Two bar magnets of the same mass,length,and b | vedclass

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(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia and $M$ is the net mag

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$3\sqrt{3} \, s$

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(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia and $M$ is the net magnetic moment.Since both magnets have the same mass,length,and breadth,their moment of inertia $I$ is the same.In the sum position,the net magnetic moment is $M_{sum} = M + 2M = 3M$. Given $T_1 = 3 \, s$,we have $3 \propto \frac{1}{\sqrt{3M}}$.In the difference position,the net magnetic moment is $M_{diff} = 2M - M = M$. Let the new time period be $T_2$.Thus,$T_2 \propto \frac{1}{\sqrt{M}}$.Taking the ratio: $\frac{T_2}{T_1} = \sqrt{\frac{3M}{M}} = \sqrt{3}$.Therefore,$T_2 = 3 	imes \sqrt{3} = 3\sqrt{3} \, s$.

Two bar magnets of the same mass,length,and breadth,but with magnetic moments $M$ and $2M$ respectively,have a time period of $3 \, s$ when placed in the same position (sum position). What will be the time period when they are placed in the different position (difference position)?

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