Chapter 7, Problem 41P

(a)

To determine

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

Answer to Problem 41P

The vector $D$ is possibly a magnetostatic field.

Explanation of Solution

Calculation:

Given field is,

$D=y^{2}z{a}_x+2\left(x+1\right)yz{a}_x-\left(x+1\right)z^2{a}_z$

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 $y^{2}z{a}_x+2\left(x+1\right)yz{a}_x-\left(x+1\right)z^2{a}_z$ for $D$ to find $\nabla \cdot D$.

$\begin{array}{c}\nabla \cdot D=\frac{\partial }{\partial x}{D}_x+\frac{\partial }{\partial y}{D}_y+\frac{\partial }{\partial z}{D}_z\\ =\frac{\partial }{\partial x}\left(y^{2}z\right)+\frac{\partial }{\partial y}\left(2\left(x+1\right)\right)+\frac{\partial }{\partial z}\left(-\left(x+1\right)\right)\\ =0+0+0\\ =0\end{array}$

Substitute $y^{2}z{a}_x+2\left(x+1\right)yz{a}_x-\left(x+1\right)z^2{a}_z$ for $D$ to find $\nabla \times D$.

$\begin{array}{c}\nabla \times D=\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ D_x& D_y& D_z\end{array}\right|\\ =\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ y^{2}z& 2\left(x+1\right)yz& -\left(x+1\right)z^2\end{array}\right|\\ =\left[\begin{array}{c}\left(\frac{\partial }{\partial y}\left(-\left(x+1\right)z^2\right)-\frac{\partial }{\partial z}\left(2\left(x+1\right)yz\right)\right)a_x-\left(\frac{\partial }{\partial x}\left(-\left(x+1\right)z^2\right)-\frac{\partial }{\partial z}\left(y^{2}z\right)\right)a_y\\ +\left(\frac{\partial }{\partial x}\left(2\left(x+1\right)yz\right)-\frac{\partial }{\partial y}\left(y^{2}z\right)\right)a_z\end{array}\right]\\ =\left[\begin{array}{c}\left(\left(0\right)-2\left(x+1\right)y\left(1\right)\right)a_x-\left(-\left(\left(1\right)z^2+0\right)-y^2\left(1\right)\right)a_y\\ +\left(\left(2\left(1\right)yz+0\right)-\left(2y\right)z\right)a_z\end{array}\right]\end{array}$

Simplify the above equation.

$\begin{array}{c}\nabla \times D=\left(-2\left(x+1\right)y\right)a_x-\left(-z^2-y^2\right)a_y+\left(2yz-2yz\right)a_z\\ =\left(-2\left(x+1\right)y\right)a_x+\left(z^2+y^2\right)a_y+\left(0\right)a_z\\ =-2\left(x+1\right)y{a}_x+\left(z^2+y^2\right)a_y\\ \ne 0\end{array}$

Therefore, for the field $D$,

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

Conclusion:

Thus, the vector $D$ is possibly a magnetostatic field.

(b)

To determine

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

Answer to Problem 41P

The vector $E$ is possibly a magnetostatic field.

Explanation of Solution

Calculation:

Given field is,

$E=\frac{\left(z+1\right)}{\rho }\cos \varphi a_{\rho }+\frac{\sin \varphi }{\rho }{a}_z$

Substitute $\frac{\left(z+1\right)}{\rho }\cos \varphi a_{\rho }+\frac{\sin \varphi }{\rho }{a}_z$ for $E$ to find $\nabla \cdot E$.

$\begin{array}{c}\nabla \cdot E=\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho E_{\rho }\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }{E}_{\varphi }+\frac{\partial }{\partial z}{E}_z\\ =\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \left(\frac{\left(z+1\right)}{\rho }\cos \varphi \right)\right)+\frac{1}{\rho }\frac{\partial }{\partial \varphi }\left(0\right)+\frac{\partial }{\partial z}\left(\frac{\sin \varphi }{\rho }\right)\\ =\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\left(z+1\right)\cos \varphi \right)+0+0\\ =0\end{array}$

Substitute $\frac{\left(z+1\right)}{\rho }\cos \varphi a_{\rho }+\frac{\sin \varphi }{\rho }{a}_z$ for $E$ to find $\nabla \times E$.

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

Simplify the above equation.

$\begin{array}{c}\nabla \times E=\frac{1}{\rho }\left[\begin{array}{c}\frac{\cos \varphi }{\rho }{a}_{\rho }-\rho \left(\sin \varphi \left(-\frac{1}{{\rho }^2}\right)-\left(\frac{\cos \varphi }{\rho }\right)\left(1+0\right)\right)a_{\varphi }\\ -\frac{\left(z+1\right)}{\rho }\left(-\sin \varphi \right)a_z\end{array}\right]\\ =\frac{1}{\rho }\left[\frac{\cos \varphi }{\rho }{a}_{\rho }-\rho \left(-\frac{\sin \varphi }{{\rho }^2}-\left(\frac{\cos \varphi }{\rho }\right)\right)a_{\varphi }+\frac{\left(z+1\right)\sin \varphi }{\rho }{a}_z\right]\\ =\frac{\cos \varphi }{{\rho }^2}{a}_{\rho }-\left(-\frac{\sin \varphi }{{\rho }^2}-\left(\frac{\cos \varphi }{\rho }\right)\right)a_{\varphi }+\frac{\left(z+1\right)\sin \varphi }{{\rho }^2}{a}_z\\ \ne 0\end{array}$

Therefore, for the field $E$,

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

Conclusion:

Thus, the vector $E$ is possibly a magnetostatic field.

(c)

To determine

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

Answer to Problem 41P

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

Explanation of Solution

Calculation:

Given field is,

$\begin{array}{c}F=\frac{1}{r^2}\left(2\cos \theta a_r+\sin \theta a_{\theta }\right)\\ =\frac{2}{r^2}\cos \theta a_r+\frac{1}{r^2}\sin \theta a_{\theta }\end{array}$

Substitute $\frac{2}{r^2}\cos \theta a_r+\frac{1}{r^2}\sin \theta a_{\theta }$ for $F$ to find $\nabla \cdot F$.

$\begin{array}{c}\nabla \cdot F=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2{F}_r\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta F_{\theta }\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }\left(F_{\varphi }\right)\\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\left(\frac{2}{r^2}\cos \theta \right)\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \left(\frac{1}{r^2}\sin \theta \right)\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \varphi }\left(0\right)\\ =\frac{1}{r^2}\frac{\partial }{\partial r}\left(2\cos \theta \right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\frac{1}{r^2}{\sin }^2\theta \right)+0\\ =\frac{1}{r^2}\left(0\right)+\frac{1}{r^3\sin \theta }\left(2\sin \theta \cos \theta \right)\end{array}$

Simplify the above equation.

$\begin{array}{c}\nabla \cdot F=0+\frac{1}{r^3}\left(2\cos \theta \right)\\ =\frac{2\cos \theta }{r^3}\\ \ne 0\end{array}$

Substitute $\frac{2}{r^2}\cos \theta a_r+\frac{1}{r^2}\sin \theta a_{\theta }$ for $F$ to find $\nabla \times F$.

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

Simplify the above equation.

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

Therefore, for the field $F$,

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

Conclusion:

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