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The area of the coil in a moving coil galvanometer is $16 c m^{2}$ and has 20 turns. The magnetic induction is 0.2 T and the couple per unit twist of the suspended wire is $10^{- 6}$ Nm per degree. If the deflection is $45^{0}$ , the current passing through it is $(\times 10^{- 3} A)$

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answer is A.

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Detailed Solution

$C θ = B I A N \Rightarrow I = \frac{C θ}{B A N} \Rightarrow I = \frac{10^{- 6} \times 45}{0.2 \times 16 \times 10^{- 4} \times 20}$

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