# Two moving coil meters, $M_1$ and $M_2$ have the following particulars: $R_1 = 10 \Omega , N_1 = 30,\\\\ A_1 = 3.6 \times 10^{-3} m^2, B_1 = 0.25 T\\\\ R_2 = 14 \Omega , N_2 = 42,\\\\ A_2 = 1.8 \times 10^{-3} m^2, B_2 = 0.50 T$ (The spring constants are identical for the two meters). Determine the ratio of  current sensitivity of $M_2\ and \ \ M_1$

The torque experienced by the moving coil M1 for a current I passing through it will be equal to $\tau =B_{1}A_{1}N_{1}I$

The coil will experience a restoring torque proportional to the twist $\phi$

$\phi k=B_{1}A_{1}N_{1}I$

The current sensitivity is therefore $\frac{B_{1}A_{1}N_{1}}{k}$

Similarly, for the coil M2, current sensitivity is $\frac{B_{2}A_{2}N_{2}}{k}$

Their ratio of current sensitivity of coil M2 to that of coil M1 is, therefore,  $\frac{B_{2}A_{2}N_{2}}{B_{1}A_{1}N_{1}}$

$=\frac{0.5\times 1.8\times 10^{-3}\times 42 }{0.25\times 3.6\times 10^{-3}\times 30}=1.4$

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