Question

The current diensty in a medium is given by

$\overrightarrow{J}=\frac{400sin\theta }{2\pi (r^2+4)}{\hat{a}}_{r}A{m}^{-2}$

The total current and the average current density flowing through the portion of a spherical surface

r = 0.8 m, $\frac{\pi }{12}\le \theta \le \frac{\pi }{4},0\le 2\pi$ are given, respectively, by

• A. 15.09 A, 12.86 $A{m}^{-2}$
• B. 7.17 A, 6.88 $A{m}^{-2}$
• C. 12.86 A, 9.23 $A{m}^{-2}$
• D. 10.28 A, 7.56 $A{m}^{-2}$

Answer 1

The correct option is B 7.17 A, 6.88 $A{m}^{-2}$

Ans : (b)

$I=\int \overrightarrow{J}.\overrightarrow{dS}$

= ${\int }_{\theta =\frac{\pi }{12}}^{\pi /4}{\int }_{\varphi =0}^{2\pi }\frac{400sin\theta }{2\pi (r^2+4)}\cdot sin\theta d\theta d\varphi$

$=\frac{400}{2\pi (r^2+4)}\cdot r^2\cdot \varphi {\mid }_0^{2\pi }{\int }_{\pi /12}^{\pi /4}si{n}^2\theta d\theta$

$=\frac{400r^2}{(r^2+4)}{\int }_{\pi /12}^{\pi /4}\left(\frac{1-cos2\theta }{2}\right)d\theta$

$=\frac{400\cdot r^2}{(r^2+4)}(\frac{(\frac{\pi }{4}-\frac{\pi }{12})}{2}-{\left(\frac{sin2\theta }{4}\right)}_{\pi /12}^{\pi /4})$

= $\frac{400\cdot r^2}{(r^2+4)}(\frac{\pi }{12}-\left(\frac{1-1/2}{4}\right)){\mid }_{r=0.8}$

= $\frac{400\times 0.8\times 0.8}{4.64}\times 0.13=7.17Amp$

Total area = $\int$ dS = $\int \int r^{2}sin\theta d\theta d\varphi$

= $r^2{\int }_{\theta =\frac{\pi }{12}}^{\pi /4}sin\theta d\theta \cdot 2\pi$

$=r^2\cdot 2\pi \cdot 0.259{\mid }_{r=0.8}$

$={0.8}^2\times 2\pi \times 0.259$

$=1.041m^2$

Average current = $\frac{7.17}{1.041}=6.88A/m^2$