Question

Given the general Vector $\overrightarrow{A}$ = $\left(sin2\varphi \right){\hat{a}}_{\varphi }$ in cylindrical system , then the curl of vector
$\overrightarrow{A}$ at (2,$\frac{\pi }{4}$,0) is

• A. 0.8 ${\hat{a}}_z$
• B. -0.5 ${\hat{a}}_z$
• C. 1.5 ${\hat{a}}_z$
• D. 0.5 ${\hat{a}}_z$

Answer 1

The correct option is D 0.5 ${\hat{a}}_z$

(d)
In cylindrical coordinate system:
$▽\times \overrightarrow{A}$ $=\frac{1}{\rho }\left|\begin{array}{ccc}{\hat{a}}_{\rho }& \rho {\hat{a}}_{\varphi }& {\hat{a}}_z\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ A_{\rho }& A_{\varphi }& A_z\end{array}\right|$

$=\frac{1}{\rho }\left|\begin{array}{ccc}{\hat{a}}_{\rho }& \rho {\hat{a}}_{\varphi }& {\hat{a}}_z\\ \frac{\partial }{\partial \rho }& \frac{\partial }{\partial \varphi }& \frac{\partial }{\partial z}\\ 0& \rho sin2\varphi & 0\end{array}\right|$

=$\frac{1}{\rho }\left[{\hat{a}}_{\rho }\right(0)-\rho {\hat{a}}_{\varphi }(0)+{\hat{a}}_z\frac{\partial }{\partial p}(\rho sin2\varphi \left)\right]$

= $\frac{1}{\rho }\left(sin2\varphi \right){\hat{a}}_z$

at point P(2, $\frac{\pi }{4}$, 0) ,
$▽\times \overrightarrow{A}$ = $=\frac{1}{2}sin(2\times \frac{\pi }{4}){\hat{a}}_z$

= 0.5${\hat{a}}_z$