A charge of 1 coul. is placed at one end of a non-conducting rod of length 0.6m. the rod is rotated in a vertical plane about a horizontal axis passing through the other end of the rod with angular frequency `10^(4) pi "radian/sec"`. Find the magnetic field at a point on the axis ofrotation at a distance of 0.8 from the centre of the path. Now half of the charge is removed from one end and placed on the other end. The rod is rotated in a vertical plane about horizontal axis passing through the mid-point of the rod with the same angnlar frequency. Calculate the magnetic field at a point on the axis at a distance of 0.4m form the centre of the rod.

Explanation

As the revolving charge q is equivalent to a current
`i=qf=qxx(omega)/(2pi)=1xx(10^(4)pi)/(2pi)=5xx10^(3)A`
Now `B=(mu_(0))/(4pi)(2piiR^(2))/((R^(2)+x^(2))^(3//2))`

`therefore B=10^(-7)xx(2pixx5xx10^(3)(0.6)^(2))/([(0.6)^(2)+(0.8)^(2)]^(3//2))=1.13xx10^(-3)T`
If half of the charge is placed at the other end and the rod is rotated at the same frequency, the equivalent current.
`i'=((q)/(2))f+((q)/(2))f=qf=i=5xx10^(3)A`

In this case, `R’ = 0.3 m` and `x’ = 0.4 m`
`therefore B'=10^(-7)xx(2pixx5xx10^(3)xx(0.3)^(2))/([(0.3)^(2)+(0.4)^(2)]^(3//2))=2.3xx10^(-3)T`