The frequency of oscillation of the magnet in an oscillation magnetometer in the earth magnetic field is $40$ oscillations per minute. A short bar magnet is placed to the north of the magnetometer, at a separation of $20 cm$ from the oscillating magnet, with its north pole pointing towards north. The frequency of oscillation is found to increase to  $60$ oscillations per minute. Calculate the magnetic moment of the short bar magnet. Horizontal component of the earth's magnetic field is $24 \mu T$.

Previously,

                        $f = \frac{1}{2\pi }\sqrt{\frac{M{B}_H}{I}}$

and then,        $f\text{'} = \frac{1}{2\pi }\sqrt{\frac{M(B_H+B)}{I}}$

                            $\frac{f{\text{'}}^2}{f^2} = \frac{B_H+B}{B_H}$

or                     ${\left(\frac{60}{40}\right)}^2 = \frac{B_H+B}{B_H}$

or                         $\frac{B}{B_H} = 1.25$

or                              $B = 1.25B_H$ 

                                     $= 1.25\times 24\mu T = 30\times {10}^{-6}T$

The oscillating magnet is in end- on position of the short magnet.
Thus,

                     $B = 30\times {10}^{-6} = \frac{{\mu }_0}{4\pi }\frac{2M\text{'}}{r^3}$

or                                         $M = \frac{30\times {10}^{-6}\times r^3}{{\displaystyle \frac{{\mu }_0}{4\pi }}\times 2}$

                                                 $= \frac{30\times {10}^{-6}\times {(0.20)}^3}{{10}^{-7}\times 2}$

                                                 $= 1.2 A-m^2$