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Charge $q$ is uniformly spread on a thin ring of radius $R$. 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 know
$$
\begin{array}{ll}
B=\frac{\mu_0 i}{2 R} \\
q=i t \Rightarrow i=\frac{q}{t}=q f \quad\left[\because f=\frac{1}{t}\right] \\
B=\frac{\mu_0 q f}{2 R}
\end{array}
$$
$$
\begin{array}{ll}
B=\frac{\mu_0 i}{2 R} \\
q=i t \Rightarrow i=\frac{q}{t}=q f \quad\left[\because f=\frac{1}{t}\right] \\
B=\frac{\mu_0 q f}{2 R}
\end{array}
$$
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