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In an iron bar (5cm×1cm×1cm) magnetic moment of am atom is 1.8×10^–23 Am². What is the torque acting on the magnet when it si normal to magnetic field 15000G (density of iron 7.8g/cc, atomic mass 56)

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10.3 Nm

12.3 Nm

9.3 Nm

11.3 Nm

answer is A.

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Detailed Solution

$τ = MBsin θ =$
$\begin{array}{l} \frac{6 .023 \times 10^{23} \times 5 \times 10^{- 6} \times 7 .8 \times 10^{3} \times 1 .8 \times 10^{- 23} \times 1500 \times 10^{- 4}}{56 \times 10^{- 3}} \\ = 11 .3 Nm \end{array}$

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