A thin ring of $10\, cm$ radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of $40\,\pi \,rad\,{s^{ - 1}}$ about its axis, perpendicular to its plane. If the magnetic field at its centre is $3.8 \times {10^{ - 9}}\,T$, then the charge carried by the ring is close to $\left( {{\mu _0} = 4\pi \times {{10}^{ - 7}}\,N/{A^2}} \right)$
JEE MAIN 2019, Medium
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$B=\frac{\mu_{0} i}{2 R}=\frac{\mu_{0} q \omega}{2 R 2 \pi}$
$\Rightarrow q=3 \times 10^{-5}\, \mathrm{C}$
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