Question

If vector current density is represented by

$\overrightarrow{J}=\left(\frac{4}{r^2}\right)\cos \theta \widehat{a_r}+20e^{-2r}\sin \theta \widehat{a_{\theta }}-r\sin \theta \cos \varphi \widehat{a_{\varphi }}A/m^2$, then total current passing through the spherical capacitor (r=3 m, 0 < $\theta$ < ${60}^{\circ }$, 0 < $\varphi$ < $2\pi$) is

• A. 12.57 A
• B. 3.14 A
• C. 6.28 A
• D. 9.42 A

Answer 1

The correct option is D 9.42 A

At the surface, $r=3,\widehat{a_r}$ is perpendicular

$\overrightarrow{ds}=r^2\sin \theta d\theta d\varphi \widehat{a_r}$

Also $\overrightarrow{J}.\overrightarrow{ds}=\left(\frac{4}{r^2}\right)cos\theta \times r^{2}sin\theta d\theta d\varphi$

$=4cos\theta sin\theta d\theta d\varphi$

Total current, $I=\oint \overrightarrow{J}.\overrightarrow{ds}={\int }_0^{{60}^{\circ }}{\int }_0^{2\pi }2sin2\theta d\theta d\varphi$

$I=2{[-\frac{cos2\theta }{2}]}_0^{{60}^{\circ }}\times (2\pi -0)$

$=[-cos{120}^{\circ }+cos{0}^{\circ }]\times 2\pi$

$=[0.5+1]\times 2\pi$

$=3\pi =9.42A$