A uniformly charged ring of radius `0.1 m` rotates at a frequency of `10^(4) rps` about its axis.Find the ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance `0.2 m` from the centre.(Use speed of light `c=3xx10^(8) m//s,pi^(2)=10`)

Answer

`9xx10^(9)J`

Explanation

Electric field at `P` is `E=(Qx)/(4piin_(0)(x^(2)-r^(2))^(3//2))`
Magnetic field at `P` is `B=mu_(0)/(4pi)(2piir^(2))/(x^(2)+r^(2))^(3//2)=mu_(0)/(4pi)(2piQf r^(2))/(x^(2)+r^(2))^(3//2)`
`f`=frequency of revolution.
Electric energy density `=1/2 epsilon _(0)E^(2)`,Magnetic energy density `B^(2)/(2mu_(0))`
`"Electric energy density" /"magnetic energy density"` `=1/2 epsilon _(0)E^(2)/B^(2)/(2mu_(0))=x^(2)/(4pi^(2)in_(0)mu_(0)f^(2)r^(4))=(x^(2)c^(2))/(4pi^(2)f^(2)r^(4))=9/pi^(2)xx10^(10)=9xx10^(9)J`