A) 0.4 A
B) 0.2 A
C) 0.04 A
D) 0.02 A
Correct Answer: B
Solution :
The flux is given by \[\text{o }\!\!|\!\!\text{ =5}{{\text{t}}^{2}}-4t+1.\] The emf induced in the resistance would be equal to \[e=-\frac{d\text{o }\!\!|\!\!\text{ }}{dt}=10t-4.\] Thus, the current induced in the loop is equal to \[i=\frac{e}{R}=-\left( \frac{10t-4}{10} \right)\]at \[t=0.2s,\]we have \[i=-\left( \frac{10\times 0.2-4}{10} \right)=\frac{2}{10}=0.2A\]You need to login to perform this action.
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