J}$ is the volume current and$\frac{d\overrightarrow{p}}{dt}$ is the rate of change of momentum with respect to the time t.

It has been observed that the total dipole moment is constant. Hence, the above equation can be expressed as:

$$\begin{gathered} \overrightarrow{A_0}=0 \\ =\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r}\frac{d\overrightarrow{p}}{dt} \end{gathered}$$

Hence, the monopole potential vanishes at${\overrightarrow{A}}_0=0$ .

Thus, the monopole expansion vanishes.

(c) Determination of the dipole moment

The equation 5.86 can be expressed as:

$\overrightarrow{m}=I\overrightarrow{a}$ …(ii)

Here,$m$is the magnetic dipole moment,$a$is the enclosed ordinary area and$I$is the current.

The equation 1.107 can be expressed as:

localid="1657623575092" $I\overrightarrow{a}=\frac{1}{2}{\oint }_c\overrightarrow{r}\times /d\overrightarrow{I}$ …(iii)

Equalling the equation (ii) and (iii).

$I\overrightarrow{a}=\frac{1}{2}{\oint }_c\overrightarrow{r}\times /d\overrightarrow{I}$

Hence, substitute $\overrightarrow{J}d\tau$for $IdI$in the above equation.

localid="1657623781646" $$\begin{gathered} I\overrightarrow{a}=\frac{1}{2}\int \left(\overrightarrow{r}\times \overrightarrow{J}\right)d\tau \\ =m \end{gathered}$$

Thus, the dipole moment is $I\overrightarrow{a}=\frac{1}{2}\int \left(\overrightarrow{r}\times \overrightarrow{J}\right)d\tau$.