At the place where the angle of dip is $60^{\circ}$,what is the relation between the vertical component $Z_{E}$ and the horizontal component $H_{E}$ of the Earth's magnetic field?

At a certain place,the angle of dip is $30^{\circ}$ and the horizontal component of the Earth's magnetic field is $0.50 \text{ Oersted}$. The Earth's total magnetic field is:

The magnetic field due to the Earth is closely equivalent to that due to:

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:

$A$ line passing through places having zero value of magnetic dip is called