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A coil of circular cross-section having 1000 turns and $4 \mathrm{~cm}^2$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{-2} \mathrm{~Wb}$ $\mathrm{m}^{-2}$ in $0.01 \mathrm{~s}$. The e.m.f. induced in the coil is:
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Verified Answer
The correct answer is:
$400 \mathrm{mV}$
$400 \mathrm{mV}$
Given: No. of turns $N=1000$
Face area, $A=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2$
Change in magnetic field,
$$
\Delta \mathrm{B}=10^{-2} \mathrm{wbm}^{-2}
$$
Time taken, $t=0.01 \mathrm{~s}=10^{-2} \mathrm{sec}$
Emf induced in the coil $e=$ ?
Applying formula,
Induced emf, $e=\frac{-d \phi}{d t}$
$$
\begin{aligned}
&=N\left(\frac{\Delta B}{\Delta t}\right) A \cos \theta \\
&=\frac{1000 \times 10^{-2} \times 4 \times 10^{-4}}{10^{-2}}=400 \mathrm{mV}
\end{aligned}
$$
Face area, $A=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2$
Change in magnetic field,
$$
\Delta \mathrm{B}=10^{-2} \mathrm{wbm}^{-2}
$$
Time taken, $t=0.01 \mathrm{~s}=10^{-2} \mathrm{sec}$
Emf induced in the coil $e=$ ?
Applying formula,
Induced emf, $e=\frac{-d \phi}{d t}$
$$
\begin{aligned}
&=N\left(\frac{\Delta B}{\Delta t}\right) A \cos \theta \\
&=\frac{1000 \times 10^{-2} \times 4 \times 10^{-4}}{10^{-2}}=400 \mathrm{mV}
\end{aligned}
$$
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