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

The initial frequency of the oscillation magnetometer is given by:

$f_1 = \dfrac{1}{2\pi} \sqrt{\dfrac{M' B_H}{I}}$

where $B_H = 25 \mu\text{T}$ is the Earth's horizontal magnetic field, which points towards the North.

A second short magnet is placed $20 \text{ cm}$ ($0.2 \text{ m}$) East of the oscillating magnet. Since its North pole is oriented towards the South, its magnetic moment $\vec{M}$ points towards the South (as magnetic moment is directed from the South pole to the North pole).

The oscillating magnet lies on the equatorial line of this second magnet. The magnetic field $\vec{B}$ produced by a short magnet at an equatorial point is given by:

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

Substituting the given values:

$B = 10^{-7} \times \dfrac{1.6}{(0.2)^3} = 10^{-7} \times \dfrac{1.6}{0.008} = 20 \times 10^{-6} \text{ T} = 20 \mu\text{T}$

At an equatorial point, the magnetic field $\vec{B}$ is directed opposite to the magnetic moment $\vec{M}$. Since $\vec{M}$ points South, $\vec{B}$ points North.

Because both the Earth's horizontal magnetic field $B_H$ and the field $B$ due to the second magnet point North, the net magnetic field $B_{net}$ is their sum:

$B_{net} = B_H + B = 25 \mu\text{T} + 20 \mu\text{T} = 45 \mu\text{T}$

The frequency of oscillation is directly proportional to the square root of the net magnetic field ($f \propto \sqrt{B_{net}}$). Therefore, the new frequency $f_2$ is:

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

$f_2 = 40 \times \sqrt{\dfrac{45}{25}} = 40 \times \dfrac{3\sqrt{5}}{5} = 24\sqrt{5}$

Using $\sqrt{5} \approx 2.236$:

$f_2 \approx 24 \times 2.236 \approx 53.66 \text{ oscillations/min}$

Rounding to the nearest integer, the new frequency is $54 \text{ oscillations/min}$.

Answer: $54 \text{ oscillations/min}$