A) 4p2 mA
B) 30 mA
C) 6 mA
D) 200 mA
Correct Answer: C
Solution :
Amplitude of the current \[{{i}_{0}}=\frac{{{e}_{0}}}{R}=\frac{\omega NBA}{R}=\frac{2\pi \nu NB(\pi {{r}^{2}})}{R}\] \[{{i}_{0}}=\frac{2\pi \times 1\times {{10}^{-2}}\times \pi {{(0.3)}^{2}}}{{{\pi }^{2}}}=6\times {{10}^{-3}}A=6mA\]You need to login to perform this action.
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