A moving coil galvanometer has $48$ $turns$ and area of coil is $4 \times {10^{ - 2}}\,{m^2}.$ If the magnetic field is $0.2\, T$, then to increase the current sensitivity by $25\%$ without changing area $(A)$ and field $(B)$ the number of turns should become
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(c) As we know
Current sensitivity ${S_i} = \frac{{NBA}}{C}$
$==>$ ${S_i} \propto N$ $==>$ $\frac{{{{({S_i})}_1}}}{{{{({S_i})}_2}}} = \frac{{{N_1}}}{{{N_2}}}$ $==>$ $\frac{{100}}{{125}} = \frac{{48}}{{{N_2}}}$
Current sensitivity ${S_i} = \frac{{NBA}}{C}$
$==>$ ${S_i} \propto N$ $==>$ $\frac{{{{({S_i})}_1}}}{{{{({S_i})}_2}}} = \frac{{{N_1}}}{{{N_2}}}$ $==>$ $\frac{{100}}{{125}} = \frac{{48}}{{{N_2}}}$
$==>$ ${N_2} = 60$.
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