A cylindrical rod magnet has a length of 5 , | vedclass

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(B) The magnetic dipole moment $M$ is given by the product of magnetization $I$ and the volume $V$ of the magnet,i.e.,$M = I 	imes V$.Given: Magnetiz

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The magnetic dipole moment $M$ is given by the product of magnetization $I$ and the volume $V$ of the magnet,i.e.,$M = I 	imes V$.Given: Magnetization $I = 5.30 	imes 10^3 \, A/m$,length $l = 5 \, cm = 0.05 \, m$,and diameter $d = 1 \, cm$,so radius $r = 0.5 \, cm = 0.005 \, m$.The volume of the cylindrical rod is $V = \pi r^2 l$.Substituting the values:$V = \pi 	imes (0.005)^2 	imes 0.05 = \pi 	imes (2.5 	imes 10^{-5}) 	imes 0.05 = 3.927 	imes 10^{-6} \, m^3$.Now,calculating the dipole moment:$M = (5.30 	imes 10^3) 	imes (3.927 	imes 10^{-6}) \approx 2.08 	imes 10^{-2} \, J/T$.

$A$ cylindrical rod magnet has a length of $5 \, cm$ and a diameter of $1 \, cm$. It has a uniform magnetization of $5.30 \times 10^3 \, A/m$. What is its magnetic dipole moment?

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