Q132 / 200 · NEET 2022 (Phase 2) · Physics
The magnetic flux linked to a circular coil of radius $R$ is:
$\phi=2 t^3+4 t^2+2 t+5 \mathrm{~Wb}$
The magnitude of induced emf in the coil at $t=5 \mathrm{~s}$ is:
- A$108 \mathrm{~V}$
- B$197 \mathrm{~V}$
- C$150 \mathrm{~V}$
- Correct answerD$192 \mathrm{~V}$
Solution
Given:
$\phi=2 t^3+4 t^2+2 t+5,$
& $t=5 \mathrm{~s}$
By Faraday's law of EMI, we have,
$\mathrm{emf}=-\frac{d \phi}{d t}$
(where, the symbols have their usual meanings)
$\begin{aligned}
e=\left|\frac{d \phi}{d t}\right| & =\left|\frac{d}{d t}\left(2 t^3+4 t^2+2 t+5\right)\right| \\
& =6 t^2+8 t+2
\end{aligned}$
$\begin{aligned}
ext{At } t & =5 s \\
e & =6(5)^2+8(5)+2 \\
& =192 \mathrm{~V}
\end{aligned}$
Hence, the correct answer is option: 'D'.