A) zero
B) \[\frac{2\pi rB}{R}\]
C) \[\frac{r\alpha }{R}\]
D) \[\frac{\pi {{r}^{2}}\alpha }{R}\]
Correct Answer: D
Solution :
The magnitude of the induced emf in the ring \[|\varepsilon |=\frac{d\phi }{dt}\] \[=\frac{d}{dt}(BA)\] \[=A\frac{d}{dt}B\] The induced current in the ring \[I=\frac{\varepsilon }{R}=\frac{\pi {{r}^{2}}\alpha }{R}\]You need to login to perform this action.
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