Solved: In the Lorenz gauge, the solution for the magnetic vector potential du

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in the lorenz gauge the solution for the magnetic vector potential due

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Electromagnetic field

In the Lorenz gauge, the solution for the magnetic vector potential due to a current density J 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 give a 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 theta=0), 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{\phi}^{\prime}}{\sqrt{r^{2}+a^{2}-2 r a \sin \theta \cos \phi^{\prime}}} \mathrm{d} \phi^{\prime}

where o' is the unit vector pointing in the direction of the current flow at a point in the loop with azimuth al angular coordinate o'. (You may find it helpful to

describe the transverse dimensions of the thin wire using Dirac delta functions.)

Finally, by using the approximation (1 – e)-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 m is a vector pointing along the positive z-axis with magnitude I × (area of loop).(5)

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