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

`B=(mu_(0)ir^(2))/(2(r^(2)+x^(2))^(3//2)),i=(qomega)/(2pi)rArrB=(mu_(0))/(4pi)(qomegar^(2))/((r^(2)+x^(2))^(3//2))`