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$
$\mathrm{V}=\mathrm{B}_{\mathrm{E}} \cos \delta, \mathrm{H}=\mathrm{B}_{\mathrm{E}} \sin \delta$
$\mathrm{V}=\mathrm{B}_{\mathrm{E}}, \mathrm{H}=\mathrm{B}_{\mathrm{E}} \tan \delta$
Two normal uniform magnetic field contain a magnetic needle making an angle $60°$ with $F$. Then the ratio of $\frac{F}{H}$ is
The angle of dip is the angle
At a point $A$ on the earth's surface the angle of $\operatorname{dip}, \delta=+25^{\circ} .$ At a point $B$ on the earth's surface the angle of dip, $\delta=-25^{\circ} .$ We can interpret that
The direction of the null points is on the equatorial line of a bar magnet, when the north pole of the magnet is pointing
The total intensity of earth's magnetic field at the magnetic equator is $5$ $units$. Its value at a magnetic latitude of $37^o $ is equal to