In a moving coil galvanometer, two moving coils $M_1$ and $M_2$ have the following particulars : $\begin{aligned} & \mathrm{R}_1=5 \Omega, \mathrm{~N}_1=15, \mathrm{~A}_1=3.6 imes 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…

Question

In a moving coil galvanometer, two moving coils $M_1$ and $M_2$ 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 $M_2$ ?

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Voltage sensitivity }=\frac{	heta}{V}=\frac{N A B}{c R} \ & 	ext { Ratio }==\left(\frac{\mathrm{N}_1 \mathrm{~A}_1 \mathrm{~B}_1}{\mathrm{~N}_2 \mathrm{~A}_2 \mathrm{~B}_2}ight) \frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{15 	imes 3.6 	imes 0.25}{21 	imes 1.8 	imes 0.5} 	imes \frac{7}{5}=\frac{1}{1}\end{aligned}$

In a moving coil galvanometer, two moving coils $M_1$ and $M_2$ have the following particulars :…

Solution