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Charge $q$ is uniformly spread on a thin ring of radius $\mathrm{R}$. The ring rotates about its axis with a uniform frequency $f \mathrm{~Hz}$. The magnitude of magnetic induction at the centre of the ring is
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Verified Answer
The correct answer is:
$\frac{\mu_{0} q f}{2 R}$
We knowmagnetic field
$$
\begin{array}{l}
\beta=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}} \\
\mathrm{q}=\mathrm{it} \Rightarrow \mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{qf} \\
\therefore \beta=\frac{\mu_{0} \mathrm{q} \mathrm{f}}{2 \mathrm{R}}
\end{array}
$$
$$
\begin{array}{l}
\beta=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{R}} \\
\mathrm{q}=\mathrm{it} \Rightarrow \mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}=\mathrm{qf} \\
\therefore \beta=\frac{\mu_{0} \mathrm{q} \mathrm{f}}{2 \mathrm{R}}
\end{array}
$$
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