Question

The period of oscillation of a freely suspended bar magnet about centre of mass is $T_0$ in a uniform magnetic field $B_0$ (Figure-I). Now the magnet are cut into three identical parts and two part again suspended in same magnetic field as shown (Figure-II). The new time period of oscillation of system is

Answer 1

T_0/3

Answer 2

The period of oscillation of a freely suspended bar magnet in a uniform magnetic field $B_0$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_0}}$ where $I$ is the moment of inertia of the magnet about its axis of oscillation, and $M$ is its magnetic moment.

Figure-I: Original Magnet Let the original bar magnet have length $L$, mass $m$, and pole strength $q$. Its magnetic moment is $M = qL$. Its moment of inertia about its center (perpendicular to its length) is $I = \frac{1}{12}mL^2$. The time period of oscillation is $T_0 = 2\pi \sqrt{\frac{I}{MB_0}}$.

Figure-II: Cut Magnets The original magnet is cut into three identical parts. Each part will have:

1. Length ($L'$): $L' = L/3$
2. Mass ($m'$): $m' = m/3$ (assuming uniform density)
3. Pole Strength ($q'$): When a magnet is cut perpendicular to its length, the pole strength of each new piece remains the same as the original magnet, so $q' = q$.
4. Magnetic Moment ($M'$): For each small part, the magnetic moment will be $M' = q'L' = q(L/3) = (qL)/3 = M/3$.
5. Moment of Inertia ($I'$): For each small part, the moment of inertia about its center will be: $I' = \frac{1}{12}m'(L')^2 = \frac{1}{12}\left(\frac{m}{3}\right)\left(\frac{L}{3}\right)^2 = \frac{1}{12}\frac{m}{3}\frac{L^2}{9} = \frac{1}{27}\left(\frac{1}{12}mL^2\right) = \frac{I}{27}$.

In Figure-II, two of these identical parts are suspended side-by-side, with their magnetic moments aligned in the same direction (N-S next to N-S). Therefore, they act as a single system with combined magnetic moment and moment of inertia.

1. Total Magnetic Moment of the system ($M_{sys}$): $M_{sys} = M' + M' = 2M' = 2\left(\frac{M}{3}\right) = \frac{2M}{3}$.
2. Total Moment of Inertia of the system ($I_{sys}$): $I_{sys} = I' + I' = 2I' = 2\left(\frac{I}{27}\right) = \frac{2I}{27}$.

New Time Period of Oscillation ($T'$): The new time period of oscillation for this system in the same magnetic field $B_0$ will be: $T' = 2\pi \sqrt{\frac{I_{sys}}{M_{sys}B_0}}$ Substitute the values of $I_{sys}$ and $M_{sys}$: $T' = 2\pi \sqrt{\frac{2I/27}{(2M/3)B_0}}$ $T' = 2\pi \sqrt{\frac{2I}{27} \times \frac{3}{2MB_0}}$ $T' = 2\pi \sqrt{\frac{6I}{54MB_0}}$ $T' = 2\pi \sqrt{\frac{I}{9MB_0}}$ $T' = 2\pi \frac{1}{3} \sqrt{\frac{I}{MB_0}}$ Since $T_0 = 2\pi \sqrt{\frac{I}{MB_0}}$, we can substitute $T_0$ into the equation: $T' = \frac{T_0}{3}$

The new time period of oscillation of the system is $T_0/3$.

The final answer is $T_0/3$.