A) 1.158 V
B) 0.58 V
C) 0.29 V
D) 5.8 V
Correct Answer: B
Solution :
\[{{e}_{0}}=\omega NBA=(2\pi \nu )NB(\pi {{r}^{2}})\]\[=2\times {{\pi }^{2}}\nu \ NB{{r}^{2}}\] \[=2\times {{(3.14)}^{2}}\times \frac{1800}{60}\times 4000\times 0.5\times {{10}^{-4}}\times {{(7\times {{10}^{-2}})}^{2}}\] \[=0.58\ V\]
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