##### Multipole Expansion of the Vector Potential
ems.ms.vp.me

Remember our expansion for the inverse distance (which we made good use of in the multipole expansion of the electrostatic field), equation 1or_Leg \[ \frac{1}{|{\bf r} - {\bf r}'|} = \frac{1}{r} \sum_{l=0}^{\infty} \left(\frac{r'}{r}\right)^l P_l (\hat{\bf r} \cdot \hat{\bf r}'). \] The expansion for the vector potential for a current loop carrying current \(I\) over path \({\cal P}\) can thus be written \[ {\bf A} ({\bf r}) = \frac{\mu_0 I}{4\pi} \sum_{l=0}^{\infty} \frac{1}{|{\bf r}|^{l+1}} \oint_{\cal P} d{\bf l}_s |{\bf r}_s|^l P_l (\hat{\bf r} \cdot \hat{\bf r}_s) \rho({\bf r}_s), \hspace{1cm} |{\bf r}| > |{\bf r}_s| ~\forall~ {\bf r}_s \in {\cal P}. \] Explicitly, the first few terms are (letting \(\hat{\bf r} \cdot \hat{\bf r}_s = \cos \theta\))

- Gr (5.80)

\begin{align} {\bf A}({\bf r}) = \frac{\mu_0 I}{4\pi} \left[ \frac{1}{r} \oint_{\cal P} d{\bf l}_s + \frac{1}{r^2} \oint d{\bf l}_s r_s \cos \theta + \frac{1}{r^3} \oint d{\bf l}_s (r_s)^2 \left( \frac{3}{2} \cos^2 \theta - \frac{1}{2} \right) + ... \right] \tag{A_Leg}\label{A_Leg} \end{align}

Again, these are known as the magnetic monopole, dipole, quadrupole terms.

Note: the magnetic monopole term always vanishes. This is simply because the total vector displacement on a closed loop is zero, \(\oint d{\bf l}_s = 0\), or in other words: there are no magnetic monopoles (also from Maxwell's equation divB0 \({\boldsymbol \nabla} \cdot {\bf B} = 0\)).

The dominant term is thus the dipole,

- Gr (5.85)

\[ {\bf A}_{di} ({\bf r}) = \frac{\mu_0 I}{4\pi} \frac{1}{r^2} \oint d{\bf l}_s (\hat{\bf r} \cdot {\bf r}_s) \tag{A_di1}\label{A_di1} \] Using the identity (see Gr Problem 1.62),

- Gr (1.108)

\[ \oint_{\cal P} d{\bf l} ({\bf c} \cdot {\bf r}) = {\bf a} \times {\bf c}, \hspace{1cm} {\bf a} = \int_{\cal S} d{\bf a} \tag{intcr}\label{intcr} \] with \({\bf c} = \hat {\bf r}\),

Parenthesis: derivation of intcr

Start from Stokes' theorem, \[ \int_{\cal S} d{\bf a} \cdot {\boldsymbol \nabla} \times {\bf v} = \oint_{\cal P} d{\bf l} \cdot {\bf v} \] Let \({\bf v} = {\bf c} T\), where \({\bf c}\) is constant. On the left-hand side: \[ LHS = \int_{\cal S} d{\bf a} \cdot (T ({\boldsymbol \nabla} \times c)) - \int_{\cal S} d{\bf a} \cdot ({\bf c} \times ({\boldsymbol \nabla} T)) \] The first term is zero since \({\bf c}\) is constant. For the second term: use the triple product vector identity \(d{\bf a} \cdot ({\bf c} \times ({\boldsymbol \nabla} T)) = {\bf c} \cdot ({\boldsymbol \nabla}T \times d{\bf a})\). The second term thus becomes \[ -\int_{\cal S} d{\bf a} \cdot ({\bf c} \times ({\boldsymbol \nabla} T)) = -{\bf c} \cdot \int_{\cal S} {\boldsymbol \nabla}T \times d{\bf a} \] Treating the right-hand side of the original equation now, \[ RHS = {\bf c} \cdot \oint_{\cal P} T d{\bf l} \] so we get (since this is valid for any \({\bf c}\)) \[ \int_{\cal S} {\boldsymbol \nabla} T \times d{\bf a} = -\oint_{\cal P} T d{\bf l}. \] Now put \(T = {\bf c} \cdot {\bf r}\) in this, yielding intcr.

Back to our problem:

\[ \oint d{\bf l}_s (\hat{\bf r} \cdot {\bf r}_s) = -\hat{\bf r} \times \int_{\cal S} d{\bf a}_s \] and defining the

magnetic dipole moment

- Gr (5.86)

\[ {\bf m} \equiv I \int_{\cal S} d{\bf a} = I {\bf a} \tag{magdim}\label{magdim} \]

we obtain the convenient expression for the

dipole term of the vector potential

- Gr (5.85)

\[ {\bf A}_{di} ({\bf r}) = \frac{\mu_0}{4\pi} \frac{{\bf m} \times \hat{\bf r}}{r^2} \tag{A_di}\label{A_di} \]

Example: bookend shape

Task: find the magnetic dipole moment of a bookend shape current-carrying loop (c.f. Gr Fig. 5.52). All sides have length \(w\) and carry current \(I\).

Solution: combine two loops, use magdim \[ {\bf m} = I w^2 \hat{\bf y} + I w^2 \hat{\bf z} \]

Note: the magnetic dipole moment is independent of the choice of origin.

Note: does there exist a pure magnetic dipole? Well, yes, but it's an infinitely small loop carrying an infinitely large current, so that the dipole term is finite.

In practice: the dipole approximation often good enough when far away from source on a scale of the source's current loops.

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42