The vertical component of the Earth's magnetic field is zero at:

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

At the place where the angle of dip is $60^{\circ}$,what is the relation between the vertical component $Z_{E}$ and the horizontal component $H_{E}$ of the Earth's magnetic field?

$A$ magnetic compass needle oscillates $30$ times per minute at a place where the angle of dip is $45^{\circ}$,and $40$ times per minute where the angle of dip is $30^{\circ}$. If $B_1$ and $B_2$ are respectively the total magnetic field due to the earth at the two places,then the ratio $B_1/B_2$ is best given by:

$A$ dip needle in a plane perpendicular to the magnetic meridian will remain:

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