A charge of `1` coulumb is placed is 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 an angular velocity `10^(4)pi rad//sec`.Find the magnetic field at a point on the axis of rotation of a distance of `0.8m` from the centre of the path is `Npixx10^(-4) T`,then find value of `N`.

Explanation

`vecB=mu_(0)/(4pi) q((vecvxxvecr)/r^(3))`
`B=mu_(0)/(4pi)(qv)/((x^(2)+R^(2)))=(10^(-7)xx1xx0.6xx10^(4)pi)/1=6pixx10^(-4) T`