The total intensity of the Earth's magnetic field at the magnetic equator is $5$ $units$. Its value at a magnetic latitude of $37^{\circ}$ is equal to:

The Earth's magnetic field at a certain place has a horizontal component of $0.3 \ G$ and the total strength of $0.5 \ G$. The angle of dip is:

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

$A$ compass needle whose magnetic moment is $60 \, A \cdot m^2$ is pointing towards the geographical north at a certain place,where the horizontal component of the Earth's magnetic field is $40 \, \mu Wb/m^2$. It experiences a torque of $1.2 \times 10^{-3} \, N \cdot m$. What is the angle of declination at this place (in $^o$)?

The angle of dip at a place on the earth gives: