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(D) The volume of the magnet $V = 	ext{length} 	imes 	ext{area} = 5 \ cm 	imes 4 \ cm^{2} = 20 \ cm^{3}$.Converting to $SI$ units: $V = 20 	imes

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$2 	imes 10^{10} \ A/m$

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(D) The volume of the magnet $V = 	ext{length} 	imes 	ext{area} = 5 \ cm 	imes 4 \ cm^{2} = 20 \ cm^{3}$.Converting to $SI$ units: $V = 20 	imes 10^{-6} \ m^{3} = 2 	imes 10^{-5} \ m^{3}$.The magnetization $M$ is defined as the magnetic moment per unit volume: $M = \frac{m}{V} = \frac{2 \ Am^{2}}{2 	imes 10^{-5} \ m^{3}} = 10^{5} \ A/m$.The relation between magnetization $M$,magnetic intensity $H$,and magnetic susceptibility $\chi$ is $M = \chi H$.Therefore,$H = \frac{M}{\chi} = \frac{10^{5}}{5 	imes 10^{-6}} = 0.2 	imes 10^{11} \ A/m = 2 	imes 10^{10} \ A/m$.

$A$ bar magnet having length $5 \ cm$ and area of cross-section $4 \ cm^{2}$ has a magnetic moment of $2 \ Am^{2}$. If the magnetic susceptibility is $5 \times 10^{-6}$,the magnetic intensity will be:

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