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A coil having effective area ' $\mathrm{A}$ ' is held with its plane normal to a magnitude field of induction ' $\mathrm{B}$ '. The magnetic induction is quickly reduced to $25 \%$ of its initial value in 1 second. The e.m.f. induced in the coil (in volt) will be
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Verified Answer
The correct answer is:
$\frac{3 \mathrm{BA}}{4}$
The formula for induced emf is $\mathrm{e}=\frac{\Delta \phi}{\Delta t}$, where $\phi=\mathrm{BA}$
Here, the area is constant and the magnetic field is changing.
$\begin{aligned}
& \therefore \quad \Delta \phi=\Delta B A \\
& \therefore \quad \Delta \phi=A \cdot \Delta B \\
& \therefore \quad \Delta B=B_1-B_2 \\
& \quad B_1=B \text { and } B_2=\frac{25}{200} B=\frac{1}{4} B \\
& \therefore \quad B=B-\frac{1}{4} B \\
& \therefore \quad B=\frac{3}{4} B
\end{aligned}$
Substituting the values,
$\begin{aligned}
\mathrm{e} & =\frac{\Delta \phi}{\Delta \mathrm{t}} \\
\mathrm{e} & =\frac{\mathrm{A} \times \frac{3}{4} \mathrm{~B}}{1} \\
\therefore \quad \mathrm{e} & =\frac{3}{4} \mathrm{AB}
\end{aligned}$
Here, the area is constant and the magnetic field is changing.
$\begin{aligned}
& \therefore \quad \Delta \phi=\Delta B A \\
& \therefore \quad \Delta \phi=A \cdot \Delta B \\
& \therefore \quad \Delta B=B_1-B_2 \\
& \quad B_1=B \text { and } B_2=\frac{25}{200} B=\frac{1}{4} B \\
& \therefore \quad B=B-\frac{1}{4} B \\
& \therefore \quad B=\frac{3}{4} B
\end{aligned}$
Substituting the values,
$\begin{aligned}
\mathrm{e} & =\frac{\Delta \phi}{\Delta \mathrm{t}} \\
\mathrm{e} & =\frac{\mathrm{A} \times \frac{3}{4} \mathrm{~B}}{1} \\
\therefore \quad \mathrm{e} & =\frac{3}{4} \mathrm{AB}
\end{aligned}$
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