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A thin rectangular bar magnet is freely suspended and executes small oscillations with period T. The magnet is then cut into two identical parts, each having half of the original length. If one of these halves is suspended freely in the same magnetic field and allowed to oscillate, its period is T'. What is the value of T'/T?

1. 1/(22)
2. 1/2
3. 2
4. 1/4

Correct answer: 1/2

Solution

Cutting into halves of half length: moment of inertia becomes I/8 and magnetic moment becomes M/2. T' = 2*pi*sqrt((I/8)/((M/2)*B)) = 2*pi*sqrt(I/(4*M*B)) = T/2, so T'/T = 1/2.

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