The magnetic field at a point due to a magnetic dipole is...

Let the magnetic moment of the dipole be $\vec{M}$ and the position vector of the point be $\vec{r}$.

The magnetic field at a point due to a magnetic dipole is given by:

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

Let the magnetic axis be along the x-axis, so $\vec{M} = M \hat{i}$.

The unit vector along the position vector is $\hat{r} = \cos\theta \hat{i} + \sin\theta \hat{j}$, where $\theta$ is the angle between $\vec{r}$ and $\vec{M}$.

Substituting these into the expression for $\vec{B}$:

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

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

It is given that the magnetic field is perpendicular to the magnetic axis. Therefore, the component of $\vec{B}$ along the x-axis must be zero.

$3\cos^2\theta - 1 = 0$

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

Using the identity $\sec^2\theta = 1 + \tan^2\theta$:

$\dfrac{1}{\cos^2\theta} = 1 + \tan^2\theta$

$3 = 1 + \tan^2\theta$

$\tan^2\theta = 2$

$\tan\theta = \sqrt{2}$

$\theta = \tan^{-1}(\sqrt{2})$