Deriving magnetic dipole moment from multipole expansion

[WeiShan Ng]

Homework Statement

This is from Griffith's Introduction to Electrodynamics, where the book is deriving the magnetic dipole moment from multipole expansion of the vector potential

The vector potential of a current loop can be written as
$$\mathbf{A(r)}=\frac{\mu_0 I}{4\pi} \left[ \frac{1}{r} \oint d\mathbf{l'} + \frac{1}{r^2} \oint r' \cos \alpha d\mathbf{l'} + \frac{1}{r^3} \oint (r')^2 \left( \frac{3}{2} \cos^2 \alpha - \frac{1}{2} \right) d\mathbf{l'} + \dots \right]$$
and the dipole is
$$\mathbf{A}_{dip}(\mathbf{r}) = \frac{\mu_o I}{4\pi r^2} \oint r' \cos \alpha d\mathbf{l'} = \frac{\mu_o I}{4\pi r^2} \oint (\mathbf{\hat{r}} \cdot \mathbf{r'} ) d\mathbf{l'}$$
The book then use the relation
$$\oint (\mathbf{\hat{r}} \cdot \mathbf{r'} )d\mathbf{l'}=-\mathbf{\hat{r}} \times \int d\mathbf{a'}$$
to get the magnetic dipole formula
$$\mathbf{A}_{dip} (\mathbf{r}) = \frac{\mu_0}{4\pi} \frac{\mathbf{m}\times \mathbf{\hat{r}}}{r^2}$$
where
$$\mathbf{m} \equiv I\int d\mathbf{a} = I \mathbf{a}$$

The problem is, I just couldn't figure out where we get the relation
$$\oint (\mathbf{\hat{r}} \cdot \mathbf{r'} )d\mathbf{l'}=-\mathbf{\hat{r}} \times \int d\mathbf{a'}$$ from? Can someone please shed some light on this? Thanks a lot.

Homework Equations

The Attempt at a Solution

 

[TSny]

The problem is, I just couldn't figure out where we get the relation
$$\oint (\mathbf{\hat{r}} \cdot \mathbf{r'} )d\mathbf{l'}=-\mathbf{\hat{r}} \times \int d\mathbf{a'}$$

Griffiths suggests using the identity (1.108) of the text which says that for any constant $\mathbf c$, $$\oint \left(\mathbf c \cdot \mathbf r' \right) d\mathbf l' = \mathbf a \times \mathbf c $$ Griffths says to use this identity with $\mathbf c = \hat{\mathbf r}$.