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 ........ $\times 10^{-4} \; Wb/m^2$.

Define geographic meridian and magnetic meridian.

At a certain place,the horizontal component of the Earth's magnetic field is $B_0$ and the angle of dip is $45^o$. The total intensity of the Earth's magnetic field at that place will be:

$A$ magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at $22^{\circ}$ with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be $0.35 \; G$. Determine the magnitude of the earth's magnetic field (in $G$) at the place.

The intensity of the magnetic field due to the Earth at a point inside a hollow steel box is:

Answer the following questions regarding earth's magnetism:
$(a)$ $A$ vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth's magnetic field.
$(b)$ The angle of dip at a location in southern India is about $18^o$. Would you expect a greater or smaller dip angle in Britain?
$(c)$ If you made a map of magnetic field lines at Melbourne in Australia,would the lines seem to go into the ground or come out of the ground?
$(d)$ In which direction would a compass free to move in the vertical plane point to,if located right on the geomagnetic north or south pole?
$(e)$ The earth's field,it is claimed,roughly approximates the field due to a dipole of magnetic moment $8 \times 10^{22} \, J \, T^{-1}$ located at its centre. Check the order of magnitude of this number in some way.
$(f)$ Geologists claim that besides the main magnetic $N-S$ poles,there are several local poles on the earth's surface oriented in different directions. How is such a thing possible at all?