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A vibration magnetometer has two identical bar magnets arranged one above the other so that they are mutually perpendicular and each bisects the other. When this system oscillates in a horizontal magnetic field, its time period is 2^(3/4) s. If one magnet is taken away and the remaining magnet is allowed to oscillate in the same field, what is its time period in seconds?

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

Correct answer: 2^(1/2)

Solution

Two perpendicular identical magnets: net moment = sqrt2*M, inertia = 2I, so T_two = 2*pi*sqrt(2I/(sqrt2*M*B)) = 2^(1/4)*T_one. Thus T_one = 2^(3/4)/2^(1/4) = 2^(1/2) s.

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