91Ó°ÊÓ

Magnetostatics is mainly used for predicting fast switching magnetic events which occur in less than a nanosecond. Moreover, magnetostatics is also a good approximation when there is no static current.

The equation 5.76 can be expressed as:

$B_{above}-B_{below}={\mathrm{μ}}_0(K×\hat{n})$ …(¾±)

Here, $B_{above}$and $B_{below}$are the magnetic field at the top and the bottom,${\mathrm{μ}}_0$ is the permeability, k is the constant and $\hat{n}$is the position vector.

At the solenoid’s surface, the magnetic field at the top is zero.

The equation of the magnetic field at the bottom at solenoid’s surface is expressed as:

$B_{below}={\mathrm{μ}}_{0}nI\hat{z}$

Here, $l$is the current and$\hat{z}$ is the position vector along the z axis.

Substitute for in the above equation.

$B_{below}={\mathrm{μ}}_{0}K\hat{z}$

Substitute $-K\hat{z}$for$(K×\hat{n})$ in the equation (i).

$B_{above}-B_{below}=-{\mathrm{μ}}_{0}K\hat{z}$

Thus, the equation 5.76 satisfies.

The equation in the example 5.11 is expressed as:

$$\begin{gathered} \mathrm{A}\left(\mathrm{r},\mathrm{θ},\mathrm{Ï•}\right)=\frac{{\mathrm{μ}}_0\mathrm{¸éÓ\neg δ}}{3}\mathrm{r}\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{Ï•}}\left(\mathrm{r}≤\mathrm{R}\right) \\ =\frac{{\mathrm{μ}}_0{\mathrm{R}}^4\mathrm{Ó\neg δ}}{3}\frac{\mathrm{²õ¾\pm ²Ôθ}}{{\mathrm{r}}^2}\widehat{\mathrm{Ï•}}\left(\mathrm{r}â‰\yen \mathrm{R}\right) \end{gathered}$$ …(¾±¾±)

The equation 5.77 is expressed as:

$A_{above}=A_{below}$

The equation 5.78 is expressed as:

$\frac{∂A_{above}}{∂n}-\frac{∂A_{below}}{∂n}=-{\mathrm{μ}}_{0}K$

In the equation of the example 5.11, the both the equations have the same values at the surface. Hence, it satisfies the equation 5.77 as$A_{above}=A_{below}$ .

Differentiating the equation (ii) with respect to the coordinate r in order to find the left side of the equation 5.78 .

$$\begin{gathered} {\left[\frac{∂A}{∂r}\right]}_{R^{+}}={\left[\frac{{\mathrm{μ}}_0{R}^4\mathrm{Ó\neg }\mathrm{δ}}{3}\left(-\frac{2\sin \mathrm{θ}}{r^3}\right)\widehat{\mathrm{Ï•}}\right]}_R \\ =\frac{2{\mathrm{μ}}_{0}R\mathrm{Ó\neg }\mathrm{δ}}{3}\sin \mathrm{θ}\widehat{\mathrm{Ï•}} \\ =\frac{{\mathrm{μ}}_{0}R\mathrm{Ó\neg }\mathrm{δ}}{3}\sin \mathrm{θ}\widehat{\mathrm{Ï•}} \end{gathered}$$

The equation of the constant K is expressed as:

$$\begin{gathered} K=\mathrm{δ}V \\ =\mathrm{δ}\left(\mathrm{Ó\neg }×r\right) \\ =\mathrm{δ}\mathrm{Ó\neg }r\sin \mathrm{θ}\widehat{\mathrm{Ï•}} \end{gathered}$$

Hence, the right and the left side of the equation 5.78 is satisfied.

Thus, the equations 5.77 and 5.78 is satisfied.