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

$A$ 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 angle of dip is $90^{\circ}$ at:

Fill in the blanks:
$(i)$ The declination is ...... at higher latitudes.
$(ii)$ The declination in India is ...... .

$A$ dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position,the angle of dip is $40^{\circ}$. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of $30^{\circ}$ with the magnetic meridian. In this position,the needle will dip by an angle:

$A$ magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at $22^{\circ}$ with the horizontal. The horizontal component of the earth's magnetic field at the place is known to be $0.35 \; G$. Determine the magnitude of the earth's magnetic field (in $G$) at the place.