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Q.

An experiment consists of two identical bar magnets, placed one over the other, such that they are mutually perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is 4 second. If one of the magnets is taken away, the period of oscillation of the other in the same field is (in second)

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answer is 3.

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Detailed Solution

$T = 2 π \sqrt{\frac{I}{M B}}$

$M^{1} = \sqrt{M^{2} + M^{2}} = \sqrt{2} M & I^{1} = I + I = 2 I, T^{1} = 2 π \sqrt{\frac{2 I}{\sqrt{2} M B}} \Rightarrow T^{1} = 2 π \sqrt{\frac{\sqrt{2} I}{M B}}$

When one of the magnets is taken away, then $M^{11} = M & I^{11} = I$

$∴ T^{11} = 2 π \sqrt{\frac{I}{M B}} & \frac{T^{11}}{T^{1}} = \frac{1}{{(2)}^{\frac{1}{4}}} \Rightarrow T^{11} = \frac{T^{1}}{{(2)}^{\frac{1}{4}}} = \frac{4}{{(2)}^{\frac{1}{4}}} = 3.34 S e c .$

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