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(B) Given that, for a moving coil galvanometer:Effective area, $A = 4 	imes 10^{-2} \, m^2$Magnetic field, $B = 5 	imes 10^{-2} \, Wb \, m^{-2}$Angl

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current sensitivity is $40 \, rad \, A^{-1}$

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(B) Given that, for a moving coil galvanometer:Effective area, $A = 4 	imes 10^{-2} \, m^2$Magnetic field, $B = 5 	imes 10^{-2} \, Wb \, m^{-2}$Angle of deflection, $\phi = 0.2 \, rad$Electric current, $I = 5 \, mA = 5 	imes 10^{-3} \, A$Assuming the number of turns $N = 1$ (as not specified, we calculate the product $NAB$):The torque balance equation is $NAB I = k \phi$, where $k$ is the torsional constant.$k = \frac{NAB I}{\phi} = \frac{(1) 	imes (4 	imes 10^{-2}) 	imes (5 	imes 10^{-2}) 	imes (5 	imes 10^{-3})}{0.2}$$k = \frac{100 	imes 10^{-7}}{0.2} = 500 	imes 10^{-7} = 5 	imes 10^{-5} \, N \, m \, rad^{-1}$.Current sensitivity $S_I = \frac{\phi}{I} = \frac{0.2}{5 	imes 10^{-3}} = 40 \, rad \, A^{-1}$.Thus, the current sensitivity is $40 \, rad \, A^{-1}$.

The coil of a moving coil galvanometer has an effective area of $4 \times 10^{-2} \, m^2$. It is suspended in a magnetic field of $5 \times 10^{-2} \, Wb \, m^{-2}$. If the deflection in the galvanometer coil is $0.2 \, rad$, when a current of $5 \, mA$ is passed through it, then

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