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(D) The magnetization $M$ is defined as the magnetic dipole moment per unit volume,$M = \frac{m_{net}}{V}$.Therefore,the net magnetic dipole moment is

Vedclass · Accepted Answer

$2.08 	imes 10^{-2} \ J/T$

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(D) The magnetization $M$ is defined as the magnetic dipole moment per unit volume,$M = \frac{m_{net}}{V}$.Therefore,the net magnetic dipole moment is $m_{net} = M 	imes V$.The volume $V$ of a cylindrical rod is given by $V = \pi r^2 l = \pi \left(\frac{d}{2}ight)^2 l$.Given: $M = 5.3 	imes 10^3 \ A/m$,$l = 5 \ cm = 0.05 \ m$,$d = 1 \ cm = 0.01 \ m$.Radius $r = \frac{d}{2} = 0.005 \ m$.$V = 3.142 	imes (0.005)^2 	imes 0.05 = 3.142 	imes 25 	imes 10^{-6} 	imes 0.05 = 3.9275 	imes 10^{-6} \ m^3$.$m_{net} = (5.3 	imes 10^3) 	imes (3.9275 	imes 10^{-6}) \approx 2.08 	imes 10^{-2} \ J/T$.

$A$ cylindrical magnetic rod has length $5 \ cm$ and diameter $1 \ cm$. It has uniform magnetization $5.3 \times 10^3 \ A/m$. Its net magnetic dipole moment is nearly

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