A deflection magnetometer is arranged with its arms oriented in the...

For the compass needle to rest in any arbitrary position, the net magnetic field at the center of the magnetometer must be zero. This means the magnetic field produced by the short magnet ($\vec{B}$) must be equal and opposite to the horizontal component of the Earth's magnetic field ($\vec{B}_H$).

Since $\vec{B}_H$ is directed towards the geographic North, the magnetic field $\vec{B}$ due to the magnet at the location of the needle must be directed towards the geographic South.

The arms of the deflection magnetometer are along the North-South direction. The magnet is placed on the arms, which means the compass needle lies on the axial line of the short magnet.

The magnetic field on the axial line of a short magnet is given by:
$\vec{B} = \dfrac{\mu_0}{4\pi} \dfrac{2\vec{M}}{d^3}$

The direction of the magnetic field on the axial line is the same as the direction of the magnetic moment $\vec{M}$ (which points from the South pole to the North pole of the magnet). For $\vec{B}$ to point towards the South, $\vec{M}$ must also point towards the South. Thus, the North pole of the magnet must point towards the South.

Equating the magnitudes of the fields:
$B_H = \dfrac{\mu_0}{4\pi} \dfrac{2M}{d^3}$

Rearranging the formula to solve for $d^3$:
$d^3 = \dfrac{\mu_0}{4\pi} \times 2 \times \left(\dfrac{M}{B_H}\right)<br><br>Substituting the given values \dfrac{\mu_0}{4\pi} =$10^{-7} \text{ T m A}^{-1} and \dfrac{M}{B_H} =$40 \text{ A m}^2 \text{ T}^{-1}$:
$d^3 = 10^{-7} \times 2 \times 40$
$d^3 = 80 \times 10^{-7} = 8 \times 10^{-6} \text{ m}^3$

Taking the cube root:
$d = 2 \times 10^{-2} \text{ m} = 2.0 \text{ cm}$

Therefore, the magnet must be placed at a distance of $2.0 \text{ cm}$ from the needle with its North pole pointing towards the South.