At a place,the earth's horizontal component of magnetic field is $0.36 \times 10^{-4} \ Wb/m^2$. If the angle of dip at that place is $60^o$,then the vertical component of the earth's magnetic field at that place in $Wb/m^2$ will be approximately:

At a certain place,the horizontal component ${B_0}$ and the vertical component ${V_0}$ of the Earth's magnetic field are equal in magnitude. The total intensity at the place will be

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:

The magnetic field due to the Earth is closely equivalent to that due to:

$A$ 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?

The correct relation is:
$B_H$ = Horizontal component of earth's magnetic field; $B_V$ = Vertical component of earth's magnetic field and $B$ = Total intensity of earth's magnetic field.