Energy Density of a Magnetic Field — Question 1

Question 1

A solenoid converts circuit energy into a local field density

A long vacuum solenoid has N turns, length ell, area A, $L=\mu_0N^2A/\ell$, and $B=\mu_0NI/\ell$. Neglect end fields. Take $\mu_0\approx4\pi\times10^{-7}$ H/m.

1. Derive magnetic energy per unit volume from the circuit energy.

2. Evaluate the density and total energy for B=.200 T and field volume .00300 cubic meters.

Question 1 - Solution

A solenoid converts circuit energy into a local field density

(a) Method and calculation

$$U=\tfrac12LI^2=\frac{\mu_0N^2A}{2\ell}\left(\frac{B\ell}{\mu_0N}\right)^2=\frac{B^2}{2\mu_0}A\ell.$$$$\boxed{u_B=U/(A\ell)=B^2/(2\mu_0)}.$$

(b) Interpretation and further calculation

$$u_B=\frac{(.200)^2}{8\pi\times10^{-7}}=1.59155\times10^4\,J/m^3.$$$$\boxed{U=u_B\mathcal V=47.7465\,J}.$$

Physical check

Energy density is nonnegative and quadratic in field magnitude.