Use the results displayed in Problem 61 to predict the (a) magnitude and (b) inclination of Earth’s magnetic field at the geomagnetic equator, the (c) magnitude and (d) inclination at geomagnetic latitude $\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$, and the (e) magnitude and (f) inclination at the north geomagnetic pole.

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,

$B=\frac{{\mu }_0\mu }{4\pi r^3}\times \sqrt{1+3{\sin }^2{\lambda }_m}$ ..... (1)

And the inclination of Earth’s magnetic field is,

$\tan \varphi =2\tan {\lambda }_m$ ..... (2)

Earth’s magnetic dipole moment,$\mu =8.00\times {10}^{22}\text{A}\cdot {\text{m}}^{\text{2}}$

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

The permeability of free space,${\mu }_0=4\pi \times {10}^{-7}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.$

On equator, the wavelength is,

${\lambda }_m=0$

So the magnetic field using equation (1) is,

$\begin{array}{rcl}B& =& \frac{{\mu }_0\mu }{4\pi r^3}\times \sqrt{1+3{\sin }^2{\lambda }_m}\\ & =& \frac{(4\pi \times {10}^{-7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.})\times (8.00\times {10}^{22}\text{A}\cdot {\text{m}}^{\text{2}})}{4\pi \times {(6.378\times {10}^6\text{m})}^3}\times \sqrt{1+3{\sin }^2\left(0\right)}\\ & =& 3.10\times {10}^{-5}\text{T}\end{array}$

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

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

$\begin{array}{rcl}\tan {\varphi }_i& =& 2\times \tan {\lambda }_m\\ & =& 2\times \tan 0°\end{array}$

${\varphi }_i=0°$

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

The wavelength is,

${\lambda }_m=60$

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

$\begin{array}{rcl}B& =& \frac{{\mu }_0\mu }{4\pi r^3}\times \sqrt{1+3{\sin }^2{\lambda }_m}\\ & =& \frac{(4\pi \times {10}^{-7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.})\times (8.00\times {10}^{22}\text{A}\cdot {\text{m}}^{\text{2}})}{4\pi \times {(6.378\times {10}^6\text{m})}^3}\times \sqrt{1+3{\sin }^2(60°)}\\ & =& 5.59\times {10}^{-5}\text{T}\end{array}$

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

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

$\begin{array}{rcl}\tan {\varphi }_i& =& 2\times \tan {\lambda }_m\\ & =& 2\times \tan 60°\end{array}$

${\varphi }_i=73.9°$

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

The wavelength,

${\lambda }_m=90$

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

$\begin{array}{rcl}B& =& \frac{{\mu }_0\mu }{4\pi r^3}\times \sqrt{1+3{\sin }^2{\lambda }_m}\\ & =& \frac{(4\pi \times {10}^{-7}{\displaystyle \raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{A}}^2$}\right.})\times (8.00\times {10}^{22}\text{A}\cdot {\text{m}}^{\text{2}})}{4\pi \times {(6.378\times {10}^6\text{m})}^3}\times \sqrt{1+3{\sin }^2(90°)}\\ & =& 6.20\times {10}^{-5}\text{T}\end{array}$

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

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

$\begin{array}{rcl}\tan {\varphi }_i& =& 2\times \tan {\lambda }_m\\ & =& 2\times \tan 90°\end{array}$

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

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