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A magnetic field of $2 \times 10^{-2} \mathrm{~T}$ acts at right angles to a coil of area $100 \mathrm{~cm}^2$ with 50 turns. The average e.m.f. induced in the coil is $0.1 \mathrm{~V}$, when it is removed from the field in time ' $t$ '. The value of ' $t$ ' is
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Verified Answer
The correct answer is:
$0.1 \mathrm{~s}$
$$
\begin{array}{ll}
& \mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\left(\phi_2-\phi_1\right)}{\mathrm{t}}=-\frac{(0-\mathrm{NBA})}{\mathrm{t}} \\
\therefore \quad & 0.1=\frac{50 \times 2 \times 10^{-2} \times 10^{-2}}{\mathrm{t}}
\end{array}
$$
$\therefore \quad$ The value of ' $t$ ' is,
$$
\mathrm{t}=\frac{10^{-2}}{0.1}=0.1 \mathrm{~s}
$$
\begin{array}{ll}
& \mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\left(\phi_2-\phi_1\right)}{\mathrm{t}}=-\frac{(0-\mathrm{NBA})}{\mathrm{t}} \\
\therefore \quad & 0.1=\frac{50 \times 2 \times 10^{-2} \times 10^{-2}}{\mathrm{t}}
\end{array}
$$
$\therefore \quad$ The value of ' $t$ ' is,
$$
\mathrm{t}=\frac{10^{-2}}{0.1}=0.1 \mathrm{~s}
$$
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