In the Lorenz gauge, the solution for the magnetic vector potential due to a current density / is given by \vec{A}(\vec{r}, t)=\frac{\mu_{0}}{4 \pi} \int \frac{\left[\vec{J}\left(\vec{r}^{\prime}, t\right)\right]}{\left|\vec{r}-\vec{r}^{\prime}\right|} \mathrm{d}^{3} \vec{r}^{\prime} Re-express the

square-bracket notation in a different mathematical form and givea physical interpretation of the expression.(2) A thin wire, formed into a circular loop with radius a, is placed horizontally at the origin of a right-handed coordinate system, such that the symmetry axis of the loop coincides with the z-axis, as shown in the figure. The loop carries a constant anti-clockwise current I. By considering (without any loss of generality) an observation point with spherical coordinates (r,theta) in the x-z plane (i.e. with o =phi), show that the magnetic vector potential is given by \vec{A}(\vec{r})=\frac{\mu_{0}}{4 \pi} I a \int_{-\pi}^{\pi} \frac{\hat{\vec{\phi}}^{\prime}}{\sqrt{r^{2}+a^{2}-2 r a \sin \theta \cos \phi^{\prime}}} \mathrm{d} \phi^{\prime} where phi' is the unit vector pointing in the direction of the current flow at a point in the loop with azimuthal angular coordinate o'. (You may find it helpful to describe the transverse dimensions of the thin wire using Dirac delta functions.) (c)Finally, by using the approximation (1 )-1/2 × 1 + €/2 with e « 1, and noting that a symmetrical integral over an odd function vanishes, show that far from the loop, with r » a, the vector potential can be expressed as \vec{A}(\vec{r})=\frac{\mu_{0}}{4 \pi} \frac{\vec{m} \times \vec{r}}{r^{3}} where the loop magnetic dipole mi is a vector pointing along the positive z-axis with magnitude I × (area of loop).(5)