Question

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

Answer 1

The magnetic field on the axis of the magnet at a distance d1 = 14 cm, can be written as:
\(B_1\) = \(\frac{\mu_0 2M}{4\pi (d_1)^3}\)=H                         ...(1)
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: 
\(B_2\) = \(\frac{\mu_0 M}{4\pi (d_2)^3}\)=H                          ...(2)
Equating equations (1) and (2), we get: 
\(\frac{2}{(d_1)^3}\)=\(\frac{1}{(d_2)^3}\)

\(\bigg(\frac{d_2}{d_1}\bigg)^3\)=\(\frac{1}{2}\)
∴ d₂ = d₁ × \(\bigg(\frac{1}{2}\bigg)^{\frac{1}{3}}\)

=14 × 0.794 = 11.1cm

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