$A$ 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:

At some location on Earth,the horizontal component of Earth's magnetic field is $18 \times 10^{-6} \ T$. At this location,a magnetic needle of length $0.12 \ m$ and pole strength $1.8 \ A \ m$ is suspended from its mid-point using a thread. It makes a $45^{\circ}$ angle with the horizontal in equilibrium. To keep this needle horizontal,the vertical force that should be applied at one of its ends is:

If ${\phi_1}$ and ${\phi_2}$ are the angles of dip observed in two vertical planes at right angles to each other and ${\phi}$ is the true angle of dip,then:

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 zero.

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

At a certain place,the horizontal component ${B_0}$ and the vertical component ${V_0}$ of the Earth's magnetic field are equal in magnitude. The total intensity at the place will be