Chapter 7, Problem 40P

(a)

To determine

Find whether the given field $A$ is electrostatic or magnetostatic field in free space.

Answer to Problem 40P

The vector $A$ is neither electrostatic nor magnetostatic field.

Explanation of Solution

Calculation:

Given field is,

$A=y\cos ax{a}_x+\left(y+e^{-x}\right)a_z$

Generally, the vector is a magnetostatic field if its dot product is zero and the vector is an electrostatic field if its cross product is zero. That is,

For magnetostatic field, $\nabla \cdot \left(\text{vector}\right)=0$

For electrostatic field, $\nabla \times \left(\text{vector}\right)=0$

Substitute $y\cos ax{a}_x+\left(y+e^{-x}\right)a_z$ for $A$ to find $\nabla \cdot A$.

$\begin{array}{c}\nabla \cdot A=\frac{\partial }{\partial x}{A}_x+\frac{\partial }{\partial y}{A}_y+\frac{\partial }{\partial z}{A}_z\\ =\frac{\partial }{\partial x}\left(y\cos ax\right)+\frac{\partial }{\partial y}\left(0\right)+\frac{\partial }{\partial z}\left(y+e^{-x}\right)\\ =\left(y\left(-\sin ax\right)\left(a\right)\right)+\frac{\partial }{\partial y}\left(0\right)+\left(0\right)\\ =-ya\sin ax+0\end{array}$

Simplify the above equation.

$\begin{array}{c}\nabla \cdot A=-ya\sin ax\\ \ne 0\end{array}$

Substitute $y\cos ax{a}_x+\left(y+e^{-x}\right)a_z$ for $A$ to find $\nabla \times A$.

$\begin{array}{c}\nabla \times A=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ A_x& A_y& A_z\end{array}\right|\\ =\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ y\cos ax& 0& \left(y+e^{-x}\right)\end{array}\right|\\ =\left[\begin{array}{c}\left(\frac{\partial }{\partial y}\left(y+e^{-x}\right)-\frac{\partial }{\partial z}\left(0\right)\right)a_x-\left(\frac{\partial }{\partial x}\left(y+e^{-x}\right)-\frac{\partial }{\partial z}\left(y\cos ax\right)\right)a_y\\ +\left(\frac{\partial }{\partial x}\left(0\right)-\frac{\partial }{\partial y}\left(y\cos ax\right)\right)a_z\end{array}\right]\\ =\left[\begin{array}{l}\frac{\partial }{\partial y}\left(y+e^{-x}\right)a_x-\left(\frac{\partial }{\partial x}\left(y+e^{-x}\right)-\frac{\partial }{\partial z}\left(y\cos ax\right)\right)a_y\\ -\frac{\partial }{\partial y}\left(y\cos ax\right)a_z\end{array}\right]\end{array}$

Simplify the above equation.

$\begin{array}{c}\nabla \times A=\left(1+0\right)a_x-\left(\left(0-e^{-x}\right)-0\right)a_y-\left(1\right)\cos ax{a}_z\\ =1a_x+e^{-x}{a}_y-\cos ax{a}_z\\ \ne 0\end{array}$

Therefore, for the field $A$,

$\begin{array}{c}\nabla \cdot A\ne 0\\ \nabla \times A\ne 0\end{array}$

Conclusion:

Thus, the vector $A$ is neither electrostatic nor magnetostatic field.

(b)

To determine

Find whether the given field $B$ is electrostatic or magnetostatic field in free space.

Answer to Problem 40P

The vector $B$ is Electrostatic field in a charge free region.

Explanation of Solution

Calculation:

Given field is,

$B=\frac{20}{\rho }{a}_{\rho }$

Substitute $\frac{20}{\rho }{a}_{\rho }$ for $B$ to find $\nabla \cdot B$.

$\begin{array}{c}\nabla \cdot B=\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho A_{\rho }\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }{A}_{\varphi }+\frac{\partial }{\partial z}{A}_z\\ =\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \left(\frac{20}{\rho }\right)\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(0\right)+\frac{\partial }{\partial z}\left(0\right)\\ =\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(20\right)+0\\ =0\end{array}$

Substitute $\frac{20}{\rho }{a}_{\rho }$ for $B$ to find $\nabla \times B$.

