A rod of length 5 , cm and diameter 1 , cm ha | vedclass

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(B) The magnetic moment $M$ is given by the product of magnetization $I$ and volume $V$ of the rod.$M = I 	imes V$Given:Magnetization $I = 5.30 	ime

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The magnetic moment $M$ is given by the product of magnetization $I$ and volume $V$ of the rod.$M = I 	imes V$Given:Magnetization $I = 5.30 	imes 10^3 \, A/m$Length $l = 5 \, cm = 5 	imes 10^{-2} \, m$Radius $r = 	ext{diameter} / 2 = 0.5 \, cm = 0.5 	imes 10^{-2} \, m$Volume $V = \pi r^2 l = \pi 	imes (0.5 	imes 10^{-2})^2 	imes (5 	imes 10^{-2}) \, m^3$$V = 3.14159 	imes 0.25 	imes 10^{-4} 	imes 5 	imes 10^{-2} \approx 3.927 	imes 10^{-6} \, m^3$Now,calculate the magnetic moment:$M = (5.30 	imes 10^3) 	imes (3.927 	imes 10^{-6})$$M \approx 20.813 	imes 10^{-3} \, J/T = 2.08 	imes 10^{-2} \, J/T$.

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

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