Planets producing larger magnetic field have larger

$A$ 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 $^o$)?

The angle between the Earth's magnetic axis and the Earth's geographical axis is approximately $... ^\circ$.

$A$ magnetic needle is free to rotate in a vertical plane which makes an angle of $60^{\circ}$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$ with the horizontal,the true dip value at that place is: (in $^{\circ}$)

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 $\tan 37^{\circ} = \frac{3}{4}$)

If ${\phi_1}$ and ${\phi_2}$ are the angles of dip observed in two vertical planes at right angles to each other and ${\phi}$ is the true angle of dip,then: