A charge of `(3.14 xx 10^-6)` C is distributed uniformly over a circular ring of radius 20.0 cm. The ring rotates about its axis with an angular velocity of 60.0 rad ` s^-1 . Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00 cm from the centre.

Explanation

Given
`q=3.14xxx10^(-6) C`
`r= 20cm =20xx10^(-2)m`
`omega =60rad//sec.`
So, time to take 1 revolution `= (2pi)/(60)`
`:. Current I = (q)/(t) = (3.14xx10^(-6)xx60)/(2pi)`
`= 30xx10^(-6)A`
Then electric field
`E=(q)/(4piepsilon_or^2)`
`=(9xx10^9xx3.14xx10^(-6)/(25xx10^(-4)`
Magnetic field
`B=(mu_O)/(2). (ir^2)/((x^2 +r^2)^(3//2)`
`=4pixx10^(-7)xx30xx10^(-6)xx20xx20xx10^(-4)/(2[5xx5xx10^(-4)+20xx20xx10^(-4)](3//2)`
`=4pixx12xx10^(-(14)/(2[425]^(3//2)xx10^(-6)`
`=4pixx12xx10^(-14)/(2(20.6)^3xx10^(-6)`
`(E)/(B) = (9xx10^9xx3.14xx10^(-6)xx2xx(20.6)^3xx10^(-6)/(25xx10^(-4)xx4pixx10^(-14)xx12)`
`=(9xx3.14xx2xx(20.6)^3)/(25xx4pixx12)`
`=1.31xx10^15).`