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(D) The magnetic field $B$ at the centre of a rotating charged ring is given by $B = \frac{\mu_0 i}{2R}$.Here,the equivalent current $i$ is $i = \frac

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$3 	imes 10^{-5} \, C$

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(D) The magnetic field $B$ at the centre of a rotating charged ring is given by $B = \frac{\mu_0 i}{2R}$.Here,the equivalent current $i$ is $i = \frac{q}{T} = \frac{q \omega}{2 \pi}$.Substituting $i$ into the magnetic field formula: $B = \frac{\mu_0 q \omega}{2R(2 \pi)} = \frac{\mu_0 q \omega}{4 \pi R}$.Given: $R = 0.1 \, m$,$\omega = 40 \pi \, rad/s$,$B = 3.8 	imes 10^{-9} \, T$,and $\mu_0 = 4 \pi 	imes 10^{-7} \, N/A^2$.Rearranging for $q$: $q = \frac{B \cdot 4 \pi R}{\mu_0 \omega}$.$q = \frac{3.8 	imes 10^{-9} 	imes 4 \pi 	imes 0.1}{4 \pi 	imes 10^{-7} 	imes 40 \pi}$.$q = \frac{3.8 	imes 10^{-10}}{40 \pi 	imes 10^{-7}} = \frac{3.8 	imes 10^{-3}}{40 \pi} \approx \frac{3.8 	imes 10^{-3}}{125.66} \approx 3.02 	imes 10^{-5} \, C$.Thus,the charge is approximately $3 	imes 10^{-5} \, C$.

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