A) 0.5A
B) 0.2A
C) 0.3A
D) 0.9A
Correct Answer: A
Solution :
Emf induced between centre and circum \[\varepsilon \,=\frac{B\omega {{a}^{2}}}{2},a\to radius\] \[\therefore \,\,i=\frac{\varepsilon }{R}=\frac{B\omega {{a}^{2}}}{2R}\] \[=\frac{0.4\times 10\times {{(5\times {{10}^{-2}})}^{2}}}{2\times 10}=0.5A\]You need to login to perform this action.
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