Question

A long wire carries a current of $4.00A$. Find the energy stored in the magnetic field inside a volume of $1.00{mm}^2$ at a distance of $10.0cm$ from the wire.

Answer 1

$I=4.00A,V=1{mm}^3,d=10cm=0.1m$
$\overrightarrow{B}=\frac{B^2}{2{\mu }_0}$
Now magnetic energy stored $=\frac{B^2}{2{\mu }_0}V=\frac{{\mu }_0^2{i}^2}{4{\pi r}^2}\times \frac{1}{2{\mu }_0}\times V=\frac{4\pi \times {10}^{-7}\times 16\times 1\times 1\times {10}^{-9}}{4\times 1\times {10}^{-2}\times 2}=\frac{8}{\pi }\times {10}^{-14}=2.55\times {10}^{-14}J$