A moving coil galvanometer has 48 turns and t | vedclass

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(C) The current sensitivity of a moving coil galvanometer is given by the formula: $S_i = \frac{NBA}{C}$ where $N$ is the number of turns,$B$ is the m

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$60$

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(C) The current sensitivity of a moving coil galvanometer is given by the formula: $S_i = \frac{NBA}{C}$ where $N$ is the number of turns,$B$ is the magnetic field,$A$ is the area,and $C$ is the torsional constant of the spring. From the formula,we see that $S_i \propto N$ when $B, A,$ and $C$ are kept constant. Let the initial sensitivity be $(S_i)_1$ and the final sensitivity be $(S_i)_2$. Given that the sensitivity increases by $25\%$,we have: $(S_i)_2 = (S_i)_1 + 0.25(S_i)_1 = 1.25(S_i)_1$ Using the proportionality $S_i \propto N$: $\frac{(S_i)_1}{(S_i)_2} = \frac{N_1}{N_2}$ $\frac{(S_i)_1}{1.25(S_i)_1} = \frac{48}{N_2}$ $\frac{1}{1.25} = \frac{48}{N_2}$ $N_2 = 48 	imes 1.25 = 60$ Therefore,the number of turns should become $60$.

$A$ moving coil galvanometer has $48$ turns and the area of the 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 the area $(A)$ and the magnetic field $(B)$,the number of turns should become:

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