A moving coil galvanometer has 100 turns and | vedclass

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(C) The torque acting on the coil in a magnetic field is given by $	au = BINA \sin \phi$,where $\phi$ is the angle between the normal to the coil and

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$4$

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(C) The torque acting on the coil in a magnetic field is given by $	au = BINA \sin \phi$,where $\phi$ is the angle between the normal to the coil and the magnetic field. In a radial magnetic field,$\phi = 90^{\circ}$,so $\sin 90^{\circ} = 1$.The restoring torque provided by the suspension wire is $	au = C 	heta$,where $C$ is the torsional constant and $	heta$ is the deflection.Equating the two torques: $C 	heta = BINA$.Given values: $N = 100$,$A = 2.0 \,cm^2 = 2.0 	imes 10^{-4} \,m^2$,$B = 0.01 \,T$,$I = 10 \,mA = 10 	imes 10^{-3} \,A$,and $	heta = 0.05 \,rad$.Substituting these values into the formula for $C$:$C = \frac{BINA}{	heta} = \frac{0.01 	imes 10 	imes 10^{-3} 	imes 100 	imes 2.0 	imes 10^{-4}}{0.05}$$C = \frac{0.01 	imes 0.01 	imes 100 	imes 2.0 	imes 10^{-4}}{0.05} = \frac{2.0 	imes 10^{-6}}{0.05} = 40 	imes 10^{-6} = 4 	imes 10^{-5} \,N-m/rad$.Comparing this with $x 	imes 10^{-5} \,N-m/rad$,we get $x = 4$.

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