Question

The magnetic field of Earth can be approximated as the magnetic field of a dipole. The horizontal and vertical components of this field at any distance r from Earth’s center are given by ${\mathbf{B}}_{\mathbf{H}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{\mu }}{\mathbf{4}{\mathbf{\pi r}}^{\mathbf{3}}}\mathbf{\times }{\mathbf{cos\lambda }}_{\mathbf{m}}\mathbf{}$,${\mathbf{B}}_{\mathbf{v}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{\mu }}{\mathbf{2}{\mathbf{\pi r}}^{\mathbf{3}}}\mathbf{\times }{\mathbf{sin\lambda }}_{\mathbf{m}}$where l_m is the magnetic latitude (this type of latitude is measured from the geomagnetic equator toward the north or south geomagnetic pole). Assume that Earth’s magnetic dipole moment has magnitude$\mathbf{\mu }\mathbf{=}\mathbf{}\mathbf{}\mathbf{8.00}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{1022}\mathbf{}\mathbf{A}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$ . (a) Show that the magnitude of Earth’s field at latitude lm is given by$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{\mu }}{\mathbf{4}{\mathbf{\pi r}}^{\mathbf{3}}}\mathbf{\times }\sqrt{\mathbf{1}\mathbf{+}\mathbf{3}{\mathbf{sin}}^{\mathbf{2}}{\mathbf{\lambda }}_{\mathbf{m}}\mathbf{}\mathbf{}}$

(b) Show that the inclination${\mathit{\varphi }}_{\mathbf{i}}$ of the magnetic field is related to the magnetic latitude${\mathbf{\lambda }}_{\mathbf{m}}$ by tan${\mathbf{\varphi }}_{\mathbf{i}}\mathbf{=}\mathbf{2}{\mathbf{tan\lambda }}_{\mathbf{m}}$ .

Short answer

a. $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\times \sqrt{1+3{\sin }^2{\mathrm{\lambda }}_{\mathrm{m}}}$

b.${\mathrm{tan\varphi }}_{\mathrm{i}}=2\tan {\mathrm{\lambda }}_{\mathrm{m}}$

Step-by-step solution

Listing the given quantities

${\mathrm{B}}_{\mathrm{H}}=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\times {\mathrm{cos\lambda }}_{\mathrm{m}}$

${\mathrm{B}}_{\mathrm{v}}=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{2{\mathrm{\pi r}}^3}\times {\mathrm{sin\lambda }}_{\mathrm{m}}$

Understanding the concepts of magnetic field

Here, we have to use Pythagoras theorem to find the magnitude of the earth’s magnetic field. The inclination of the magnetic field is found using the equation of the tangent ratio and the vertical and the horizontal component of the magnetic field.

Formula:

$\mathrm{B}=\sqrt{{\mathrm{B}}_{\mathrm{h}}^2+{\mathrm{B}}_{\mathrm{v}}^2}$

$\mathrm{tan\varphi }=\frac{{\mathrm{B}}_{\mathrm{v}}}{{\mathrm{B}}_{\mathrm{H}}}$

(a) Calculations of the B

$\begin{array}{c}\mathrm{B}=\sqrt{{\left(\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\times {\mathrm{cos\lambda }}_{\mathrm{m}}\right)}^2+{\left(\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{2{\mathrm{\pi r}}^3}\times {\mathrm{sin\lambda }}_{\mathrm{m}}\right)}^2}\\ =\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\times \sqrt{{\left({\mathrm{cos\lambda }}_{\mathrm{m}}\right)}^2+{\left(2{\mathrm{sin\lambda }}_{\mathrm{m}}\right)}^2}\\ =\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\times \sqrt{(1-{\sin }^2{\mathrm{\lambda }}_{\mathrm{m}})+4{\sin }^2{\mathrm{\lambda }}_{\mathrm{m}}}\\ =\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\times \sqrt{1+3{\sin }^2{\mathrm{\lambda }}_{\mathrm{m}}}\end{array}$

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\times \sqrt{1+3{\sin }^2{\mathrm{\lambda }}_{\mathrm{m}}}$

(b) Calculations of the inclination  ϕi of the magnetic field

$\begin{array}{c}{\mathrm{tan\varphi }}_{\mathrm{i}}=\frac{{\mathrm{B}}_{\mathrm{v}}}{{\mathrm{B}}_{\mathrm{h}}}\\ =\frac{\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{2{\mathrm{\pi r}}^3}\times {\mathrm{sin\lambda }}_{\mathrm{m}}}{\frac{{\mathrm{\mu }}_0\mathrm{\mu }}{4{\mathrm{\pi r}}^3}\times {\mathrm{cos\lambda }}_{\mathrm{m}}}\\ =2{\mathrm{tan\lambda }}_{\mathrm{m}}\end{array}$

${\mathrm{tan\varphi }}_{\mathrm{i}}=2\tan {\mathrm{\lambda }}_{\mathrm{m}}$