$\begin{array}{c}\nabla \times B=\frac{1}{\rho }\left|\begin{array}{ccc}{a}_{\rho }& \rho a_{\varphi }& a_z\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ A_{\rho }& \rho A_{\varphi }& A_z\end{array}\right|\\ =\frac{1}{\rho }\left|\begin{array}{ccc}{a}_{\rho }& \rho a_{\varphi }& a_z\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ \frac{20}{\rho }& 0& 0\end{array}\right|\\ =\frac{1}{\rho }\left[\begin{array}{c}\left(\frac{\partial }{\partial \varphi }\left(0\right)-\frac{\partial }{\partial z}\left(0\right)\right)a_{\rho }-\left(\frac{\partial }{\partial \rho }\left(0\right)-\frac{\partial }{\partial z}\left(\frac{20}{\rho }\right)\right)\rho a_{\varphi }\\ +\left(\frac{\partial }{\partial \rho }\left(0\right)-\frac{\partial }{\partial \varphi }\left(\frac{20}{\rho }\right)\right)a_z\end{array}\right]\\ =\frac{1}{\rho }\left[0+\frac{\partial }{\partial z}\left(\frac{20}{\rho }\right)\rho a_{\varphi }-\frac{\partial }{\partial \varphi }\left(\frac{20}{\rho }\right)a_z\right]\end{array}$

Simplify the above equation.

$\begin{array}{c}\nabla \times B=\frac{1}{\rho }\left[\rho \frac{\partial }{\partial z}\left(\frac{20}{\rho }\right)a_{\varphi }-\frac{\partial }{\partial \varphi }\left(\frac{20}{\rho }\right)a_z\right]\\ =\frac{1}{\rho }\left[\rho \left(0\right)a_{\varphi }-\left(0\right)a_z\right]\\ =\frac{1}{\rho }\left(0\right)\\ =0\end{array}$

Therefore, for the field $B$,

$\begin{array}{c}\nabla \cdot B=0\\ \nabla \times B=0\end{array}$

Conclusion:

Thus, the vector $B$ is Electrostatic field in a charge free region.

(c)

To determine

Find whether the given field $C$ is electrostatic or magnetostatic field in free space.

Answer to Problem 40P

The vector $C$ is the magnetostatic field.

Explanation of Solution

Calculation:

Given field is,

$C=r^2\sin \theta a_{\varphi }$

Substitute $r^2\sin \theta a_{\varphi }$ for $C$ to find $\nabla \cdot C$.

$\begin{array}{c}\nabla \cdot C=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2{A}_r\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta A_{\theta }\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }\left(A_{\varphi }\right)\\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\left(0\right)\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \left(0\right)\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }\left(r^2\sin \theta \right)\\ =0+0+\frac{1}{r\sin \theta }\left(0\right)\\ =0\end{array}$

Substitute $r^2\sin \theta a_{\varphi }$ for $C$ to find $\nabla \times C$.

$\begin{array}{c}\nabla \times C=\frac{1}{r^2\sin \theta }\left|\begin{array}{ccc}{a}_r& r{a}_{\theta }& r\sin \theta a_{\varphi }\\ \frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ 0& r\left(0\right)& r\sin \theta \left(r^2\sin \theta \right)\end{array}\right|\\ =\frac{1}{r^2\sin \theta }\left|\begin{array}{ccc}{a}_r& r{a}_{\theta }& r\sin \theta a_{\varphi }\\ \frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ 0& 0& r^3{\sin }^2\theta \end{array}\right|\\ =\frac{1}{r^2\sin \theta }\left[\begin{array}{l}\left(\frac{\partial }{\partial \theta }\left(r^3{\sin }^2\theta \right)-\frac{\partial }{\partial \varphi }\left(0\right)\right)a_r\\ -\left(\frac{\partial }{\partial r}\left(r^3{\sin }^2\theta \right)-\frac{\partial }{\partial \varphi }\left(0\right)\right)r{a}_{\theta }\\ +\left(\frac{\partial }{\partial r}\left(0\right)-\frac{\partial }{\partial \theta }\left(0\right)\right)r\sin \theta a_{\varphi }\end{array}\right]\\ =\frac{1}{r^2\sin \theta }\left[\frac{\partial }{\partial \theta }\left(r^3{\sin }^2\theta \right)a_r-\frac{\partial }{\partial r}\left(r^3{\sin }^2\theta \right)r{a}_{\theta }+\left(0\right)r\sin \theta a_{\varphi }\right]\end{array}$

Simplify the above equation.

$\begin{array}{c}\nabla \times C=\frac{1}{r^2\sin \theta }\left[r^3\left(2\sin \theta \right)\left(\cos \theta \right)a_r-3r^2{\sin }^2\theta a_{\theta }\right]\\ =\frac{1}{r^2\sin \theta }\left[2r^3\sin \theta \cos \theta a_r-3r^2{\sin }^2\theta a_{\theta }\right]\\ =\left(\frac{1}{r^2\sin \theta }\right)2r^3\sin \theta \cos \theta a_r-\left(\frac{1}{r^2\sin \theta }\right)3r^2{\sin }^2\theta a_{\theta }\\ =2r\cos \theta a_r-3\sin \theta a_{\theta }\\ \ne 0\end{array}$

Therefore, for the field $C$,

$\begin{array}{c}\nabla \cdot C=0\\ \nabla \times C\ne 0\end{array}$

Conclusion:

Thus, the vector $C$ is the magnetostatic field.