The angles of dip are $30^{\circ}$ and $45^{\circ}$ at two different places. The ratio of the horizontal components of the Earth's magnetic field at these places will be:

The value of horizontal component of earth's magnetic field at a place is $0.35 \times 10^{-4} \,T$. If the angle of dip is $60^{\circ}$,the value of vertical component of earth's magnetic field is nearly ............. $\times 10^{-4} \,T$.

$A$ 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 $^o$)?

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

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

The Earth's magnetic field at a certain place has a horizontal component of $0.3 \ G$ and the total strength of $0.5 \ G$. The angle of dip is: