The angle of dip at a certain place on earth is $60^{\circ}$ and the magnitude of earth's horizontal component of magnetic field is $0.26 \, G$. The magnetic field at the place on earth is.....$G$

At a place,the magnitudes of the horizontal component and total intensity of the magnetic field of the earth are $0.3 \; Oersted$ and $0.6 \; Oersted$ respectively. The value of the angle of dip at this place will be.....$^o$

The angle of dip is the angle

$A$ dip needle lies initially in the magnetic meridian when it shows an angle of dip $\theta$ at a place. The dip circle is rotated through an angle $x$ in the horizontal plane and then it shows an angle of dip $\theta'$. Then $\frac{\tan \theta'}{\tan \theta}$ is

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.

$A$ 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