The value of horizontal component of earth's magnetic field at a place is $0.35 \times 10^{-4} \,T$. If the angle of dip is $60^{\circ}$,the value of vertical component of earth's magnetic field is nearly ............. $\times 10^{-4} \,T$.

The value of the horizontal component of the earth's magnetic field and the angle of dip are $1.8 \times 10^{-5} \, Wb/m^2$ and $30^{\circ}$ respectively at a certain place. The total intensity of the earth's magnetic field at that place will be:

The angle of dip is the angle

At a certain location in Africa,a compass points $12^{\circ}$ west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points $60^{\circ}$ above the horizontal. The horizontal component of the earth's field is measured to be $0.16 \; G$. Specify the direction and magnitude of the earth's field at the location.

If the angle of dip at places $A$ and $B$ are $30^{\circ}$ and $45^{\circ}$ respectively,the ratio of the horizontal component of the Earth's magnetic field at $A$ to that at $B$ will be.
$[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \quad \sin 30^{\circ}=\frac{1}{2}, \quad \cos 30^{\circ}=\frac{\sqrt{3}}{2}]$

The plane of a dip circle is set in the geographic meridian and the apparent dip is $\delta_1$. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is $\delta_2$. The declination $\theta$ at the place is