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(D) The torque on the coil in a magnetic field is given by $	au = N I A B \sin(	heta)$. Since the magnetic field is parallel to the plane of the coi

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$10^{-2}$

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(D) The torque on the coil in a magnetic field is given by $	au = N I A B \sin(	heta)$. Since the magnetic field is parallel to the plane of the coil,the angle between the normal to the coil and the magnetic field is $90^{\circ}$,so $\sin(90^{\circ}) = 1$.The restoring torque provided by the torsion band is $	au = C \phi$,where $C = 10^{-6} \, N \cdot m/rad$ and $\phi = 10^{\circ} = 10 	imes \frac{\pi}{180} \, rad$.Equating the torques: $C \phi = N I A B$.Given values: $N = 175$,$I = 1 \, mA = 10^{-3} \, A$,$A = 1 \, cm^2 = 10^{-4} \, m^2$,$C = 10^{-6} \, N \cdot m/rad$,and $\phi = \frac{\pi}{18} \, rad$.Substituting the values: $10^{-6} 	imes \frac{\pi}{18} = 175 	imes 10^{-3} 	imes 10^{-4} 	imes B$.$B = \frac{10^{-6} 	imes \pi}{18 	imes 175 	imes 10^{-7}} = \frac{10 	imes \pi}{18 	imes 175} \approx \frac{31.4}{3150} \approx 0.0099 \approx 10^{-2} \, T$.

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