Two normal uniform magnetic field contain a magnetic needle making an angle ${60^ \circ }$ with F. Then the ratio of $\dfrac{F}{H}$​ is ?
A. $1:2$
B. $2:1$
C. $\sqrt 3 :1$
D. $1:\sqrt 3 $

Hint: We are aware that the Earth's magnetosphere is made up of the geomagnetic field, also known as the Earth's magnetic field, which effectively reaches into space for thousands of kilometres. Near the equator, when the field is horizontal, its strength is at its lowest. Gauss units are used to express the magnetic field's intensity.

Formula used:
$H = F\tan \theta $
where $\theta $ defines the angle through which the needle deflects.

Complete step by step solution:
We are aware that the portion of a material's magnetic field that results from an external current and is not intrinsic to the material itself is known as magnetic field strength, also known as magnetic intensity or magnetic field intensity.

We can also draw the conclusion that an angle is a measurement of how far an object is moving from its intended course. Measured between the line of sight to the target and the line of sight to the aiming point, the angle of a deflection shot in gunnery.

We know that $H = F\tan \theta $ where $\theta $ defines the angle through which the needle deflects.
So, at the equilibrium of needle we can say that putting$\theta = {60^ \circ }$
$\tan {60^ \circ } = \dfrac{H}{F}$
$ \Rightarrow \sqrt 3 = \dfrac{H}{F}$
$\therefore \dfrac{F}{H} = \dfrac{1}{{\sqrt 3 }}$

Hence the correct option is D.

Note: It is important to keep in mind the following facts about the magnetic field:
- There are two different types of magnetic fields: the B-field and the H-field.
- Teslas are used to measure B-field (T). A/m units are used to express the H-field.
- The earth's magnetic field is oriented so that it extends from its magnetic north pole to its magnetic south pole. The inclined bar magnet is merely a fictitious magnet; it does not exist.