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A magnetic needle of pole strength $20 \sqrt{3}$ Am is pivoted at its centre. Its N-pole is pulled eastward by a string. The horizontal force required to produce a deflection of 30° from magnetic meridian (take B_H = 10^–4T) is

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$4 \times 10^{- 3} N$

$2 \times 10^{- 3} N$

$4 \sqrt{3} \times 10^{- 3} N$

$\frac{2}{\sqrt{3}} \times 10^{- 3} N$

answer is A.

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Detailed Solution

$\begin{array}{l} F = 2 {mB}_{H} Tan θ \\ = 2 \times 20 \sqrt{3} \times 10^{- 4} \times \frac{1}{\sqrt{3}} = 4 \times 10^{- 3} N \end{array}$

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