The dip angle in a vertical plane at an angle of $\cos^{-1} \left( \frac{1}{\sqrt{2}} \right)$ from the magnetic meridian is $60^\circ$. Find the actual dip angle at that place.

At a place,if the earth's horizontal and vertical components of magnetic fields are equal,then the angle of dip will be.......$^o$

$A$ 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:

The relation between $Z_E, H_E$,and $B_E$ 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 $G$)

At a certain place,the angle of dip is $30^{\circ}$ and the horizontal component of the Earth's magnetic field is $0.5 \ G$. The Earth's total magnetic field (in $G$) at that place is: