Moving coil galvanometers $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ have resistance, number of turns, area of…
Moving coil galvanometers $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ have resistance, number of turns, area of
coil and magnetic field as follows.
$\mathrm{R}_{1}=10 \Omega, \mathrm{R}_{2}=14 \Omega, \mathrm{N}_{1}=30, \mathrm{~N}_{2}=42$
$\mathrm{A}_{1}=3 \cdot 6 \times 10^{-3} \mathrm{~m}^{2}, \mathrm{~A}_{2}=1 \cdot 8 \times 10^{-2} \mathrm{~m}^{2}, \mathrm{~B}_{1}=0 \cdot 25 \mathrm{~T}, \mathrm{~B}_{2}=0 \cdot 50 \mathrm{~T}$
(Spring constants are same for both materials)
The ratio of (i) current sensitivity and (ii) voltage sensitivity for galvanometer
$\left(\mathrm{M}_{2}\right.$ to $\left.\mathrm{M}_{1}\right)$ is respectively
$1: 1, \quad 1 \cdot 4: 1$
$1: 1 \cdot 4$, $1: 1$
$4: 1$, $1: 1$
$1 \cdot 4: 1$, $1: 1$
Solution
(a)
Current sensitivity of \(M_1\) is given as,
\(\begin{aligned}
& I_{51}=N_1 B_1 A_1 / k_1 \\
& I_{52}=N_2 B_2 A_2 / k_2
\end{aligned}\)
where,
\(\mathrm{N}=\) turns,
\(\mathrm{B}=\) Magnetic field,
\(\mathrm{k}=\) spring constant,
\(\mathrm{A}=\) area
\(\mathrm{R}=\) Electric resistance.
\(\mathrm{I}_{51} / \mathrm{I}_{52}=1.4\)
(b)
Voltage sensitivity relation is given as,
\(\begin{aligned}
& V_{s 1}=N_1 B_1 A_1 / k_1 R_1 \\
& V_{s 2}=N_2 B_2 A_2 / k_2 R_2
\end{aligned}\)
where,
\(\mathrm{N}=\) turns,
\(\mathrm{B}=\) Magnetic field,
\(\mathrm{k}=\) spring constant,
\(\mathrm{A}=\) area
\(\mathrm{R}=\) Electric resistance.
\(V_{s 1} N_{52}=1\)