Q136 / 160 · AP EAMCET 2020 (22 Sep Shift 2) · Physics
A rod of length $1.0 \mathrm{~m}$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 \mathrm{~T}$ with a frequency of $12 \mathrm{rev} / \mathrm{s}$. The induced emf across the ends to the rod is
- A$18.89 \mathrm{~V}$
- B$3 \mathrm{~V}$
- C$15 \mathrm{~V}$
- Correct answerD$9.42 \mathrm{~V}$
Solution
Length of rod, $l=1.0 \mathrm{~m}$
$$
\begin{aligned}
B=0.25 \mathrm{~T}, f & =12 \mathrm{rev} / \mathrm{s} \\
\therefore \quad \text { Induced emf, } e & =\frac{B \omega l^2}{2}=\frac{0.25 \times 2 \pi f}{2} \times 1^2 \\
& =0.25 \pi f=0.25 \pi \times 12 \\
& =3 \pi=9.42 \mathrm{~V}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.