A rectangular coil of effective area 0.05 m^ | vedclass

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(B) The torque on a current-carrying coil in a magnetic field is given by $	au = NIAB \sin 	heta$. For a radial magnetic field,the plane of the coil

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$30$

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(B) The torque on a current-carrying coil in a magnetic field is given by $	au = NIAB \sin 	heta$. For a radial magnetic field,the plane of the coil is always parallel to the magnetic field,so $	heta = 90^{\circ}$ and $\sin 90^{\circ} = 1$. Thus,the magnetic torque is $	au_m = NIAB$. The restoring torque provided by the suspension fiber is $	au_r = K \phi$,where $K$ is the torsional constant and $\phi$ is the angle of rotation. Equating the two torques for equilibrium: $NIAB = K \phi$. Given: $A = 0.05 \ m^2$,$B = 0.01 \ Wb/m^2$,$K = 5 	imes 10^{-9} \ Nm/	ext{degree}$,$I = 300 \ \mu A = 300 	imes 10^{-6} \ A$,and assuming $N = 1$. Substituting the values: $\phi = \frac{NIAB}{K} = \frac{1 	imes 300 	imes 10^{-6} 	imes 0.01 	imes 0.05}{5 	imes 10^{-9}}$. $\phi = \frac{300 	imes 10^{-6} 	imes 5 	imes 10^{-4}}{5 	imes 10^{-9}} = \frac{300 	imes 10^{-10}}{5 	imes 10^{-9}} = 60 	imes 10^{-1} = 6^{\circ}$. Wait,re-calculating: $\phi = \frac{300 	imes 10^{-6} 	imes 0.0005}{5 	imes 10^{-9}} = \frac{150000 	imes 10^{-12}}{5 	imes 10^{-9}} = 30000 	imes 10^{-3} = 30^{\circ}$. Therefore,the angle of rotation is $30^{\circ}$.

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