The relations amongst the three elements elements earth's magnettic fleld, namely horizontal component $H$, vertical component $V$ and dip $\delta$ are, $( B_{E}=$ total magnetic fleld)
$\mathrm{V}=\mathrm{B}_{\mathrm{E}} \tan \delta, \mathrm{H}=\mathrm{B}_{\mathrm{E}}$
$ \mathrm{V}=\mathrm{B}_{\mathrm{E}} \sin \delta, \mathrm{H}=\mathrm{B}_{\mathrm{E}} \cos \delta$
$\mathrm{V}=\mathrm{B}_{\mathrm{E}} \cos \delta, \mathrm{H}=\mathrm{B}_{\mathrm{E}} \sin \delta$
$\mathrm{V}=\mathrm{B}_{\mathrm{E}}, \mathrm{H}=\mathrm{B}_{\mathrm{E}} \tan \delta$
At a place, the magnitudes of the horizontal component and total intensity of the magnetic field of the earth are $0.3 $ and $ 0.6$ Oersted respectively. The value of the angle of dip at this place will be.....$^o$
A current carrying coil is placed with its axis perpendicular to $N-S$ direction. Let horizontal component of earth's magnetic field be $H_0$ and magnetic field inside the loop be $H$. If a magnet is suspended inside the loop, it makes angle $\theta $ with $H$. Then $\theta =$
The vertical component of the earth's magnetic field is zero at a place where the angle of dip is.....$^o$
At a place, if the earth's horizontal and vertical components of magnetic fields are equal, then the angle of dip will be.......$^o$
If $\theta _1$ and $\theta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\theta$ is given by