The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} \; Wb/m^2$. If the angle of dip at that place is $60^{\circ}$,then the value of the vertical component of the earth's magnetic field will be ........ $\times 10^{-4} \; Wb/m^2$.

The lines joining the places of the same horizontal intensity are known as

$A$ compass needle is free to rotate in a horizontal plane. Its magnetic moment is $60 \ Am^2$. When it is pointing geographical north,it experiences a torque of $1.2 \times 10^{-3} \ Nm$ due to the Earth's magnetic field. If the Earth's magnetic field in the horizontal direction is $40 \times 10^{-6} \ T$,what is the declination in degrees at that place?

The horizontal component of the Earth's magnetic field is $0.22 \ G$ and the total magnetic field is $0.4 \ G$. The angle of dip is:

Choose the correct option regarding the relationship between true dip $(\phi)$ and apparent dip $(\phi^{\prime})$:

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