A) \[\frac{Bvr}{2}\]
B) \[\frac{3Bvr}{2}\]
C) \[\frac{Bvr}{2}\]
D) \[\frac{-3Bvr}{2}\]
Correct Answer: C
Solution :
[c] Considering pure rolling of OA about A: the induced emf across OA will be: \[\left| \left. {\vec{e}} \right| \right.=\frac{B\omega {{(r)}^{2}}}{2}\] From lenz law, 0 will be the negative end, while A will be the positive end. Hence \[{{V}_{0}}-{{V}_{A}}=-\frac{B\omega {{r}^{2}}}{2}\]You need to login to perform this action.
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