Q.From separate magnet method of vibrational magnetometer, describe the method of comparison of magnetic moment of two magnets under the following points:

Each magnet is oscillated (in turn) in the vibration magnetometer's earth-field box; since the period T ∝ 1/√M, the ratio of the two magnets' magnetic moments equals the inverse-square ratio of their periods.

1. Derivation of the formula: A small bar magnet, free to oscillate in a horizontal plane under the earth's horizontal magnetic field $H$, executes simple harmonic oscillations (for small angular amplitude) with time period: $T = 2\pi\sqrt{\frac{I}{MH}}$ where $I$ is the magnet's moment of inertia about the vertical suspension axis and $M$ its magnetic moment. Squaring and rearranging: $M = \frac{4\pi^2 I}{T^2 H}$ Separate magnet method: Place magnet 1 (moment $M_1$) in the vibration magnetometer and measure its period $T_1$. Remove it and place magnet 2 (moment $M_2$, same size/mass so moment of inertia $I$ is (approximately) the same) and measure $T_2$. Since $I$ and $H$ are common to both measurements: $M_1 = \frac{4\pi^2 I}{T_1^2 H}, \qquad M_2 = \frac{4\pi^2 I}{T_2^2 H}$ Dividing: $\boxed{\frac{M_1}{M_2} = \frac{T_2^2}{T_1^2}}$ So the magnet with the shorter period has the larger magnetic moment.
2. Observation table (typical form):

(Repeat each timing 2–3 times and take the mean for accuracy.)

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