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A metal rod of length $1 \mathrm{~m}$ is rotated about one of its ends in a plane right angles to a field of inductance $2.5 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}$. If it makes 1800 revolutions/min. Calculate induced e.m.f. between its ends.
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Verified Answer
The correct answer is:
$0.471 \mathrm{~V}$
Given : $\ell=1 \mathrm{~m}, \mathrm{~B}=5 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}$
$$
\mathrm{f}=\frac{1800}{60}=30 \text { rotations } / \mathrm{sec}
$$
In one rotation, the moving rod of the metal traces a circle of radius $\mathrm{r}=\ell$
$\therefore$ Area swept in one rotation $=\pi r^{2}$
$$
\begin{array}{l}
\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})=\mathrm{B} \cdot \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\mathrm{B} \pi \mathrm{r}^{2}}{\mathrm{~T}} \\
=\mathrm{B} \mathrm{f} \pi \mathrm{r}^{2}=\left(5 \times 10^{-3}\right) \times 3.14 \times 30 \times 1=0.471 \mathrm{~V}
\end{array}
$$
$$
\mathrm{f}=\frac{1800}{60}=30 \text { rotations } / \mathrm{sec}
$$
In one rotation, the moving rod of the metal traces a circle of radius $\mathrm{r}=\ell$
$\therefore$ Area swept in one rotation $=\pi r^{2}$
$$
\begin{array}{l}
\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{BA})=\mathrm{B} \cdot \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\frac{\mathrm{B} \pi \mathrm{r}^{2}}{\mathrm{~T}} \\
=\mathrm{B} \mathrm{f} \pi \mathrm{r}^{2}=\left(5 \times 10^{-3}\right) \times 3.14 \times 30 \times 1=0.471 \mathrm{~V}
\end{array}
$$
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