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(B) The torque on a current-carrying coil in a magnetic field is given by $	au = N I A B \sin 	heta$. For small oscillations, $\sin 	heta \approx \

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$20$

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(B) The torque on a current-carrying coil in a magnetic field is given by $	au = N I A B \sin 	heta$. For small oscillations, $\sin 	heta \approx 	heta$, so $	au = - (N I A B) 	heta$. Comparing this with the equation for simple harmonic motion $	au = -k 	heta$, we get the restoring torque constant $k = N I A B$. The time period of oscillation is $T = 2 \pi \sqrt{\frac{I_{moment}}{k}}$, where $I_{moment}$ is the moment of inertia. Given $T = \frac{1}{3} \, s$, $I_{moment} = 9 	imes 10^{-5} \, kg \cdot m^2$, $B = \pi 	imes 10^{-2} \, T$, $I = 2 \, A$, and $r = 9 \, cm = 0.09 \, m$. The area $A = \pi r^2 = \pi (0.09)^2 = 81 \pi 	imes 10^{-4} \, m^2$. Substituting these values: $\frac{1}{3} = 2 \pi \sqrt{\frac{9 	imes 10^{-5}}{N 	imes 2 	imes 81 \pi 	imes 10^{-4} 	imes \pi 	imes 10^{-2}}}$. Squaring both sides: $\frac{1}{9} = 4 \pi^2 \frac{9 	imes 10^{-5}}{N 	imes 162 \pi^2 	imes 10^{-6}}$. Simplifying: $\frac{1}{9} = \frac{36 \pi^2 	imes 10^{-5}}{N 	imes 162 \pi^2 	imes 10^{-6}} = \frac{36 	imes 10}{162 N} = \frac{360}{162 N}$. Thus, $N = \frac{360 	imes 9}{162} = \frac{3240}{162} = 20$. Therefore, the number of turns is $20$.

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