$A$ magnetic needle is free to rotate in a vertical plane which makes an angle of $60^{\circ}$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$ with the horizontal,the true dip value at that place is: (in $^{\circ}$)

Explain the angle of dip.

At an angle of $30^{\circ}$ to the magnetic meridian,the apparent dip is $45^{\circ}$. Find the true dip.

The angle of dip at a place is $40.6^\circ$ and the intensity of the vertical component of the earth's magnetic field $V = 6 \times 10^{-5} \text{ Tesla}$. The total intensity of the earth's magnetic field $(I)$ at this place is:

$A$ magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at $30^{\circ}$ with the horizontal. The horizontal component of the earth's magnetic field at the place is $0.3 \ G$. Then the magnitude of the earth's magnetic field at the location is

$A$ 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.)