A) 2.28 V
B) 4.28 V
C) 6.28 V
D) 2.5 V
Correct Answer: C
Solution :
\[e=\frac{1}{2}B{{l}^{2}}\omega =B{{l}^{2}}\pi \nu \] \[\Rightarrow e=0.5{{(20\times {{10}^{-2}})}^{2}}\times 3.14\times 100=6.28\ V\]
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