In a moving coil galvanometer, two moving coils M 1 and M 2 have the following particulars : R 1 =5 Ω, N 1 =15, A 1 =3.6 × 10 -3 m 2 , B 1 =0.25 T R 2 =7 Ω, N 2 =21, A 2 =1.8 × 10 -3 m 2 , B 2 =0.50 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 ?

Physics · Current Electricity

JEE Main 2025 — 2 April, Evening Shift — Question 61

In a moving coil galvanometer, two moving coils $M_{1}$ and $M_{2}$ have the following particulars :

$R_{1}=5 \Omega, N_{1}=15, A_{1}=3.6 \times 10^{-3} \mathrm{~m}^{2}, B_{1}=0.25 \mathrm{~T}$

$R_{2}=7 \Omega, N_{2}=21, A_{2}=1.8 \times 10^{-3} \mathrm{~m}^{2}, B_{2}=0.50 \mathrm{~T}$

Assuming that torsional constant of the springs are same for both coils, what will be the ratio

of voltage sensitivity of $M_{1}$ and $M_{2}$ ?

$K \theta=N I A B$

$\frac{\theta}{l}=\frac{N A B}{K}$

$\frac{\theta}{V}=\frac{\theta}{R I}=\frac{N A B}{K R}$

Ratio of voltage sensitivity $=\left(\frac{N_{1} A_{1} B_{1}}{R_{1}}\right)\left(\frac{R_{2}}{N_{2} A_{2} B_{2}}\right)$

$=\frac{5}{7} \times \frac{2}{1} \times \frac{1}{2} \times \frac{7}{5}=1$

Answer key and solution verified before publishing.

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