A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip $$30^\circ$$. The number of oscillations per minute become 10 at another place Q of $$60^\circ$$ dip. The ratio of the total magnetic field at the two places $$(B_Q : B_P)$$ is:

A compass needle oscillates 20 times/min at place P (dip $$30^\circ$$) and 10 times/min at place Q (dip $$60^\circ$$). Find $$B_Q : B_P$$.

Recall that the frequency of oscillation of a magnetic needle is given by $$f = \frac{1}{2\pi}\sqrt{\frac{MB_H}{I}}$$, where $$B_H$$ is the horizontal component of the magnetic field, $$M$$ is the magnetic moment, and $$I$$ is the moment of inertia. Since $$f \propto \sqrt{B_H}$$, it follows that $$f^2 \propto B_H$$.

The horizontal component of the magnetic field relates to the total magnetic field by $$B_H = B\cos\delta$$, where $$\delta$$ is the angle of dip. Therefore, at place P, $$B_{H_P} = B_P \cos 30^\circ = B_P \cdot \frac{\sqrt{3}}{2}$$ and at place Q, $$B_{H_Q} = B_Q \cos 60^\circ = B_Q \cdot \frac{1}{2}$$.

Using the ratio of squared frequencies, we write $$\frac{f_P^2}{f_Q^2} = \frac{B_{H_P}}{B_{H_Q}}$$. Substituting $$f_P = 20$$ and $$f_Q = 10$$ yields $$\frac{20^2}{10^2} = \frac{B_P \cdot \frac{\sqrt{3}}{2}}{B_Q \cdot \frac{1}{2}}$$, which simplifies to $$4 = \frac{B_P \sqrt{3}}{B_Q}$$. From this it follows that $$\frac{B_Q}{B_P} = \frac{\sqrt{3}}{4}$$.

Hence, the required ratio of the magnitudes of the magnetic fields is $$B_Q : B_P = \sqrt{3} : 4$$. Therefore, the answer is Option A: $$\sqrt{3} : 4$$.