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Galvanometer

A moving coil...

Question

A moving coil galvanometer has 48 turns and area of coil is $4 \times 10^{- 2} m^{2}$ . If the magnetic field is 0.2T, then to increase the current sensitivity by 25% without changing area (A) and field (B) the number of turns should become :

A

24

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B

36

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C

60

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D

54

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Solution

The correct option is C 60
Current sensitivity $S_{1} = \frac{N B A}{C}$
$S_{1} \propto N$
$∴ \frac{S_{1}}{S_{2}} = \frac{N_{1}}{N_{2}} \Rightarrow \frac{100}{125} = \frac{48}{N_{2}}$
$N_{2} = 60$

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