At two different places,the angles of dip are respectively $30^{\circ}$ and $45^{\circ}$. At these two places,the ratio of the horizontal component of Earth's magnetic field is:

$A$ 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 $4 \pi \times 10^{-7} \, N/A^2$):

At a certain place,the horizontal component of the Earth's magnetic field is $B_0$ and the angle of dip is $45^o$. The total intensity of the Earth's magnetic field at that place 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 $R = 6.4 \times 10^6 \, m$)

If the angle of dip at places $A$ and $B$ are $30^{\circ}$ and $45^{\circ}$ respectively,the ratio of the horizontal component of the Earth's magnetic field at $A$ to that at $B$ will be.
$[\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \quad \sin 30^{\circ}=\frac{1}{2}, \quad \cos 30^{\circ}=\frac{\sqrt{3}}{2}]$

$A$ 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 $^o$)?