Solved: Calculate Magnetic Energy of Nonmagnetic Wire

[jesuslovesu]

[SOLVED] Magnetic Energy

Whoops, never mind, i didn't use the right B field

Homework Statement

Show that the magnetic energy per unit length of a nonmagnetic wire is $$\frac{\mu_0 I^2 }{ 16 \pi}$$ (inside the wire)

Homework Equations

$$W = 1/2 L I^2$$
$$W = \int \int \int \frac{|B|^2} {2 \mu}$$

The Attempt at a Solution

Well, At first I was going to use W = 1/2 Li^2, but then I realized that I don't know how to find the self inductance of a wire. (inside it at least)

I know the B field of a wire, but when I try to integrate it, I get something completely different.

$$W = \int_0^L \int_0^{2\pi} \int_0^a \frac{(\mu_0 I)^2} {(2 * 2\pi r)^2} rdr d \phi dz$$ (i'll worry about the per unit length part later)
But as you can see that gives me a ln(a/0) which can't happen, so I'm stumped

 

[jbsteele]

. Any help would be appreciated. SOLVED:Actually, the magnetic field inside a wire is zero, so the energy is also zero. The energy comes from the B field outside of the wire. The correct equation is W = \int_0^L \int_0^{2\pi} \int_a^b \frac{B^2}{2\mu} rdr d\phi dz Where B is the B field outside the wire and a and b are the inner and outer radius of the wire. So the answer is W = \frac{\mu_0 I^2 }{ 16 \pi}

 

[Moetasim]

.
Great job on trying to solve this problem! It's always important to double check your equations and make sure you're using the right values and units. In this case, the magnetic energy per unit length of a nonmagnetic wire is actually given by the equation W = \frac{\mu_0 I^2 }{ 16 \pi} (inside the wire). This is because the self inductance of a wire is negligible and can be ignored in this calculation.

To find the magnetic energy per unit length, you can use the equation W = \int \int \int \frac{|B|^2} {2 \mu} and integrate over the volume of the wire. This will give you the same result as the equation given in the problem.

Keep up the good work!