A) 0.5A
B) 0.1A
C) 2A
D) 1A
Correct Answer: A
Solution :
\[\phi =50{{t}^{2}}+4\] \[\operatorname{emf}=\frac{d\phi }{dt}=100t\] \[\operatorname{emf}\left( t=2s \right)=100\times 2=200V\] \[\operatorname{I}=\frac{E}{R}=\frac{200}{400}=0.5A\]You need to login to perform this action.
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