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

`3.67\cdot 10^(-5) J/m^3`

Step-by-step explanation:

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

`B=(\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=((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 = (B^2)/(2\mu_0)`

And substituting, we find

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