A bar magnet of length 10 cm and pole strength 20 A-m is deflected through 30^o from the magnetic meridian. If earth's horizontal component field is $\frac{320}{4 π} A / m$ . The moment of the deflecting couple is

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160 x10^-7Nm

320 $π$ x10⁷Nm

320 x10^-7Nm

160 $π$ x10^-7Nm

answer is C.

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Detailed Solution

$\begin{array}{rcl} B & = & μ_{0} H = 4 π x 10^{- 7} x \frac{320}{4 π} \\ τ & = & MB \sin (θ) = 20 x 10^{- 1} x 320 x 10^{- 7} x \frac{1}{2} = 320 x 10^{- 7} Nm \end{array}$

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