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 as its centre. Check the order of magnitude of this number in some way.

Hint: Use the formula of the magnetic field given below, substitute all the known parameters in it to find the value of the magnetic field due to the dipole magnetic moment. Compare this value of the magnetic field to the magnitude of the main number of the universe.

Formula used:
The formula of the magnetic field is given by
$B = \dfrac{{\mu _0}M}{4 \pi d^3}$
Where $B$ is the magnetic field, ${\mu _0}$ is the permeability of the free space, $M$is the mass and $d$ is the distance of the electric field.

Complete step by step solution:
It is given that the
Magnetic moment of the earth, $M = 8 \times {10^{22}}\,J{T^{ - 1}}$
The distance of the electric field, $d = 6.4 \times {10^6}\,m$
By using the formula of the magnetic moment of the electric dipole,
$B = \dfrac{{\mu _0}M}{4 \pi d^3}$
Substituting the values of known parameters in the above formula to determine the value of the magnetic field, we get
$\Rightarrow$ $B = \dfrac{{10^{ - 7}} \times 8 \times {10^{22}}}{\left( {6.4 \times {{10}^6}} \right)^3}$
By performing the simple arithmetic operation in the right hand side of the equation, we get
$\Rightarrow$ $B = 0.31 \times {10^{ - 4}}\,T$

Hence the above values are equal to that of the earth magnetic field.

Note: The earth’s magnetic field signifies the earth’s gravitational force that is present on the surface of the earth. This is a weak force and it is available on any surface of the earth. This earth’s magnetic force varies in the northern and the southern hemisphere. This is maximum at the inner part of the earth.