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

Explanation

If the bar magnet is turned around by `180^@` , the neutral point would now be on equatorial line. Let `d_1` be the distance of new neutral points from the centre. Then we have:
` (mu_0)/(4pi ) (M)/(d_1^3) = B_H` ...(i)
From equation we have
` (mu_0)/(4pi) (2M)/(d^3) = B_H`
From eqn. (1) and (2) , we have
`(mu_0)/(4pi ) (2M)/(d^3) = (mu_0)/(4pi) (M)/(d_1^3)`
Or `d_1^3 = (d^3)/(2)`
Or ` d_1 = d xx 10^(-1//3)`
Or ` d_1 = 14 xx 2^(-1//3) = 11.1 cm`