At a certain place,the angle of dip is $60^{\circ}$ and the horizontal component of the earth's magnetic field $(B_H)$ is $0.8 \times 10^{-4} \,T$. The earth's overall magnetic field is

Explain the angle of dip.

The vertical component of the earth's magnetic field is $6 \times 10^{-5} \text{ T}$ at a place where the angle of dip is $37^{\circ}$. The earth's resultant magnetic field at that place will be (Given $\tan 37^{\circ} = \frac{3}{4}$)

The angle of dip is the angle

Let $m$ and $r$ be the dipole moment and radius of the Earth,respectively. Then,the Earth's magnetic field at the equator 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 $G$)