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 : dip angle is $ \pm $ $45^{°}$

At a point $A$ on the earth's surface the angle of $\operatorname{dip}, \delta=+25^{\circ} .$ At a point $B$ on the earth's surface the angle of dip, $\delta=-25^{\circ} .$ We can interpret that

A magnetic needle oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^{\circ}$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

The value of angle of dip is zero at the magnetic equator because on it

The horizontal component of the earth's magnetic field at any place is $0.36 \times 10^{-4} Wb / m ^{2}$. If the angle of dip at that place is $60^{\circ}$, then the value of vertical component of the earth's magnetic field will be ........ $\times 10^{-4}\;W b / m^{2}$

The angle of dip is the angle