Answer:
[tex]3.67\cdot 10^{-5} J/m^3[/tex]
Explanation:
First of all, we need to calculate the magnetic field magnitude at 25 cm from the wire, which is given by
[tex]B=\frac{\mu_0 I}{2\pi r}[/tex]
where
μ0 = 4π × 10-7 t · m/a is the vacuum permeability
I = 12 A is the current in the wire
r = 25 cm = 0.25 m is the distance from the wire
Substituting,
[tex]B=\frac{(4\pi \cdot 10^{-7})(12)}{2\pi(0.25)}=9.6\cdot 10^{-6} T[/tex]
Now we can calculate the energy density of the magnetic field, which is given by
[tex]u = \frac{B^2}{2\mu_0}[/tex]
And substituting, we find
[tex]u = \frac{(9.6\cdot 10^{-6})^2}{2(4\pi \cdot 10^{-7})}=3.67\cdot 10^{-5} J/m^3[/tex]
