A magnetic dipole having a magnetic moment of 0.72 2 A m^2 is...

Let the east direction be along the positive x-axis ($\hat{i}$) and the north direction be along the positive y-axis ($\hat{j}$).

The magnetic moment of the dipole is $\vec{M} = M \hat{i}$, where $M = 0.72\sqrt{2} \text{ A m}^2$.

The horizontal component of the earth's magnetic field is $\vec{B}_H = B_H \hat{j}$, where $B_H = 18 \mu\text{T} = 18 \times 10^{-6} \text{ T}$.

At the neutral point, the magnetic field due to the dipole $\vec{B}_d$ must exactly cancel the earth's magnetic field, so $\vec{B}_d = -B_H \hat{j}$.

The magnetic field of a dipole at a position vector $\vec{r}$ is given by:

$\vec{B}_d = \dfrac{\mu_0}{4\pi r^3} [3(\vec{M} \cdot \hat{r})\hat{r} - \vec{M}]$

Let the position vector of the neutral point be $\vec{r} = r(\cos\theta \hat{i} + \sin\theta \hat{j})$, where $\theta$ is the angle measured from the east.

Substituting $\vec{M} = M \hat{i}$ and $\hat{r} = \cos\theta \hat{i} + \sin\theta \hat{j}$:

$\vec{B}_d = \dfrac{\mu_0 M}{4\pi r^3} [3\cos\theta(\cos\theta \hat{i} + \sin\theta \hat{j}) - \hat{i}]$

$\vec{B}_d = \dfrac{\mu_0 M}{4\pi r^3} [(3\cos^2\theta - 1)\hat{i} + (3\sin\theta\cos\theta)\hat{j}]$

For $\vec{B}_d$ to have no x-component:

$3\cos^2\theta - 1 = 0 \Rightarrow \cos^2\theta = \dfrac{1}{3}$

This implies $\sin^2\theta = 1 - \cos^2\theta = \dfrac{2}{3}$, so $\tan^2\theta = 2 \Rightarrow \tan\theta = \pm \sqrt{2}$.

Equating the y-component to $-B_H$:

$\dfrac{\mu_0 M}{4\pi r^3} (3\sin\theta\cos\theta) = -B_H$

Since $B_H$ is positive, the product $\sin\theta\cos\theta$ must be negative, which means $\theta$ lies in the fourth quadrant (south of east). Thus, $\tan\theta = -\sqrt{2}$, corresponding to an angle of $\tan^{-1}(\sqrt{2})$ south of east.

Substituting $\cos\theta = \dfrac{1}{\sqrt{3}}$ and $\sin\theta = -\dfrac{\sqrt{2}}{\sqrt{3}}$ into the y-component equation:

$\dfrac{\mu_0 M}{4\pi r^3} \left[3 \left(-\dfrac{\sqrt{2}}{\sqrt{3}}\right) \left(\dfrac{1}{\sqrt{3}}\right)\right] = -B_H$

$\dfrac{\mu_0 M \sqrt{2}}{4\pi r^3} = B_H$

Substituting the given values:

$\dfrac{10^{-7} \times 0.72\sqrt{2} \times \sqrt{2}}{r^3} = 18 \times 10^{-6}$

$\dfrac{1.44 \times 10^{-7}}{r^3} = 18 \times 10^{-6}$

$r^3 = \dfrac{1.44 \times 10^{-7}}{18 \times 10^{-6}} = 0.008 \text{ m}^3$

$r = \sqrt[3]{0.008} = 0.2 \text{ m} = 20 \text{ cm}$

The neutral point is at a distance of $20 \text{ cm}$ from the dipole, at an angle $\tan^{-1}(\sqrt{2})$ south of east.

Answer: $20 \text{ cm}$ from the dipole, $\tan^{-1}(\sqrt{2})$ south of east