The magnetic field due to the Earth is closely equivalent to that due to:

The angle of dip at a place is $60^{\circ}$. At this place,the total intensity of the Earth's magnetic field is $0.64 \text{ units}$. The horizontal intensity of the Earth's magnetic field at this place is . . . . . . $\text{units}$.

The earth's magnetic field lines resemble that of a dipole at the centre of the earth. If the magnetic moment of this dipole is close to $8 \times 10^{22} \text{ Am}^2$,the value of earth's magnetic field near the equator is close to $.... \text{ Gauss}$ (radius of the earth $= 6.4 \times 10^6 \text{ m}$)

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

$A$ magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at $30^{\circ}$ with the horizontal. The horizontal component of the earth's magnetic field at the place is $0.3 \ G$. Then the magnitude of the earth's magnetic field at the location is

The angles of dip are $30^{\circ}$ and $45^{\circ}$ at two different places. The ratio of the horizontal components of the Earth's magnetic field at these places will be: