At the magnetic poles of the earth,a compass needle will be

The correct relation is:
$B_H$ = Horizontal component of earth's magnetic field; $B_V$ = Vertical component of earth's magnetic field and $B$ = Total intensity of earth's magnetic field.

If $B_V$ and $B_H$ are respectively the vertical and horizontal components of the earth's magnetic field at a place where the angle of dip is $60^{\circ}$, then the total magnetic field at that place 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:

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 $\pm 45^{\circ}$.

$A$ dip needle lies initially in the magnetic meridian when it shows an angle of dip $\theta$ at a place. The dip circle is rotated through an angle $x$ in the horizontal plane and then it shows an angle of dip $\theta'$. Then $\frac{\tan \theta'}{\tan \theta}$ is