Question

In a moving coil galvanometer, two moving coils $ M_1 $ and $ M_2 $ have the following particulars: $ R_1 = 5 \, \Omega $, $ N_1 = 15 $, $ A_1 = 3.6 \times 10^{-3} \, \text{m}^2 $, $ B_1 = 0.25 \, \text{T} $ $ R_2 = 7 \, \Omega $, $ N_2 = 21 $, $ A_2 = 1.8 \times 10^{-3} \, \text{m}^2 $, $ B_2 = 0.50 \, \text{T} $ 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 $?

• A. 1 : 1
• B. 1 : 4
• C. 1 : 3
• D. 1 : 2

Hint:

In moving coil galvanometers, the voltage sensitivity is directly proportional to the number of turns, the area of the coil, and the magnetic field strength, and inversely proportional to the resistance.

Answer 1

The Correct Option is A

The voltage sensitivity is given by: \[ \frac{\theta}{V} = \frac{NAB}{cR} \] where \( N \) is the number of turns, \( A \) is the area, \( B \) is the magnetic field, \( c \) is the torsional constant, and \( R \) is the resistance. Thus, the ratio of the voltage sensitivity of \( M_1 \) and \( M_2 \) is: \[ \text{Ratio} = \frac{N_1 A_1 B_1}{N_2 A_2 B_2} = \frac{15 \times 3.6 \times 10^{-3} \times 0.25}{21 \times 1.8 \times 10^{-3} \times 0.50} = \frac{1}{1} \] Thus, the ratio of voltage sensitivity of \( M_1 \) and \( M_2 \) is 1 : 1.