A small cube of volume $V=1 \mathrm{~mm}^3$ is placed at the center of a circular loop of radius $R=

Question

A small cube of volume $V=1 \mathrm{~mm}^3$ is placed at the center of a circular loop of radius $R=$ 10 cm carrying a current of $I=4 \mathrm{~A}$. The magnetic energy stored inside the cube is:

Professor PC Thomas Classes · Answer

Magnetic Field at Center $(B)$ : For a circular loop, $B=\frac{\mu_0 I}{2 R}$. $$ B=\frac{4 \pi 	imes 10^{-7} 	imes 4}{2 	imes 0.1}=8 \pi 	imes 10^{-6} \mathrm{~T} . $$ Magnetic Energy Density ( $u_B$ ): The energy stored per unit volume is $u_B=\frac{B^2}{2 \mu_0}$. $$ u_B=\frac{\left(8 \pi 	imes 10^{-6}ight)^2}{2 	imes 4 \pi 	imes 10^{-7}}=\frac{64 \pi^2 	imes 10^{-12}}{8 \pi 	imes 10^{-7}}=8 \pi 	imes 10^{-5} \mathrm{~J} / \mathrm{m}^3 . $$ Total Energy ( $U$ ): $$ U=u_B 	imes 	ext { Volume } . $$ Given Volume $=1 \mathrm{~mm}^3=10^{-9} \mathrm{~m}^3$. $$ U=\left(8 \pi 	imes 10^{-5}ight) 	imes 10^{-9} \approx 25.1 	imes 10^{-14} \mathrm{~J}=2.5 	imes 10^{-13} \mathrm{~J} . $$