Energy Density of a Magnetic Field — Question 5

Question 5

A radial field profile concentrates energy near the outer boundary

Inside a cylindrical vacuum region of radius a and length ell, the prescribed axial field is $B(r)=B_0r/a$. Consider only energy in this region.

1. Find its total energy and mean energy density.

2. Find the radius containing 90 percent of that energy.

Question 5 - Solution

A radial field profile concentrates energy near the outer boundary

(a) Method and calculation

$$U=\frac{B_0^2}{2\mu_0a^2}\int_0^a r^2(2\pi r\ell)dr=\frac{\pi a^2\ell B_0^2}{4\mu_0}.$$$$\boxed{\langle u_B\rangle=B_0^2/(4\mu_0)=u_B(a)/2}.$$

(b) Interpretation and further calculation

$$U(r<c)/U=(c/a)^4=.900.$$$$\boxed{c=a(.900)^{1/4}=.974004a}.$$

Physical check

A field increasing linearly with radius produces a fourth-power cumulative energy, not an area fraction.