A point charge Q (= 3 times 10^{-12} , C) rot | vedclass

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(A) The net magnetic field at the centre of the loop is zero when the magnetic field due to the rotating charge is exactly equal and opposite to the h

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$10^{11}$

Vedclass · Answer

(A) The net magnetic field at the centre of the loop is zero when the magnetic field due to the rotating charge is exactly equal and opposite to the horizontal component of the earth's magnetic field $(B_H)$.Given: $Q = 3 	imes 10^{-12} \, C$,$R = 1 \, mm = 10^{-3} \, m$,$B_H = 30 \, \mu T = 30 	imes 10^{-6} \, T$.The magnetic field produced by a rotating charge at the centre of the circle is $B_q = \frac{\mu_0 i}{2R}$.Since $i = \frac{Q}{T}$ and $T = \frac{2\pi}{\omega}$,we have $i = \frac{Q\omega}{2\pi}$.Substituting $i$ into the formula for $B_q$:$B_q = \frac{\mu_0 (Q\omega / 2\pi)}{2R} = \frac{\mu_0 Q \omega}{4\pi R}$.Equating $B_q$ to $B_H$:$\frac{\mu_0 Q \omega}{4\pi R} = B_H$Using $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$:$10^{-7} 	imes \frac{3 	imes 10^{-12} 	imes \omega}{10^{-3}} = 30 	imes 10^{-6}$$3 	imes 10^{-16} 	imes \omega = 30 	imes 10^{-6} 	imes 10^{-3}$$3 	imes 10^{-16} 	imes \omega = 30 	imes 10^{-9}$$\omega = \frac{30 	imes 10^{-9}}{3 	imes 10^{-16}} = 10 	imes 10^7 = 10^8 \, rad/s$.Wait,re-calculating: $\omega = \frac{30 	imes 10^{-6} 	imes 10^{-3}}{3 	imes 10^{-12} 	imes 10^{-7}} = \frac{30 	imes 10^{-9}}{3 	imes 10^{-19}} = 10^{11} \, rad/s$.Thus,the correct option is $A$.

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