A) X, Aa
B) X pR2a
C) Y, pA2a
D) Y, pR2a
Correct Answer: A
Solution :
Since the current is increasing, so inward magnetic flux linked with the ring also increasing (as viewed from left side). Hence induced current in the ring is anticlockwise, so end x will be positive. Induced emf \[|e|=A\frac{dB}{dt}=A\frac{d}{dt}({{B}_{o}}+\alpha t)\]\[\Rightarrow |e|=A\alpha \]You need to login to perform this action.
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