Assume the dipole model for earth’s magnetic field $\mathrm{B}$ which is given by

${{\rm{B}}_{\rm{v}}} = $ vertical component of magnetic field

$ = \frac{{{\mu _0}}}{{4\pi }}\frac{{2m\,\cos \theta }}{{{r^3}}}$

${{\rm{B}}_H} = $ Horizontal component of magnetic field

${{\rm{B}}_H} = \frac{{{\mu _0}}}{{4\pi }}\frac{{m\,\sin \theta }}{{{r^3}}}$

$\theta $ $= 90^{°}$ -latitude as measured from magnetic equator.

$(a)$ Find loci of points for which : $\left| {{\rm{\vec B}}} \right|$ is minimum;

$(a)$ Given that, $\mathrm{B}_{\mathrm{V}}=\frac{\mu_{0}}{4 \pi} \frac{2 m \cos \theta}{r^{3}}$

$\mathrm{B}_{\mathrm{H}}=\frac{\mu_{0}}{4 \pi} \frac{m \sin \theta}{r^{3}}$

- Squaring and adding equation $(1)$ and $(2)$,

$B_{\mathrm{V}}^{2}+B_{\mathrm{H}}^{2}=\left(\frac{\mu_{0}}{4 \pi}\right)^{2} \frac{m^{2}}{r^{6}}\left[4 \cos ^{2} \theta+\sin ^{2} \theta\right]$

$\therefore B^{2}=\left(\frac{\mu_{0}}{4 \pi}\right)^{2} \frac{m^{2}}{r^{6}}\left[4 \cos ^{2} \theta+1-\cos ^{2} \theta\right]$

$\therefore B=\sqrt{B_{\mathrm{V}}^{2}+B_{\mathrm{H}}^{2}}$

$=\sqrt{\left(\frac{\mu_{0}}{4 \pi}\right)^{2} \frac{m^{2}}{r^{6}}\left[3 \cos ^{2} \theta+1\right]}$

$=\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}\left[3 \cos ^{2} \theta+1\right]^{1 / 2}$

From above equation, the value of $B$ is minimum if $\cos \theta=0$, hence $\theta=\frac{\pi}{2}$ Thus, $\mathrm{B}$ is minimum at the magnetic equator.

$(b)$ For angle of dip,

$\tan \delta=\frac{B_{\mathrm{V}}}{B_{\mathrm{H}}}=\frac{2\left(\frac{\mu_{0}}{4 \pi} \frac{m \cos \theta}{r^{3}}\right)}{\left(\frac{\mu_{0}}{4 \pi} \frac{m \sin \theta}{r^{3}}\right)}$

$=2 \frac{\cos \theta}{\sin \theta}$

$\frac{B_{\mathrm{V}}}{\mathrm{B}_{\mathrm{H}}}=\tan \delta=2 \cot \theta$

For dip angle $\delta=0$, then

$\cot \theta=0$

$\therefore \quad \theta=\frac{\pi}{2}$

Hence locus is on magnetic equator.