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A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in $0.5$ second. The magnetic flux between the pole pieces is known to be $8 \times 10^{-4} \mathrm{~Wb}$. The emf induced in the wire is
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$1.6 \mathrm{mV}$
Induced emf $\mathrm{e}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}$. Assuming, small change in flux $\mathrm{d} \phi=8 \times 10^{-4} \mathrm{~Wb}$ change in time $\mathrm{dt}=0.5 \mathrm{~s}$
$\begin{aligned}
|\mathrm{e}| &=\frac{8 \times 10^{-4}}{0.5}=\frac{80 \times 10^{-4}}{5} \\
&=16 \times 10^{-4}=1.6 \times 10^{-3} \mathrm{~V}=1.6 \mathrm{mV}
\end{aligned}$
$\begin{aligned}
|\mathrm{e}| &=\frac{8 \times 10^{-4}}{0.5}=\frac{80 \times 10^{-4}}{5} \\
&=16 \times 10^{-4}=1.6 \times 10^{-3} \mathrm{~V}=1.6 \mathrm{mV}
\end{aligned}$
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