A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth's magnetic field at the place is 0.36 G and the angle of dip is zero.If the bar magnet is turned around by `180^(@)`, where will the new null points be located ?

Explanation

When the bar magnet is turned `180^(@)`, neutral points would lie on equitorial line, so that
`B_(2) = (mu_(0))/(4pi) (M)/(d_(2)^(3)) = H`
On the equitorial line of magnet at same direction (d), field due to the magnet is
`B_(2) = (mu_(0))/(4pi)(M)/(d_(2)^(3)) = H`
In the previous question
`B_(1) = (mu_(0))/(4pi)(2M)/(d_(1)^(3))`
`d_(2)^(3) = (d_(1)^(3))/(2) = ((14)^(3))/(2)`
`d_(2) = (14)/(2^(1//3)) = 11.1 cm`