Question

The magnetic field of earth at the equator is approximately $4 \times {10^{ - 5}}T$. The radius of earth is $6.4 \times {10^6}m$. Then the dipole moment of the earth will be nearly of the order of:
(A) ${10^{23}}A{m^{^2}}$
(B) ${10^{20}}A{m^{^2}}$
(C) ${10^{16}}A{m^{^2}}$
(D) ${10^{10}}A{m^{^2}}$

Answer 1

Hint
To solve this question, we need to use the formula for the equatorial magnetic field for a bar magnet. The magnetic field of the earth can be approximated to be a bar magnet, whose parameters are already given in the question.
Formula Used: The formula which is used to solve this question is given by
$\Rightarrow {B_E} = \dfrac{{{\mu _0}m}}{{4\pi {r^3}}}$
Here $m$ is the magnetic moment, $r$ is the distance from the centre of the magnet, and ${\mu _0}$ is the magnetic permeability in vacuum.

Complete step by step answer
The magnetic field of the earth can be approximated to be an equivalent bar magnet. So, we can use the formula for the magnetic field of a bar magnet at the equatorial position.
We know that the magnetic field at the equatorial position is approximately given by
$\Rightarrow {B_E} = \dfrac{{{\mu _0}m}}{{4\pi {r^3}}}$
This gives the magnetic moment as
$\Rightarrow m = \dfrac{{4\pi {B_E}{r^3}}}{{{\mu _0}}}$ (1)
According to the question, ${B_E} = 4.5 \times {10^{ - 5}}T$ and $r = 6.4 \times {10^6}m$.
Also, we know that ${\mu _0} = 4\pi \times {10^{ - 7}}$.
Substituting these in (1) we get
$\Rightarrow m = \dfrac{{4\pi \left( {4.5 \times {{10}^{ - 5}}} \right){{\left( {6.4 \times {{10}^6}} \right)}^3}}}{{4\pi \times {{10}^{ - 7}}}}$
On solving, we get
$\Rightarrow m = 1.179 \times {10^{23}}A{m^2}$
Thus, we see that the magnetic moment comes out to be equal to $1.179 \times {10^{23}}A{m^2}$ which is clearly of the order of ${10^{23}}A{m^2}$.
Hence, the correct answer is option A.

Note
If we do not remember the formula for the equatorial magnetic field, then we can take the help of the electrostatic analogy. We know that the electric field due to a dipole at its equatorial position is given by $E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{p}{{{r^3}}}$. Replacing the electric field $E$ with the magnetic field $B$, the electric dipole moment $p$ with the magnetic dipole moment $m$ and the constant $\dfrac{1}{{4\pi {\varepsilon _0}}}$ with the constant $\dfrac{{{\mu _0}}}{{4\pi }}$ we can get the corresponding expression for the equatorial magnetic field.