Question

The earth's magnetic field at the equator is approximately $0.4\times {10}^{-4}T$. What is the earth's dipole moment?$R_e=6.4\times {10}^{6}m$.

Answer 1

Step 1: Given data

The magnetic field at the equator is $B=0.4\times {10}^{-4}T$

The radius of the earth is $R_e=6.4\times {10}^{6}m$

Absolute permeability of free space ${\mu }_o=4\pi \times {10}^{-7}H/m$

Step 2: Formula used

$B=\frac{{\mu }_{o}m}{4\pi {R_e}^3}$

Step 3: Calculation of dipole moment:

Let, $m$ be the pole strength of earth's magnet.

Now, the magnetic field is, $B=\frac{{\mu }_{o}m}{4\pi {R_e}^3}$

$$\begin{gathered} \Rightarrow m=\frac{\mathrm{B}\times 4\pi {R_e}^3}{{\mathrm{\mu }}_{\mathrm{o}}} \\ \Rightarrow m=\frac{0.4\times {10}^{-4}}{{10}^{-7}}\times {\left(6.4\times {10}^6\right)}^3 \\ \Rightarrow m=1.05\times {10}^{23}A{m}^2 \end{gathered}$$

Hence, the dipole moment of the earth is $1.05\times {10}^{23}A{m}^2$.