In a moving coil galvanometer, two moving coils $\mathrm{M}_1$ and $\mathrm{M}_2$ have the following particulars :

$$ \begin{aligned} & \mathrm{R}_1=5 \Omega, \mathrm{~N}_1=15, \mathrm{~A}_1=3.6 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_1=0.25 \mathrm{~T} \\ & \mathrm{R}_2=7 \Omega, \mathrm{~N}_2=21, \mathrm{~A}_2=1.8 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_2=0.50 \mathrm{~T} \end{aligned} $$

Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_1$ and $M_2$ ?

<p>In a moving coil galvanometer, the voltage sensitivity is given by the formula:</p> <p>$ \text{Voltage sensitivity} = \frac{\theta}{V} = \frac{N \cdot A \cdot B}{c \cdot R} $</p> <p>Here, $ \theta $ is the deflection angle, $ V $ is the voltage, $ N $ is the number of turns, $ A $ is the coil area, $ B $ is the magnetic field, $ c $ is the torsional constant of the spring, and $ R $ is the resistance of the coil.</p> <p>For the two moving coils $ \mathrm{M}_1 $ and $ \mathrm{M}_2 $, we want to determine the ratio of their voltage sensitivities. Given that the torsional constant $ c $ is the same for both coils, we can express the ratio as:</p> <p>$ \text{Ratio of voltage sensitivities} = \left(\frac{N_1 \cdot A_1 \cdot B_1}{N_2 \cdot A_2 \cdot B_2}\right) \cdot \frac{R_2}{R_1} $</p> <p>Plugging in the given values:</p> <p><p>$ N_1 = 15 $</p></p> <p><p>$ A_1 = 3.6 \times 10^{-3} \, \text{m}^2 $</p></p> <p><p>$ B_1 = 0.25 \, \text{T} $</p></p> <p><p>$ R_1 = 5 \, \Omega $</p></p> <p><p>$ N_2 = 21 $</p></p> <p><p>$ A_2 = 1.8 \times 10^{-3} \, \text{m}^2 $</p></p> <p><p>$ B_2 = 0.50 \, \text{T} $</p></p> <p><p>$ R_2 = 7 \, \Omega $</p></p> <p>We calculate:</p> <p>$ \text{Ratio} = \left(\frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5}\right) \times \frac{7}{5} = \frac{1}{1} $</p> <p>Thus, the ratio of the voltage sensitivities of $ M_1 $ to $ M_2 $ is 1:1.</p>