Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of $$3 \text{ s}$$ and $$4 \text{ s}$$ respectively. If their moments of inertia are in the ratio of $$3:2$$ then the ratio of their magnetic moments will be

Two bar magnets oscillate in Earth's magnetic field with time periods $$T_1 = 3$$ s and $$T_2 = 4$$ s, and their moments of inertia are in the ratio $$I_1 : I_2 = 3 : 2$$. We need the ratio of their magnetic moments.

Since the time period of a bar magnet oscillating in a magnetic field $$B$$ is given by $$T = 2\pi\sqrt{\frac{I}{MB}},$$ where $$I$$ is the moment of inertia and $$M$$ is the magnetic moment, squaring this expression yields $$T^2 = 4\pi^2 \frac{I}{MB} \implies M = \frac{4\pi^2 I}{T^2 B}.$$

Therefore, the ratio of the magnetic moments becomes $$\frac{M_1}{M_2} = \frac{I_1}{I_2} \times \frac{T_2^2}{T_1^2}.$$

Substituting the given ratios into the above expression yields $$\frac{M_1}{M_2} = \frac{3}{2} \times \frac{16}{9} = \frac{48}{18} = \frac{8}{3}.$$

Hence, the correct answer is Option B: $$8:3$$.