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

\[U = \frac{B^2 V}{2 \mu_0} = \frac{\mu_0^2 i^2}{4 \pi^2 r^2} \times \frac{1}{2 \mu_0} \times V\]

\[U = \frac{\mu_0 i^2}{4 \pi^2 r^2} \times \frac{1}{2} \times V\]

\[U = \frac{(4\pi \times {10}^{- 7} ) \times (4 )^2 \times (1 \times {10}^{- 9} )}{(4 \pi^2 \times {10}^{- 2} ) \times 2}\]

\[U = 2 . 55 \times  {10}^{- 14}   J\]