If the angles of dip at two places are $30^{\circ}$ and $45^{\circ}$ respectively,then the ratio of horizontal components of earth's magnetic field at the two places will be

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.

$A$ line passing through places having zero value of magnetic dip is called

$A$ 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 $^o$)?

$A$ long straight horizontal cable carries a current of $2.5\;A$ in the direction $10^{\circ}$ south of west to $10^{\circ}$ north of east. The magnetic meridian of the place happens to be $10^{\circ}$ west of the geographic meridian. The earth's magnetic field at the location is $0.33\;G,$ and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points,magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth's magnetic field.)

At a place,the earth's horizontal component of magnetic field is $0.36 \times 10^{-4} \ Wb/m^2$. If the angle of dip at that place is $60^o$,then the vertical component of the earth's magnetic field at that place in $Wb/m^2$ will be approximately: