If the bar magnet in the above problem is turned around by `180^@`, where will the new null points be located?

Explanation

The magnetic field on the axis of the magnet at a distance `d_(1)=14cm`, can be written as:
`B_(1)=(mu_(@)2M)/(4pi(d_(1))^(3))=H` ..............(1)
where,
M=Magnetic moment
`mu_(0)=` Permaeability of free space
H=Horizontal component of the magnetic field at `d_(1)`
if the bar magnet is turned through `180^(@)` , then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance `d_(2)`, on the equatorial line of the magnet can be written as:
`B_(2)=(mu_(@)M)/(4pi(D_(2))^(3))=H`
Equating equations (1) and (2), we get:
`(2)/((d_(1))^(3))=(1)/((d_(2))^(3))`
`((d_(2)/(d_(1)))^(3)=(1)/(2)`
`therefored_(2)=d_(1)xx((1)/(2))^((1)/(3))`
`=14xx0.794=11.1cm`
The new null points will be located 11.1 cm on the normal bisector,