Q108 / 180 · AIIMS 2001 · Physics
A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in $0.4 \mathrm{sec}$. If magnetic flux between the pole pieces is known to be $8 \times 10^{-4} \mathrm{~Wb}$, then induced emf in the wire, is
- A$4 \times 10^{-3} \mathrm{~V}$
- B$8 \times 10^{-3} \mathrm{~V}$
- Correct answerC$2 \times 10^{-3} \mathrm{~V}$
- D$6 \times 10^{-3} \mathrm{~V}$.
Solution
The charge of flux through the metal wire
Time taken
$$
\begin{aligned}
\Delta \phi & =8 \times 10^{-4} \mathrm{~Wb} \\
\Delta t & =0.4 \mathrm{sec}
\end{aligned}
$$
$\therefore$ Induced emf.
$$
\begin{aligned}
e & =\frac{\Delta \phi}{\Delta t} \\
& =\frac{8 \times 10^{-4}}{0.4} \\
& =20 \times 10^{-3} \mathrm{~V}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.