A thin magnet has small-oscillation time period T in a uniform magnetic field. It is bent sharply at its midpoint so that the magnetic-moment vectors of its two equal halves make an angle 120° with each other. Assume its moment of inertia about the suspension axis remains unchanged.
The new time period is:

Question

A thin magnet has small-oscillation time period T in a uniform magnetic field. It is bent sharply at its midpoint so that the magnetic-moment vectors of its two equal halves make an angle 120° with each other. Assume its moment of inertia about the suspension axis remains unchanged.

The new time period is:

G1 Physics · Accepted Answer

Each half has magnetic moment M/2. The two half-moment vectors make angle 120°. Therefore, M' = √[(M/2)²+(M/2)²+2(M/2)²cos120°]. Since cos120° = -1/2, M' = √[M²/4+M²/4-M²/4] = M/2. For small oscillations, T ∝ 1/√M when I and B are unchanged. Thus, T'/T = √(M/M') = √(M/(M/2)) = √2. Hence, T' = √2T.