(a) By whatever means you can think of (short of looking it up), find the vector potential a distance from an infinite straight wire carrying a current . Check that $\mathbf{\nabla }\mathbf{.}\mathbf{A}\mathbf{=}\mathbf{0}$and $\mathbf{\nabla }\mathbf{\times }\mathbf{A}\mathbf{=}\mathbf{B}$.

(b) Find the magnetic potential inside the wire, if it has radius R and the current is uniformly distributed.

The given data is listed below as:

• The current carries by the wire is $I$,
• The radius of the wire is $R$,

The magnetostatics is described as the study of the magnetic fields where the currents do not change with time. In the study of magnetostatics, the charges are kept stationary.

It has been observed that the vector potential$A$is parallel to the current $I$, and it is a function of the wire’s distance that is $s$.

The equation of the cylindrical coordinates is expressed as:

$\mathrm{A}=\mathrm{A}\left(\mathrm{s}\right)\stackrel{⏜}{\mathrm{z}}$

Here,$z$is the unit vector in the z axis.

The equation of the magnetic field is expressed as:

$B=\nabla \times A$ …(i)

Here, $B$is the magnetic field and $\nabla$is the curl.

The equation of the magnetic field can also be expanded as:

$$\begin{gathered} B=-\frac{\partial A}{\partial s}\stackrel{⏜}{\varphi } \\ =\frac{{\mu }_{0}I}{2\mathrm{\pi s}}\stackrel{⏜}{\varphi } \end{gathered}$$

Substitute the above value in the equation (i).

$A\left(r\right)=-\frac{{\mu }_{0}I}{2\mathrm{\pi }}In\left(s/a\right)\stackrel{⏜}{z}$

Hence, the equation (i) can also be written as:

$$\begin{gathered} \nabla \times A=-\frac{\partial A_z}{\partial s}\stackrel{⏜}{\varphi } \\ =\frac{{\mu }_{0}I}{2\mathrm{\pi s}}\stackrel{⏜}{\varphi } \\ =B \end{gathered}$$

The equation of the dot product of the curl and the vector potential can be expressed as:

$$\begin{gathered} \nabla .A=\frac{\partial A_z}{\partial z} \\ =0 \end{gathered}$$

Thus, the vector potential is $-\frac{{\mu }_{0}I}{2\mathrm{\pi }}In\left(s/a\right)\stackrel{⏜}{z}$and$\nabla .A=0$ and$\nabla \times A=B$ is proved.

The equation of the magnetic field is expressed as:

$\oint B.dl=B2\mathrm{\pi s}$ …(ii)

Here,$B$is the magnetic field,$s$is the distance from the wire and $dl$is the increase in the length.

The above equation can also be written as:

$$\begin{gathered} B2\mathrm{\pi s}={\mathrm{\mu }}_0{\mathrm{I}}_{\mathrm{enc}} \\ ={\mathrm{\mu }}_0{\mathrm{J\pi s}}^2 \\ ={\mathrm{\mu }}_0\frac{1}{{\mathrm{\pi R}}^2}{\mathrm{\pi s}}^2 \\ =\frac{{\mathrm{\mu }}_0{\mathrm{Is}}^2}{{\mathrm{R}}^2} \end{gathered}$$

Hence, the equation (ii) can be written as:

$$\begin{gathered} \oint B.dl=\frac{{\mu }_{0}I{s}^2}{R^2} \\ B=\frac{{\mu }_{0}I{s}^2}{R^2}\stackrel{⏜}{\varphi } \end{gathered}$$

The above equation can be written in terms of the magnetic potential.

$$\begin{gathered} \frac{\partial A}{\partial s}=-\frac{{\mu }_{0}I}{2\mathrm{\pi }}\frac{s}{R^2} \\ A=-\frac{{\mu }_{0}I}{4{\mathrm{\pi R}}^2}\left(s^2-b^2\right)\stackrel{⏜}{z} \end{gathered}$$

Here, $A$is the magnetic potential.

As the magnetic potential must be continuous at the radius of the wire, then the equation can be expressed as:

$-\frac{{\mu }_{0}I}{\mathrm{\pi }}In\left(R/a\right)=-\frac{{\mu }_{0}I}{4{\mathrm{\pi R}}^2}\left(R^2-b^2\right)$

Hence, the magnetic potential has two values such as $-\frac{{\mu }_{0}I}{4{\mathrm{\pi R}}^2}\left(s^2-R^2\right)\stackrel{⏜}{z}$for $s\le R$and$-\frac{{\mu }_{0}I}{2\mathrm{\pi }}In\left(s/R^2\right)\stackrel{⏜}{z}$ for $s\ge R$.

Thus, the magnetic potential inside the wire are$-\frac{{\mu }_{0}I}{4{\mathrm{\pi R}}^2}\left(s^2-R^2\right)\stackrel{⏜}{z}$ for$s\le R$ and$-\frac{{\mu }_{0}I}{2\mathrm{\pi }}In\left(s/R^2\right)\stackrel{⏜}{z}$ for$s\ge R$ respectively.