Question

(a) Check Eq. 5.76 for the configuration in Ex. 5.9.

(b) Check Eqs. 5.77 and 5.78 for the configuration in Ex. 5.11.

Short answer

(a) The equation 5.76 satisfies.

(b) The equations 5.77 and 5.78 is satisfied.

Step-by-step solution

Significance of magnetostatics

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.

(a) Checking the equation 5.76

The equation 5.76 can be expressed as:

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

Here, $B_{above}$and $B_{below}$are the magnetic field at the top and the bottom,${\mu }_0$ 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:

$B_{below}={\mu }_{0}nI\hat{z}$

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

Substitute for in the above equation.

$B_{below}={\mu }_{0}K\hat{z}$

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

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

Thus, the equation 5.76 satisfies.

(b) Checking the equation 5.77 and 5.78

The equation in the example 5.11 is expressed as:

$$\begin{gathered} \mathrm{A}\left(\mathrm{r},\mathrm{\theta },\mathrm{\varphi }\right)=\frac{{\mathrm{\mu }}_0\mathrm{R\omega \delta }}{3}\mathrm{r}\mathrm{sin\theta }\hat{\mathrm{\varphi }}\left(\mathrm{r}\le \mathrm{R}\right) \\ =\frac{{\mathrm{\mu }}_0{\mathrm{R}}^4\mathrm{\omega \delta }}{3}\frac{\mathrm{sin\theta }}{{\mathrm{r}}^2}\hat{\mathrm{\varphi }}\left(\mathrm{r}\ge \mathrm{R}\right) \end{gathered}$$ …(ii)

The equation 5.77 is expressed as:

$A_{above}=A_{below}$

The equation 5.78 is expressed as:

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

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 .

$$\begin{gathered} {\left[\frac{\partial A}{\partial r}\right]}_{R^{+}}={\left[\frac{{\mu }_0{R}^4\omega \delta }{3}\left(-\frac{2\sin \theta }{r^3}\right)\hat{\varphi }\right]}_R \\ =\frac{2{\mu }_{0}R\omega \delta }{3}\sin \theta \hat{\varphi } \\ =\frac{{\mu }_{0}R\omega \delta }{3}\sin \theta \hat{\varphi } \end{gathered}$$

The equation of the constant K is expressed as:

$$\begin{gathered} K=\delta V \\ =\delta \left(\omega \times r\right) \\ =\delta \omega r\sin \theta \hat{\varphi } \end{gathered}$$

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

Thus, the equations 5.77 and 5.78 is satisfied.