If the bar magnet in exercise 5.13 is turned around by `180^@,` where will the new null points be located?

Explanation

When bar magnet given in above example is turned around by 180° such that its poles get interchanged, we obtain new null points `N_1 and N_2` on the equator of bar magnet in the plane of figure, suppose at distance y from the centre of magnet. See the figure given below.

At null points `N_1 and N_2`, we have
`therefore (mu_0)/( 4 pi) ((m) /( y^3) )= (mu_0)/( 4pi) ((2m)/( r^3))" "`[From equation (1) of above example]
`therefore y^(3) = (r^3)/( 2)`
`therefore y= (r)/( 2^(1//3)) `
`therefore y= (14)/( 1.2599) = 11.112` cm
Note : Here all the points located on the circle of radius y in the equatiorial plane of a bar magnet are also null points along with points `N_(1) and N_(2)`.