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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
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Verified Answer
The correct answer is:
$9.42 \mathrm{~V}$
Length of rod, $l=1.0 \mathrm{~m}$
$$
\begin{aligned}
B=0.25 \mathrm{~T}, f & =12 \mathrm{rev} / \mathrm{s} \\
\therefore \quad \text { Induced emf, } e & =\frac{B \omega l^2}{2}=\frac{0.25 \times 2 \pi f}{2} \times 1^2 \\
& =0.25 \pi f=0.25 \pi \times 12 \\
& =3 \pi=9.42 \mathrm{~V}
\end{aligned}
$$
$$
\begin{aligned}
B=0.25 \mathrm{~T}, f & =12 \mathrm{rev} / \mathrm{s} \\
\therefore \quad \text { Induced emf, } e & =\frac{B \omega l^2}{2}=\frac{0.25 \times 2 \pi f}{2} \times 1^2 \\
& =0.25 \pi f=0.25 \pi \times 12 \\
& =3 \pi=9.42 \mathrm{~V}
\end{aligned}
$$
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