Question

A vector magnetic potential is given by,

$\overrightarrow{A}=10\sin \theta {\delta }_{\theta }Wb/m$

$Themagneticfluxdensityat$ $\overrightarrow{B}(2,\frac{\pi }{2},0)willbe$

• A. 0
• B. $10{\hat{a}}_{r}Wb/m^2$
• C. $5{\hat{a}}_{\varphi }Wb/m^2$
• D. $-10{\hat{a}}_{\varphi }Wb/m^2$

Answer 1

The correct option is C $5{\hat{a}}_{\varphi }Wb/m^2$

The magnetic flux density is,

$\overrightarrow{B}$= $▽\times \overrightarrow{A}$

$=\frac{1}{r^2\sin \theta }\left|\begin{array}{ccc}{\hat{a}}_r& r{\hat{a}}_{\theta }& r\sin \theta {\hat{a}}_{\varphi }\\ \frac{\delta }{\delta r}& \frac{\delta }{\delta \theta }& \frac{\delta }{\delta \varphi }\\ 0& r\left(10\sin \theta \right)& 0\end{array}\right|$

$=\frac{1}{r^2\sin \theta }[-\frac{\delta }{\delta \varphi }(10r\sin \theta \left)\right]{\hat{a}}_r-\frac{r}{r^2\sin \theta }\left[\frac{\delta }{{\delta }_r}\right(0\left)\right]{\hat{a}}_{\theta }+\frac{10r{\sin }^2\theta }{r^2\sin \theta }\left[\frac{\delta }{\delta r}\right(r\left)\right]{\hat{a}}_{\varphi }$

$=\frac{10\sin \theta }{r}{\hat{a}}_{\varphi }$

$\overrightarrow{B}(2,\frac{\pi }{2},0)=\frac{10\sin \pi /2}{2}{\hat{a}}_{\varphi }=5{\hat{a}}_{\varphi }Wb/m^2$