# Potential energy stored in the field of an inductor

The potential energy of the magnetic field inside an air-core inductor can be found using $u_{air}=\frac{L \pi R^2 B^2}{2\mu_0}$, which makes sense to me.

The potential energy inside an inductor with a core that fills the entire interior volume is $u_{core} = \frac{L \pi R^2 B^2}{2\mu_0}(\frac{\mu_k}{\mu_0})$ where $\mu_k$ is the relative permeability of the core. This also makes logical sense.

What happens if the core doesn't fill the entire interior volume of the inductor windings? I would expect the potential energy to be something along the lines of

$$u_{\text{partially filled}} = \frac{B^2}{2\mu_0}(\frac{\mu_k}{\mu_0})(\frac{\text{volume of core}}{\text{ volume of interior }})$$

However, I haven't been able to find much on the subject of inductor partial cores. (I'm more interested in this problem in the context of a solenoid than a transformer.)

Perhaps I could play with some of the things I've run across for transformers, but I'd be working with DC so I'm not sure how much is applicable.

I found the equations I'm using here.

The site that you cite is correct. You just have to find the energy stored in the air part, the energy in the core part and then add them.

If the field in the air part is $B_0$ then the energy of the field in air is energy density$$\frac{B^{2}_0}{2\mu_0}$$

multiplied by volume occupied by air (this depends on the radius and length of the core, let's say radius of inductor is R, length is L and radius and length of core are r, and l)

Then $$U_{air}=\frac{B^{2}_0*\pi(R^2L-r^2l)}{2\mu_0}$$

and energy density in the core is $$\frac{B^{2}_0}{2\mu_0}\frac{\mu_{core}}{\mu_0}$$

and energy in the core is $$\frac{B^{2}_0}{2\mu_0}\frac{\mu_{core}}{\mu_0}*\pi r^2l$$

and net energy is $$\frac{B^{2}_0}{2\mu_0}\pi (R^2L+(\mu_r-1)r^2l)$$

• Perfect! Just what I was looking for. Thank you very much. – CoilKid Apr 10 '17 at 12:09
• My pleasure to help! – GeeJay Apr 10 '17 at 16:43