In a moving coil galvanometer, two moving coils M_1 and M_2 have the following particulars : aligned & R_1=5 , ~N_1=15, ~A_1=3.6 × 10^ -3 ~m^2, ~B_1=0.25 ~T & R_2=7 , ~N_2=21, ~A_2=1.8 × 10^ -3 ~m^2, ~B_2=0.50 ~T

Correct answer: A. 1: 1 aligned & Voltage sensitivity =(θ)/(V)=(N A B)/(c R) & Ratio ==( N_1 A_1 B_1 N_2 A_2 B_2 ) R_2 R_1 =(15 × 3.6 × 0.25)/(21 × 1.8 × 0.5) × (7)/(5)=(1)/(1) aligned

JEE Main 20252 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Evening Shift
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Main 2025 · 2 April · Evening Shift · Physics · Magnetic Effects of Current

In a moving coil galvanometer, two moving coils M_1 and M_2 have...

In a moving coil galvanometer, two moving coils

M_1 and M_2

have the following particulars :

\begin{aligned}<br/>& \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}<br/>\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

Options

A $1: 1$
B $1: 4$
C $1: 3$
D $1: 2$

Correct answer

A. $1: 1$

Step-by-step solution

\begin{aligned} & \text{ Voltage sensitivity }=\frac{\theta}{V}=\frac{N A B}{c R} \\ & \text{ Ratio }==\left(\frac{\mathrm{N}_1 \mathrm{ A}_1 \mathrm{ B}_1}{\mathrm{ N}_2 \mathrm{ A}_2 \mathrm{ B}_2}\right) \frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5} \times \frac{7}{5}=\frac{1}{1}\end{aligned}

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