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A thin ring of radius ' $R$ ' meter has charge ' $q$ ' coulomb uniformly spread on it. The ring rotates about its axis with a constant frequency of $f$ revolution/s. The value of magnetic induction in $\mathrm{Wb} \mathrm{m}{ }^{-2}$ at the center of the ring is ( $\mu_0=$ Permeability of free space)
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Verified Answer
The correct answer is:
$\frac{\mu_0 \mathrm{qf}}{2 \mathrm{R}}$
For circular coil of radius $\mathrm{R}$ carrying a current $\mathrm{I}$, the magnetic field at the center is given by
$$
B=\frac{\mu_0 I}{2 R}
$$
Here $\mathrm{I}=\mathrm{qf}$
$$
\therefore \mathrm{B}=\frac{\mu_0 \mathrm{qf}}{2 \mathrm{R}}
$$
$$
B=\frac{\mu_0 I}{2 R}
$$
Here $\mathrm{I}=\mathrm{qf}$
$$
\therefore \mathrm{B}=\frac{\mu_0 \mathrm{qf}}{2 \mathrm{R}}
$$
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