The Earth's magnetic field at the equator is approximately $0.4 \, G$. Calculate the Earth's magnetic dipole moment (nearly). (Radius of Earth $R = 6.4 \times 10^6 \, m$)

$A$ compass needle is free to rotate in a horizontal plane. Its magnetic moment is $60 \ Am^2$. When it is pointing geographical north,it experiences a torque of $1.2 \times 10^{-3} \ Nm$ due to the Earth's magnetic field. If the Earth's magnetic field in the horizontal direction is $40 \times 10^{-6} \ T$,what is the declination in degrees at that place?

$A$ long vertical wire carries a steady current of $5.0 \, A$. $A$ sensitive magnetic compass is placed in a plane perpendicular to the wire and $10.0 \, cm$ south of it. It registers a deflection of $60^{\circ}$ north of east. The magnitude of the horizontal component of the earth's magnetic field is (permeability of free space is $4 \pi \times 10^{-7} \, N/A^2$):

Give the location of magnetic:
$(i)$ north pole
$(ii)$ south pole

Explain magnetic declination.

The vertical component of the earth's magnetic field is $6 \times 10^{-5} \text{ T}$ at a place where the angle of dip is $37^{\circ}$. The earth's resultant magnetic field at that place will be (Given $\tan 37^{\circ} = \frac{3}{4}$)