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 \[18^\circ \]. 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 ground geomagnetic north or south pole?
(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment \[8 \times {10^{22}}{\rm{ J}}{{\rm{T}}^{ - 1}}\] 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 surface oriented in different directions. How is such a thing possible at all?

Hint: We know that the magnetic field of the ground is directly proportional to the magnetic dipole and inversely proportional to the radius of earth. We will explain each question in a sequential manner.

Complete step by step answer:
(a) Magnetic declination, dip angle and value of magnetic field of earth in x-direction are used to describe the magnetic field of the ground.
(b) We know that the angle of dip at any country is directly proportional to the distance between that country and the north pole. We also know that Britain is situated at a distance near to the north pole, therefore we can say that the dip angle in Britain is greater than that in India.
(c) The location of Melbourne (Australia) is in the southern hemisphere of the earth so the lines of magnetic field will appear to come out of the ground.
(d) We know that the value of the magnetic field of earth at the north or south pole is zero; therefore, we can say that the compass can move in any direction.
(e) We know that the expression for magnetic field of earth is given by:
\[B = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{M}{{{r^3}}}\]
Here B is the magnetic field of earth, \[{\mu _0}\] is the magnetic permeability, M is the magnetic moment on the equatorial line and r is the radius of earth.
We know that the value of magnetic permeability is \[4\pi \times {10^{ - 7}}{{\rm{H}} {\left/
{\vphantom {{\rm{H}} {\rm{m}}}} \right.
} {\rm{m}}}\], the magnetic moment of earth is \[7.95 \times {10^{22}}{\rm{ }}{{\rm{A}} {\left/
{\vphantom {{\rm{A}} {{{\rm{m}}^2}}}} \right.
} {{{\rm{m}}^2}}}\] and radius of earth is \[6.4 \times {10^4}{\rm{ m}}\].Therefore, we will substitute \[4\pi \times {10^{ - 7}}{{\rm{H}} {\left/
{\vphantom {{\rm{H}} {\rm{m}}}} \right.
} {\rm{m}}}\] for \[\mu \], \[7.95 \times {10^{22}}{\rm{ }}{{\rm{A}} {\left/
{\vphantom {{\rm{A}} {{{\rm{m}}^2}}}} \right.
} {{{\rm{m}}^2}}}\] for M and \[6.4 \times {10^4}{\rm{ m}}\] for r in the above expression.
\[
\Rightarrow B = \dfrac{{4\pi \times {{10}^{ - 7}}{{\rm{H}} {\left/
{\vphantom {{\rm{H}} {\rm{m}}}} \right.
} {\rm{m}}}}}{{4\pi }} \times \dfrac{{7.95 \times {{10}^{22}}{\rm{ }}{{\rm{A}} {\left/
{\vphantom {{\rm{A}} {{{\rm{m}}^2}}}} \right.
} {{{\rm{m}}^2}}}}}{{{{\left( {6.4 \times {{10}^4}{\rm{ m}}} \right)}^3}}}\\
\therefore B= 0.3{\rm{ G}}
\]
(f) We can consider earth’s magnetic field as a dipole field. The local value of the north pole is higher than the south pole therefore they can be considered to have different positions of magnetised materials.

Note: We can remember the value of magnetic permeability, magnetic moment and radius of earth to solve similar problems. We can note that the magnetic property of earth is due to the small particles having magnetic properties.