$A$ 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 $^o$)?

Planets producing larger magnetic field have larger

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 the vertical component of the earth's magnetic field will be ........ $\times 10^{-4} \; Wb/m^2$.

$A$ dip needle in a plane perpendicular to the magnetic meridian will remain:

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 the dip angle is $\pm 45^{\circ}$.

The earth's magnetic field lines resemble that of a dipole at the centre of the earth. If the magnetic moment of this dipole is close to $8 \times 10^{22} \text{ Am}^2$,the value of earth's magnetic field near the equator is close to $.... \text{ Gauss}$ (radius of the earth $= 6.4 \times 10^6 \text{ m}$)