Two moving coil meters,$M_{1}$ and $M_{2}$ have the following particulars:
$R_{1}=10 \,\Omega, \quad N_{1}=30$
$A_{1}=3.6 \times 10^{-3} \,m^{2}, \quad B_{1}=0.25 \,T$
$R_{2}=14 \,\Omega, \quad N_{2}=42$
$A_{2}=1.8 \times 10^{-3} \,m^{2}, \quad B_{2}=0.50 \,T$
(The spring constants are identical for the two meters). Determine the ratio of
$(a)$ current sensitivity and
$(b)$ voltage sensitivity of $M_{2}$ and $M_{1}$.

(A) For moving coil meter $M_{1}:$
Resistance,$R_{1}=10 \,\Omega, \quad N_{1}=30, \quad A_{1}=3.6 \times 10^{-3} \,m^{2}, \quad B_{1}=0.25 \,T, \quad K_{1}=K$
For moving coil meter $M_{2}:$
Resistance,$R_{2}=14 \,\Omega, \quad N_{2}=42, \quad A_{2}=1.8 \times 10^{-3} \,m^{2}, \quad B_{2}=0.50 \,T, \quad K_{2}=K$
$(a)$ Current sensitivity $I_{s} = \frac{NBA}{K}$.
Ratio of current sensitivity $\frac{I_{s2}}{I_{s1}} = \frac{N_{2} B_{2} A_{2}}{K_{2}} \times \frac{K_{1}}{N_{1} B_{1} A_{1}} = \frac{42 \times 0.50 \times 1.8 \times 10^{-3}}{30 \times 0.25 \times 3.6 \times 10^{-3}} = \frac{37.8}{27} = 1.4$.
$(b)$ Voltage sensitivity $V_{s} = \frac{I_{s}}{R} = \frac{NBA}{KR}$.
Ratio of voltage sensitivity $\frac{V_{s2}}{V_{s1}} = \frac{I_{s2}}{R_{2}} \times \frac{R_{1}}{I_{s1}} = \frac{I_{s2}}{I_{s1}} \times \frac{R_{1}}{R_{2}} = 1.4 \times \frac{10}{14} = 1.4 \times \frac{5}{7} = 1$.