Question

What is the energy density in the magnetic field 25 cm from a long straight wire carrying a current of 12 a? (μ0 = 4π × 10-7 t · m/a) what is the energy density in the magnetic field 25 cm from a long straight wire carrying a current of 12 a? (μ0 = 4π × 10-7 t · m/a)?

Answer 1

Answer:

$3.67\cdot 10^{-5} J/m^3$

Explanation:

First of all, we need to calculate the magnetic field magnitude at 25 cm from the wire, which is given by

$B=\frac{\mu_0 I}{2\pi r}$

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,

$B=\frac{(4\pi \cdot 10^{-7})(12)}{2\pi(0.25)}=9.6\cdot 10^{-6} T$

Now we can calculate the energy density of the magnetic field, which is given by

$u = \frac{B^2}{2\mu_0}$

And substituting, we find

$u = \frac{(9.6\cdot 10^{-6})^2}{2(4\pi \cdot 10^{-7})}=3.67\cdot 10^{-5} J/m^3$