In an oscillation magnetometer, a short magnet completes 40...

The initial frequency of the oscillating magnet is $f_1 = 40 \text{ oscillations/min}$ in the earth's horizontal magnetic field $B_H = 25 \mu\text{T}$.

The second magnet is placed $20 \text{ cm}$ east of the oscillating magnet, with its north pole pointing north. This means its magnetic moment $\vec{M}$ points north, and the oscillating magnet lies on its equatorial line.

The magnetic field produced by the second magnet at the position of the oscillating magnet is given by the equatorial field formula:

$B_{eq} = \dfrac{\mu_0}{4\pi} \dfrac{M}{d^3}$

Substituting the given values ($M = 1.6 \text{ A m}^2$ and $d = 0.2 \text{ m}$):

$B_{eq} = 10^{-7} \times \dfrac{1.6}{(0.2)^3} = 10^{-7} \times \dfrac{1.6}{0.008} = 200 \times 10^{-7} \text{ T} = 20 \mu\text{T}$

On the equatorial line, the magnetic field direction is opposite to the magnetic moment. Since $\vec{M}$ points north, $B_{eq}$ points south.

The net magnetic field at the location of the oscillating magnet is the vector sum of $B_H$ (which points north)$and B_{eq}$ (which points south):

$B_{net} = B_H - B_{eq} = 25 \mu\text{T} - 20 \mu\text{T} = 5 \mu\text{T}$

The frequency of oscillation of a magnetometer is given by $f = \dfrac{1}{2\pi} \sqrt{\dfrac{MB}{I}}$, which implies $f \propto \sqrt{B}$.

$\dfrac{f_2}{f_1} = \sqrt{\dfrac{B_{net}}{B_H}}$

$f_2 = 40 \times \sqrt{\dfrac{5}{25}} = 40 \times \dfrac{1}{\sqrt{5}} = \dfrac{40}{2.236} \approx 17.88 \text{ oscillations/min}$

Rounding to the nearest integer, the new frequency is $18 \text{ oscillations/min}$.