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A thin semi-circular conducting ring of radius R is falling with its plane verticle in horizontal magnetic induction `(vec B)`. At the position MNQ the speed of the ring is v, and the potential difference developed across the ring is A. zeroB. `Bv pi R^(2)//2` and M is higher potencialC. `pi RBv ans Q is at higher potentialD. 2RBv ans Q is at higher potential |
Answer» Correct Answer - D Induced emf produced across MNQ will be same as the induced emf produced in straight wire MQ. `:. E=Bvl=Bvxx2R` with Q at higher potential. |
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