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A magnet 10 cm long and having a pole strength of 2 amp-m is deflected through 30^o from the magnetic meridian. The horizontal component of earth's induction is 0.32 x 10^-4 T. The value of deflecting couple is

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a

32 x 10^-7 N-m

b

64 x 10^-7 N-m

c

48 x 10^-7 N-m

d

6 x 10^-7 N-m

answer is B.

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Detailed Solution

$\begin{array}{l} Here M = 10 \times 10^{- 2} \times 2 = 20 \times 10^{- 2} \\ = 0 \cdot 2 {amp-m}^{2} \\ ∴ τ = MBsin θ \\ = 0 \cdot 2 \times (0 \cdot 32 \times 10^{- 4}) \sin 30^{\circ} \\ = 0 \cdot 2 \times (0 \cdot 32 \times 10^{- 4}) \times (1 / 2) \\ = 32 \times 10^{- 7} N - m . \end{array}$

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