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A magnetic field of flux denisty $1.0 \mathrm{~Wb} \mathrm{~m}^{-2}$ acts normal to a 80 turn coil of $0.01 \mathrm{~m}^2$ area. If this coil is removed from the field in $0.2 \mathrm{~s}$, then the emf induced in it is
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Verified Answer
The correct answer is:
$4 \mathrm{~V}$
Given, magnetic flux density,
$$
B=1 \mathrm{~Wb} / \mathrm{m}^2
$$
Number of turns, $N=80$
Area of coil, $A=0.01 \mathrm{~m}^2$
$$
\Delta t=0.2 \mathrm{~s}
$$
According to Faraday's law of electromagnetic induction, induced emf
$$
\begin{aligned}
|e| & =\frac{N \Delta \phi}{\Delta t}=N \cdot \frac{B A}{\Delta t} \quad[\because \Delta \phi=B A] \\
& =\frac{80 \times 1 \times 0.01}{0.2}=4 \mathrm{~V}
\end{aligned}
$$
$$
B=1 \mathrm{~Wb} / \mathrm{m}^2
$$
Number of turns, $N=80$
Area of coil, $A=0.01 \mathrm{~m}^2$
$$
\Delta t=0.2 \mathrm{~s}
$$
According to Faraday's law of electromagnetic induction, induced emf
$$
\begin{aligned}
|e| & =\frac{N \Delta \phi}{\Delta t}=N \cdot \frac{B A}{\Delta t} \quad[\because \Delta \phi=B A] \\
& =\frac{80 \times 1 \times 0.01}{0.2}=4 \mathrm{~V}
\end{aligned}
$$
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