The angles of dip are $30^{\circ}$ and $45^{\circ}$ at two different places. The ratio of the horizontal components of the Earth's magnetic field at these places will be:

$A$ compass needle oscillates $20$ times per minute at a place where the dip is $30^{\circ}$ and $30$ times per minute where the dip is $60^{\circ}$. The ratio of total magnetic field due to the earth at the two places respectively is $\frac{4}{\sqrt{x}}$. The value of $x$ is $............$.

At a point on the surface of the Earth,the value of the horizontal component of the Earth's magnetic field is equal to the value of the vertical component of the Earth's magnetic field. The angle of dip is:

The angle of dip at a place is $30^{\circ}$,and the horizontal component of the Earth's magnetic field is $5 \times 10^{-5} \, T$. Calculate the vertical component and the total magnetic field at that place.

Assume the dipole model for Earth's magnetic field $B$,which is given by:
$B_v = \text{vertical component of magnetic field} = \frac{\mu_0}{4\pi} \frac{2m \cos \theta}{r^3}$
$B_H = \text{horizontal component of magnetic field} = \frac{\mu_0}{4\pi} \frac{m \sin \theta}{r^3}$
where $\theta = 90^{\circ} - \text{latitude}$ as measured from the magnetic equator.
$(a)$ Find the loci of points for which the dip angle is $\pm 45^{\circ}$.

$A$ magnet is suspended in such a way that it oscillates in the horizontal plane. It makes $20$ oscillations per minute at a place where the dip angle is $30^{\circ}$ and $15$ oscillations per minute at a place where the dip angle is $60^{\circ}$. The ratio of the total Earth's magnetic field at the two places is: