If the angle of dip at places $A$ and $B$ are $30^{\circ}$ and $45^{\circ}$ respectively,the ratio of the horizontal component of the Earth's magnetic field at $A$ to that at $B$ will be.
$[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \quad \sin 30^{\circ}=\frac{1}{2}, \quad \cos 30^{\circ}=\frac{\sqrt{3}}{2}]$

The value of the horizontal component of the earth's magnetic field and the angle of dip are $1.8 \times 10^{-5} \, Wb/m^2$ and $30^{\circ}$ respectively at a certain place. The total intensity of the earth's magnetic field at that place will be:

$A$ long straight horizontal cable carries a current of $2.5\;A$ in the direction $10^{\circ}$ south of west to $10^{\circ}$ north of east. The magnetic meridian of the place happens to be $10^{\circ}$ west of the geographic meridian. The earth's magnetic field at the location is $0.33\;G,$ and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points,magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth's magnetic field.)

The ratio between the total intensity of the magnetic field at the equator to the poles is:

The earth's magnetic field at the equator is approximately $0.4 \; G$. Estimate the earth's dipole moment.

Assertion: If a compass needle is kept at the magnetic north pole of the Earth,the compass needle may stay in any direction.
Reason: $A$ dip needle will stay vertical at the north pole of the Earth.