Find the vector potential above and below the plane surface current in Ex. 5.8.

The vector potential is described as a particular vector field that has a curl in order to identify the required vectors. Moreover, the vector potential involves a gradient in a particular vector field.

The equation in the example 5.8 is expressed as:

$K=K\hat{x}$

Here, $\mathit{K}$is the uniform surface current, K is the constant and $\hat{x}$is the unit vector in the x direction.

The equation of the magnetic field in example 5.8 is expressed as:

$$\begin{gathered} B=\frac{{\mu }_{0}K}{2}\hat{y}\left(z<0\right) \\ =-\frac{{\mu }_{0}K}{2}\hat{y}\left(z>0\right) \end{gathered}$$

Here, B is the magnetic field,${\mu }_0$is the permeability, z is the coordinate on the z axis and $\hat{y}$is the unit vector in the y direction.

The vector potential is parallel to the constant K and it is dependent on the function z .

The equation of the vector potential is expressed as:

$A=A\left(z\right)\hat{x}$

The equation of the magnetic field is expressed as:

$B=\nabla \times A$

Here, $\nabla$is the curl and A is the vector potential.

Substitute the values in the above equation.

$$\begin{gathered} B=\left|\begin{array}{ccc}\hat{x}& \hat{y}& \hat{z}\\ \partial l\partial x& \partial l\partial y& \partial l\partial z\\ A\left(z\right)& 0& 0\end{array}\right| \\ =\frac{\partial A}{\partial z}\hat{y} \\ =\pm \frac{{\mu }_{0}K}{2}\hat{y} \end{gathered}$$

From the above equation, the vector potential can be identified.

The vector potential can be expressed as $-\frac{{\mu }_{0}K}{2}\left|z\right|\hat{x}$.

Thus, the vector potential above and below the plane surface current is role="math" localid="1657522544831" $-\frac{{\mu }_{0}K}{2}\left|z\right|\hat{x}$.