A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 30° with the horizontal. The horizontal component of the earth's magnetic field at the place is B. The strength of the earth's magnetic field at the place is:

$\frac{2 B}{\sqrt{3}}$ $\frac{B}{\sqrt{3}}$ $B \sqrt{2}$ 2 B

$\frac{2 B}{\sqrt{3}}$

The correct answer is Option (1) → $\frac{2 B}{\sqrt{3}}$