Question

A long wire carries a current of 4.00 A. Find the energy stored in the magnetic field inside a volume of 1.00 mm³ at a distance of 10.0 cm from the wire.

Answer 1

Current flowing through the wire, i = 4.00 A
Volume of the region, V = 1 mm³
Distance of the region from the wire, d = 10 cm = 0.1 m
Magnetic field due to the current-carrying straight wire, B = $\frac{{\mu }_{0}i}{2\pi r}$
The magnetic energy stored is given by
$$\begin{gathered} U=\frac{B^{2}V}{2{\mu }_0}=\frac{{{\mu }_0}^2{i}^2}{4{\pi }^2{r}^2}\times \frac{1}{2{\mu }_0}\times V \\ U=\frac{{\mu }_0{i}^2}{4{\pi }^2{r}^2}\times \frac{1}{2}\times V \\ U=\frac{(4\mathrm{\pi }\times {10}^{-7})\times (4{)}^2\times (1\times {10}^{-9})}{(4{\pi }^2\times {10}^{-2})\times 2} \\ U=2.55\times {10}^{-14}\mathrm{J} \end{gathered}$$