At two different places, the angles of dip are respectively $30^{\circ}$ and $45^{\circ}$. At these two places, the ratio of the horizontal component of Earth's magnetic field is:

Define geographic meridian and magnetic meridian.

The angle of dip at a place is $40.6^\circ$ and the intensity of the vertical component of the earth's magnetic field $V = 6 \times 10^{-5} \text{ Tesla}$. The total intensity of the earth's magnetic field $(I)$ at this place is:

$A$ short bar magnet placed in a horizontal plane has its axis aligned along the north-south direction. Null points are found on the axis of the magnet at $20 \ cm$ from the centre of the magnet. The Earth's magnetic field at the place is $B$ and the angle of dip is $0^{\circ}$. If the total magnetic field on the normal bisector of the magnet at $20 \ cm$ from the centre of the magnet is $0.6 \ G$,then the magnitude of $B$ is: (in $G$)

$A$ magnetic needle is free to rotate in a vertical plane which makes an angle of $60^{\circ}$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$ with the horizontal,the true dip value at that place is: (in $^{\circ}$)

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