91Ó°ÊÓ

A magnetic field is a vector field that describes the magnetic effect on moving electric charges, electric currents, and magnetic materials. A moving charge in a magnetic field exerts a force perpendicular to its own velocity and to the magnetic field.

Here, you have to use the formula for Earth’s magnetic field,

$B=\frac{{\mathrm{μ}}_0\mathrm{μ}}{4\mathrm{π}{r}^3}×\sqrt{1+3{\sin }^2{\mathrm{λ}}_m}$ ..... (1)

And the inclination of Earth’s magnetic field is,

$\tan \mathrm{ϕ}=2\tan {\mathrm{λ}}_m$ ..... (2)

Earth’s magnetic dipole moment,$\mathrm{μ}=8.00×{10}^{22}\text{A}â‹\dots {\text{m}}^{\text{2}}$

The radius of the Earth, $r=6.378×{10}^6\text{m}$

The permeability of free space,${\mathrm{μ}}_0=4\mathrm{π}×{10}^{−7}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.$

On equator, the wavelength is,

${\mathrm{λ}}_m=0$

So the magnetic field using equation (1) is,

$\begin{array}{rcl}B& =& \frac{{\mathrm{μ}}_0\mathrm{μ}}{4\mathrm{Ï€}{r}^3}×\sqrt{1+3{\sin }^2{\mathrm{λ}}_m}\\ & =& \frac{(4\mathrm{Ï€}×{10}^{−7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.})×(8.00×{10}^{22}\text{A}â‹\dots {\text{m}}^{\text{2}})}{4\mathrm{Ï€}×{(6.378×{10}^6\text{m})}^3}×\sqrt{1+3{\sin }^2\left(0\right)}\\ & =& 3.10×{10}^{-5}\text{T}\end{array}$

Hence, the magnitude of Earth’s magnetic field at the geomagnetic equator is $\begin{array}{rcl}& & 3.10×{10}^{-5}\text{T}\end{array}$.

Use equation (2) to define the inclined of Earth's magnetic field at the equator.

$\begin{array}{rcl}\tan {\mathrm{ϕ}}_i& =& 2×\tan {\mathrm{λ}}_m\\ & =& 2×\tan 0°\end{array}$

${\mathrm{ϕ}}_i=0°$

Hence, the inclination of Earth’s magnetic field at the geomagnetic equator is $0°$.

The wavelength is,

${\mathrm{λ}}_m=60$

Define the magnitude of Earth's magnetic field as below,

$\begin{array}{rcl}B& =& \frac{{\mathrm{μ}}_0\mathrm{μ}}{4\mathrm{Ï€}{r}^3}×\sqrt{1+3{\sin }^2{\mathrm{λ}}_m}\\ & =& \frac{(4\mathrm{Ï€}×{10}^{−7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.})×(8.00×{10}^{22}\text{A}â‹\dots {\text{m}}^{\text{2}})}{4\mathrm{Ï€}×{(6.378×{10}^6\text{m})}^3}×\sqrt{1+3{\sin }^2(60°)}\\ & =& 5.59×{10}^{-5}\text{T}\end{array}$

Hence, the magnitude of Earth’s magnetic field at the geomagnetic latitude is $\begin{array}{rcl}& & 5.59×{10}^{-5}\text{T}\end{array}$.

Define the inclination of Earth’s magnetic field at the geomagnetic latitude is,

$\begin{array}{rcl}\tan {\mathrm{ϕ}}_i& =& 2×\tan {\mathrm{λ}}_m\\ & =& 2×\tan 60°\end{array}$

${\mathrm{ϕ}}_i=73.9°$

Hence, the inclination of Earth’s magnetic field at the geomagnetic latitude $60°$is $73.9°$.

The wavelength,

${\mathrm{λ}}_m=90$

The magnitude of the Earth's magnetic field at north pole is,

$\begin{array}{rcl}B& =& \frac{{\mathrm{μ}}_0\mathrm{μ}}{4\mathrm{Ï€}{r}^3}×\sqrt{1+3{\sin }^2{\mathrm{λ}}_m}\\ & =& \frac{(4\mathrm{Ï€}×{10}^{−7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.})×(8.00×{10}^{22}\text{A}â‹\dots {\text{m}}^{\text{2}})}{4\mathrm{Ï€}×{(6.378×{10}^6\text{m})}^3}×\sqrt{1+3{\sin }^2(90°)}\\ & =& 6.20×{10}^{-5}\text{T}\end{array}$

Hence, the magnitude of Earth’s magnetic field at the north geomagnetic pole is $\begin{array}{rcl}& & 6.20×{10}^{-5}\text{T}\end{array}$.

Again use equation (2) to determine the inclination as below.

$\begin{array}{rcl}\tan {\mathrm{ϕ}}_i& =& 2×\tan {\mathrm{λ}}_m\\ & =& 2×\tan 90°\end{array}$

$\begin{array}{rcl}{\mathrm{ϕ}}_i& =& 90°\end{array}$

Hence, the inclination of Earth’s magnetic field at the north geomagnetic pole is$90°$.