A uniformly charged ring of radius `R` is rotated about its axis with constant linear speed `v` of each of its particle. The ratio of electric field to magnetic field at a point `P` on the axis of the ring distant `x=R` from centre of ring is `( c` is speed of light )

Explanation

Electric field at a point on axis of the uniformly charged ring is given by the following equation.
`E = (Qx)/(4piepsi_(0)(R^(2) + x^(2))^(3//2) )` ....(i)
When the ring rotates with an angular velocity to then the equivalent amount of current can be written as follows:
`i = Q/(2 pi//omega) = (Qomega)/(2 pi) `
Now, magnetic field at a point on axis of current carrying loop is given by the following equation:
`B = (mu_(0)iR^(2))/(2(R^(2) + x^(2))^(3//2)) = (mu_(0)(Qomega)/(2pi) R^(2))/(2(R^(2)+x^(2))^(3//2))`
` B = (mu_(0)Q omega R^(2))/(4 pi (R^(2) + x^(2))^(3//2)) ` .....(iii)
From equations (i) and (ii) we get the ratio of electric field and magnetic field.
`E/B = ((Qx)/(4piepsi_(0)(R^(2) + x^(2))^(3//2)))/((mu_(0)QomegaR^(2))/(4pi (R^(2)+x^(2))^(3//2)))=(x/epsi_(0))/((mu_(0)omegaR^(2))/1) = x/(mu_(0)epsi_(0)omegaR^(2))`
`E/B = x/(mu_(0)epsi_(0)omegaR^(2)) ` .....(iii)
We should be aware that speed of light can be written as
`c = 1/sqrt(mu_(0)epsi_(0)) rArr 1/(mu_(0)epsi_(0)) = c^(2) `
Substituting in equation (iii) we can modify the result as follows in terms of speed of light.
`E/B = (c^(2)x)/(omegaR^(2)) `