A thin uniformly charged ring of radius `a = 10 cm` rotates about its axsi with an angluaar velocity `omega = 100 rad//s`. Find the ratio of volume energy densities of magnetic and electric fields on the axis of the ring at a point removed from its centre by a distance `l = a`

Explanation

The electric field at `P` is
`E_(p) = (ql)/(4pi epsilon_(0) (a^(2) + l^(2))^(3//2))`
To get the magnetic field note that the rotating ring consitutes a current `I = q omega//2pi`, and the corresponding magnetic field at `P` ,
`B_(p) = (mu_(0) a^(2) i)/(2(a^(2) + l^(2))^(3//2))`.
Thus, `(w_(E))/(w_(M)) = (epsion_(0) mu_(0) E^(2))/(B^(2)) = epsilon_(0) mu_(0) ((ql xx 2)/(4pi epsilon_(0) mu_(0) a^(2) i))^(2)`
`= (1)/(epsilon_(0) mu_(0))((l)/(a^(2) omega))^(2)`
or, `(w_(M))/(w_(E)) = epsilon_(0) mu_(0) omega^(2) a^(4)//l^(2)`