A) 0.125 Nm2/C
B) 0.02 Nm2/C
C) 0.005 Nm2/C
D) 3.14 Nm2/C
Correct Answer: A
Solution :
\[\overset{\to }{\mathop{E}}\,={{E}_{0}}\hat{i}+2{{E}_{0}}\hat{j}\] Given, \[{{E}_{0}}=100N/c\] So, \[\overset{\to }{\mathop{E}}\,=100\hat{i}+200\hat{j}\] Radius of circular surface = 0.02 m Area \[=\pi {{r}^{2}}=\frac{22}{7}\times 0.02\times 0.02\] \[=1.25\times {{10}^{-3}}\hat{i}\,{{m}^{2}}\][Loop is parallel to Y-Z plane] Now, flux \[(\phi )=EA\cos \theta \] \[=\left( 100\hat{i}+200\hat{j} \right).1.25\times {{10}^{-3}}\hat{i}\cos {{\theta }^{o}}[\theta ={{0}^{o}}]\] \[=1.25\times {{10}^{-3}}N\,{{m}^{2}}/c\] \[=0.125N\,{{m}^{2}}/c\]
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