The angle of dip at a place is $30^{\circ}$,and the horizontal component of the Earth's magnetic field is $5 \times 10^{-5} \, T$. Calculate the vertical component and the total magnetic field at that place.

$A$ 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:

$A$ long straight horizontal cable carries a current of $2.5\;A$ in the direction $10^{\circ}$ south of west to $10^{\circ}$ north of east. The magnetic meridian of the place happens to be $10^{\circ}$ west of the geographic meridian. The earth's magnetic field at the location is $0.33\;G,$ and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points,magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth's magnetic field.)

What is magnetic declination?

At a certain place,the angle of dip is $30^{\circ}$ and the horizontal component of the Earth's magnetic field is $0.50 \text{ Oersted}$. The Earth's total magnetic field is:

$A$ compass needle oscillates $20$ times per minute at a place where the dip is $30^{\circ}$ and $30$ times per minute where the dip is $60^{\circ}$. The ratio of total magnetic field due to the earth at the two places respectively is $\frac{4}{\sqrt{x}}$. The value of $x$ is $............$.