If the bar magnet in exercise 5.13 is turned around by 180º,

Solution:

The magnetic field on the axis of the magnet at a distance $d_{1}=14 \mathrm{~cm}$, can be written as:

$B_{1}=\frac{\mu_{0} 2 M}{4 \pi\left(d_{1}\right)^{3}}=H$    ...(i)

Where,

M = Magnetic moment

$\mu_{0}=$ Permeability of free space

H = Horizontal component of the magnetic field at d1

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 d2, on the equatorial line of the magnet can be written as:

$B_{2}=\frac{\mu_{0} M}{4 \pi\left(d_{2}\right)^{3}}=H$    ..(ii)

Equating equations (1) and (2), we get:

$\frac{2}{\left(d_{1}\right)^{3}}=\frac{1}{\left(d_{2}\right)^{3}}$

$\left(\frac{d_{2}}{d_{1}}\right)^{3}=\frac{1}{2}$

$\therefore d_{2}=d_{1} \times\left(\frac{1}{2}\right)^{\frac{1}{3}}$

$=14 \times 0.794=11.1 \mathrm{~cm}$

The new null points will be located 11.1 cm on the normal bisector.