# 5.14. If the bar magnet in exercise 5.13 is turned around by $180 \degree$, where will the new null points be located?

Given, d = 14 cm

The magnetic field at a distance d from the centre of the magnet on its axis:

$B = \mu_{0}m/2\pi d^{3}$

If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial (perpendicular bisector) line.

The magnetic field at a distance d' from the centre of the magnet on the normal bisector is:

$B = \mu_{0}m/4\pi d'^3$

Equating these two, we get:

$\dpi{100} \\ \frac{1}{2d^3} = \frac{1}{4d\ '^3}\\ \\ \implies \frac{d\ '^3}{d^3} = \frac{1}{2}$

d' = 14 x 0.794 = 11.1cm

The new null points will be at a distance of 11.1 cm on the normal bisector.

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