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(C) The magnetic dipole moment $m$ is given by the product of magnetization $M$ and the volume $V$ of the rod.$m = M 	imes V$Given:Magnetization $M =

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$2 	imes 10^{-2} \ J/T$

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(C) The magnetic dipole moment $m$ is given by the product of magnetization $M$ and the volume $V$ of the rod.$m = M 	imes V$Given:Magnetization $M = 5.3 	imes 10^{3} \ A/m$Length $l = 5 \ cm = 5 	imes 10^{-2} \ m$Diameter $d = 1 \ cm = 1 	imes 10^{-2} \ m$,so radius $r = 0.5 	imes 10^{-2} \ m$Volume $V = \pi r^2 l = \frac{22}{7} 	imes (0.5 	imes 10^{-2})^2 	imes (5 	imes 10^{-2})$$V = \frac{22}{7} 	imes 0.25 	imes 10^{-4} 	imes 5 	imes 10^{-2} \approx 3.14 	imes 1.25 	imes 10^{-6} \approx 3.925 	imes 10^{-6} \ m^3$Now,$m = (5.3 	imes 10^3) 	imes (3.925 	imes 10^{-6}) \approx 20.8 	imes 10^{-3} \approx 2.08 	imes 10^{-2} \ J/T$Rounding to the nearest option,the value is $2 	imes 10^{-2} \ J/T$.

$A$ cylindrical magnetic rod has length $5 \ cm$ and diameter $1 \ cm$. It has uniform magnetization $M = 5.3 \times 10^{3} \ A/m$. Its net magnetic dipole moment is nearly (take $\pi = 22/7$):

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