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A metal disc of radius $100 \mathrm{~cm}$ is rotated at a constant angular speed of $60 \mathrm{rad} / \mathrm{s}$ in a plane at right angles to an external field of magnetic induction $0.05 \mathrm{~Wb} / \mathrm{m}^{2}$. The emf induced between the centre and a point on the rim will be
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Verified Answer
The correct answer is:
$1.5 \mathrm{~V}$
Induced emf produced between the centre and a point on the disc is given by
$\begin{array}{l}
e=\frac{1}{2} \omega \mathrm{BR}^{2} \\
\omega=60 \mathrm{rad} / \mathrm{s}, \mathrm{B}=0.05 \mathrm{~Wb} / \mathrm{m}^{2} \\
\text { and } \mathrm{R}=100 \mathrm{~cm}=1 \mathrm{~m} \\
\text { We get } e=\frac{1}{2} \times 60 \times 0.05 \times(1)^{2}=1.5 \mathrm{~V}
\end{array}$
$\begin{array}{l}
e=\frac{1}{2} \omega \mathrm{BR}^{2} \\
\omega=60 \mathrm{rad} / \mathrm{s}, \mathrm{B}=0.05 \mathrm{~Wb} / \mathrm{m}^{2} \\
\text { and } \mathrm{R}=100 \mathrm{~cm}=1 \mathrm{~m} \\
\text { We get } e=\frac{1}{2} \times 60 \times 0.05 \times(1)^{2}=1.5 \mathrm{~V}
\end{array}$
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