At a certain place in the magnetic meridian the Earth's magnetic field is twice its vertical component. The ratio of horizontal component of Earth's magnetic field and the total magnetic field of the Earth at the place is

A short bar magnet of magnetic moment $2.5 \mathrm{Am}^2$ is kept in a uniform magnetic field of $4 \times 10^{-5} \mathrm{~T}$. The work done in moving the magnet from its most stable position to most unstable position is

The magnetic field at a point $P$ on the axis of a short bar magnet of magnetic moment $M$ is $B$. If another short bar magnet of magnetic moment 2 M is placed on the first magnet such that their axes are perpendicular and their centres coincide. The resultant magnetic field at the point $P$ due to both the magnets is

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 $\operatorname{dip}$ is $60^{\circ}$, then the total magnetic field at that place is