Question 24: Magnetism and Matter

Question

In a moving coil galvanometer, two moving coils $\mathrm{M}_1$ and $\mathrm{M}_2$ have the following particulars :
$$ \begin{aligned} & \mathrm{R}_1=5 \Omega, \mathrm{~N}_1=15, \mathrm{~A}_1=3.6 	imes 10^{-3} \mathrm{~m}^2, \mathrm{~B}_1=0.25 \mathrm{~T} \ & \mathrm{R}_2=7 \Omega, \mathrm{~N}_2=21, \mathrm{~A}_2=1.8 	imes 10^{-3} \mathrm{~m}^2, \mathrm{~B}_2=0.50 \mathrm{~T} \end{aligned} $$
Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_1$ and $M_2$ ?

Accepted Answer

In a moving coil galvanometer, the voltage sensitivity is given by the formula:
$ 	ext{Voltage sensitivity} = \frac{	heta}{V} = \frac{N \cdot A \cdot B}{c \cdot R} $
Here, $ 	heta $ 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.
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:
$ 	ext{Ratio of voltage sensitivities} = \left(\frac{N_1 \cdot A_1 \cdot B_1}{N_2 \cdot A_2 \cdot B_2}ight) \cdot \frac{R_2}{R_1} $
Plugging in the given values:
$ N_1 = 15 $
$ A_1 = 3.6 	imes 10^{-3} \, 	ext{m}^2 $
$ B_1 = 0.25 \, 	ext{T} $
$ R_1 = 5 \, \Omega $
$ N_2 = 21 $
$ A_2 = 1.8 	imes 10^{-3} \, 	ext{m}^2 $
$ B_2 = 0.50 \, 	ext{T} $
$ R_2 = 7 \, \Omega $
We calculate:
$ 	ext{Ratio} = \left(\frac{15 	imes 3.6 	imes 0.25}{21 	imes 1.8 	imes 0.5}ight) 	imes \frac{7}{5} = \frac{1}{1} $
Thus, the ratio of the voltage sensitivities of $ M_1 $ to $ M_2 $ is 1:1.

Answer

$1: 1$

Answer

$1: 3$

Answer

$1: 2$

Answer

$1: 4$

Magnetism and Matter

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