Question

Prove Eq. 5.78, using Eqs. 5.63, 5.76, and 5.77. [Suggestion: I'd set up Cartesian coordinates at the surface, with Z perpendicular to the surface and X parallel to the current.]

Short answer

The equation 5.78 is proved.

Step-by-step solution

Significance of the magnetostatics

Magnetostatics is described as the subfield of electromagnetics that describes a static field of the magnet. Moreover, magnetostatics also appears around the magnetized bodies’ surface.

Proving the equation 5.78

The equation 5.78 is expressed as:

$\frac{\partial A_{above}}{\partial n}-\frac{\partial A_{below}}{\partial n}=-{\mu }_{0}K$

Here, $\frac{\partial A_{above}}{\partial n}$ is the derivative of the potential of $A_{above}$, $\frac{\partial A_{below}}{\partial n}$ is the derivative of the potential of $A_{below}$, ${\mu }_0$ is the permeability and K is the constant.

The equation 5.77 is expressed as:

$A_{above}=A_{below}$

The equation 5.76 is expressed as:

$B_{above}-B_{below}={\mu }_0(K\times \hat{n})$ …(i)

Here, $B_{above}$ is one component of the magnetic field, $\hat{n}$ is the position vector and role="math" localid="1657534284873" $B_{below}$ is another component of the magnetic field.

The equation 5.63 is expressed as:

$\nabla \cdot A=0$

Here, $\nabla$ is the curl.

As $A_{above}=A_{below}$, then the value of $\frac{\partial a}{\partial y}$ and $\frac{\partial a}{\partial x}$ are also same in below and also in above.

The equation of the magnetic field can be expressed as:

$B_{above}-B_{below}=\left(-\frac{\partial {\mathrm{A}}_{\mathrm{y}\mathrm{above}}}{\partial \mathrm{z}}+\frac{\partial {\mathrm{A}}_{\mathrm{y}\mathrm{below}}}{\partial \mathrm{z}}\right)\hat{x}+\left(-\frac{\partial {\mathrm{A}}_{\mathrm{x}\mathrm{above}}}{\partial \mathrm{z}}+\frac{\partial {\mathrm{A}}_{\mathrm{x}\mathrm{below}}}{\partial \mathrm{z}}\right)\hat{\mathrm{y}}$ …(iii)

Comparing the equation (ii) and (iii).

role="math" localid="1657533984204" $\left(-\frac{\partial {\mathrm{A}}_{\mathrm{y}\mathrm{above}}}{\partial \mathrm{z}}+\frac{\partial {\mathrm{A}}_{\mathrm{y}\mathrm{below}}}{\partial \mathrm{z}}\right)\hat{\mathrm{x}}+\left(-\frac{\partial {\mathrm{A}}_{\mathrm{x}\mathrm{above}}}{\partial \mathrm{z}}+\frac{\partial {\mathrm{A}}_{\mathrm{x}\mathrm{below}}}{\partial \mathrm{z}}\right)\hat{\mathrm{y}}={\mathrm{\mu }}_0(K\times \hat{n})$

As the equation 5.76 is along the axis, then the above equation can be expressed as:

role="math" localid="1657534042142" $\left(-\frac{\partial {\mathrm{A}}_{\mathrm{y}\mathrm{above}}}{\partial \mathrm{z}}+\frac{\partial {\mathrm{A}}_{\mathrm{y}\mathrm{below}}}{\partial \mathrm{z}}\hat{\mathrm{x}}\right)+\left(-\frac{\partial {\mathrm{A}}_{\mathrm{x}\mathrm{above}}}{\partial \mathrm{z}}+\frac{\partial {\mathrm{A}}_{\mathrm{x}\mathrm{below}}}{\partial \mathrm{z}}\right)\hat{\mathrm{y}}={\mathrm{\mu }}_0\mathrm{K}(-\hat{\mathrm{y}})$

Due to the x and the y components, the above equation can be reduced to two values such as:

$$\begin{gathered} \left(-\frac{\partial A_{yabove}}{\partial z}-\frac{\partial {A_y}_{below}}{\partial z}\right)\hat{x}=\hat{x} \\ \frac{\partial A_{yabove}}{\partial z}-\frac{\partial {A_y}_{below}}{\partial z} \\ \left(-\frac{\partial A_{xabove}}{\partial z}-\frac{\partial {A_x}_{below}}{\partial z}\right)\hat{y}=-{\mu }_{0}K\hat{y} \\ \left(\frac{\partial A_{xabove}}{\partial z}-\frac{\partial A_{x_{}below}}{\partial z}\right)=-{\mu }_{0}K \end{gathered}$$

According to the equation 5.63, the normal derivative of the A component is parallel to the product of the permeability and constant K .

Hence, the above equation can be expressed as:

$\frac{\partial A_{above}}{\partial n}-\frac{\partial A_{below}}{\partial n}=-{\mu }_{0}K$

Thus, the equation 5.78 is proved.