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The magnetic flux through a loop of resistance $10 \Omega$ is given by $\phi=5 t^2-4 t+1$ Weber. How much current is induced in the loop after $0.2 \mathrm{sec}$ ?
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Verified Answer
The correct answer is:
$0.2 \mathrm{~A}$
Hints : $\phi=5 t^2-4 t+1$
$$
\frac{d \phi}{d t}=10 t-4
$$
$$
\mathrm{I}=\frac{e}{\mathrm{R}}=\frac{-d \phi / d t}{\mathrm{R}}=-\frac{10 t-4}{10}
$$
At $t=0.2 \mathrm{sec}$
$$
\mathrm{I}=\frac{-(10 \times 0.2-4)}{10}=-\frac{(2-4)}{10}=+\frac{2}{10}=+0.2 \mathrm{~A}=0.2 \mathrm{~A}
$$
$$
\frac{d \phi}{d t}=10 t-4
$$
$$
\mathrm{I}=\frac{e}{\mathrm{R}}=\frac{-d \phi / d t}{\mathrm{R}}=-\frac{10 t-4}{10}
$$
At $t=0.2 \mathrm{sec}$
$$
\mathrm{I}=\frac{-(10 \times 0.2-4)}{10}=-\frac{(2-4)}{10}=+\frac{2}{10}=+0.2 \mathrm{~A}=0.2 \mathrm{~A}
$$
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