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(C) The magnetic field $B$ at the center of a circular loop of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.The magnetic energ

Vedclass · Accepted Answer

$1.57 	imes 10^{-14} \ J$

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(C) The magnetic field $B$ at the center of a circular loop of radius $r$ carrying current $i$ is given by $B = \frac{\mu_0 i}{2r}$.The magnetic energy density is $u_d = \frac{B^2}{2\mu_0}$.The total magnetic energy $U$ stored in the volume $V$ of the cube is $U = u_d 	imes V = \frac{B^2}{2\mu_0} 	imes a^3$,where $a$ is the side of the cube.Substituting $B = \frac{\mu_0 i}{2r}$,we get $U = \frac{1}{2\mu_0} \left( \frac{\mu_0 i}{2r} ight)^2 	imes a^3 = \frac{\mu_0 i^2 a^3}{8r^2}$.Given: $\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$,$i = 2 \ A$,$r = 0.2 \ m$,$a = 10^{-3} \ m$.$U = \frac{4\pi 	imes 10^{-7} 	imes (2)^2 	imes (10^{-3})^3}{8 	imes (0.2)^2} = \frac{4\pi 	imes 10^{-7} 	imes 4 	imes 10^{-9}}{8 	imes 0.04} = \frac{16\pi 	imes 10^{-16}}{0.32} = 50\pi 	imes 10^{-16} \approx 1.57 	imes 10^{-14} \ J$.

$A$ small cube of side $1 \ mm$ is placed at the center of a circular loop of radius $20 \ cm$. If the current in the loop is $2 \ A$,then the magnetic energy stored inside the cube is (Assume $\mu_0 = 4 \pi \times 10^{-7} \ SI$ units).

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