A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field
at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field
at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.

Answer –

Horizontal component of earth’s magnetic field is given by BH = 0.35 G

Angle made by the needle with the horizontal plane is known as the Angle of dip, given by δ=22°

Earth’s magnetic field strength is given by B

We can relate B and BH in following manner –

${{B}_{H}}=B\cos \delta $

Rearranging the above relation, we get –

$B=\frac{{{B}_{H}}}{\cos \delta }=\frac{0.35}{\cos {{22}^{\circ }}}$$$

$B=0.38G$

Hence, 0.38 G is the strength of the earth’s magnetic field at the given location.