The dip angle in a vertical plane at an angle of $\cos^{-1} \left( \frac{1}{\sqrt{2}} \right)$ from the magnetic meridian is $60^\circ$. Find the actual dip angle at that place.

$A$ compass needle whose magnetic moment is $60 \, A \cdot m^2$ is pointing towards the geographical north at a certain place,where the horizontal component of the Earth's magnetic field is $40 \, \mu Wb/m^2$. It experiences a torque of $1.2 \times 10^{-3} \, N \cdot m$. What is the angle of declination at this place (in $^o$)?

Assume the dipole model for Earth's magnetic field $B$,which is given by:
$B_{V} = \text{vertical component of magnetic field} = \frac{\mu_{0}}{4\pi} \frac{2m \cos \theta}{r^{3}}$
$B_{H} = \text{horizontal component of magnetic field} = \frac{\mu_{0}}{4\pi} \frac{m \sin \theta}{r^{3}}$
where $\theta = 90^{\circ} - \text{latitude}$ as measured from the magnetic equator.
$(a)$ Find the loci of points for which $|\vec{B}|$ is minimum.

The correct relation is:
$B_H$ = Horizontal component of earth's magnetic field; $B_V$ = Vertical component of earth's magnetic field and $B$ = Total intensity of earth's magnetic field.

At a given place on the Earth,the angle between the Magnetic Meridian and the Geographic Meridian is called . . . . . . .

At the magnetic poles of the earth,a compass needle will be