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A 16cm² coil has 20 turns. Its suspended by a phosphor bronze wire of 10^–6 Nm per degree as couple per unit twist. When ‘i’ ampere is passing through it, in a 0.2T field, the wire twisted by 45°. The value of ‘i’ is?

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8 × 10^–3 A

7 × 10^–3 A

7 × 10^–2 A

answer is A.

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Detailed Solution

$\begin{array}{l} niAB = Cθ; 10^{- 6} \times 45 = 0 .2 \times i \times 16 \times 10^{- 4} \times 20 \\ i = 7 \times 10^{- 3} A \end{array}$

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