Magnetic dipole moment

Remarks

Because $\nabla \cdot J = 0$ and $J$ has compact support, $\int J \, dy = 0$, and so $m$ does not depend on the choice of origin. For a plane loop enclosing area $a$ with unit normal $n$ fixed by the right-hand rule, $m = I a \, n$. Far from the currents, the vector potential is $A(x) \approx \frac{\mu_0}{4\pi} \frac{m \times x}{\abs{x}^3}$, and the magnetic field is approximately $\frac{\mu_0}{4 \pi} \frac{3 (m \cdot \hat x) \hat x - m}{\abs{x}^3}$, with $\hat x = x / \abs{x}$. In a uniform field $B$, the torque on the distribution is $m \times B$ and its energy is $-m \cdot B$. Example: a circular loop of radius $R$ in the plane $x_3 = 0$ with counterclockwise current $I$ has $m = (0, 0, I \pi R^2)$. Non-example: a current flowing straight along a line from one point to another is not a closed loop and has $\nabla \cdot J \ne 0$ at its ends, so the definition does not apply. The unit of $m$ is the ampere square metre.