A bar-magnet of moment of inertia 49 times 10 | vedclass

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(D) The time period $T$ of a vibrating bar magnet is given by the formula:$T = 2 \pi \sqrt{\frac{I}{MH}}$where $I$ is the moment of inertia,$M$ is the

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$5000$

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(D) The time period $T$ of a vibrating bar magnet is given by the formula:$T = 2 \pi \sqrt{\frac{I}{MH}}$where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the magnetic field induction.Squaring both sides,we get:$T^2 = 4 \pi^2 \frac{I}{MH}$Rearranging to solve for $M$:$M = \frac{4 \pi^2 I}{T^2 H}$Given values:$I = 49 	imes 10^{-2} \,kg-m^2$$H = 0.5 	imes 10^{-4} \,T$$T = 8.8 \,s$Substituting these values:$M = \frac{4 	imes (3.14)^2 	imes 49 	imes 10^{-2}}{(8.8)^2 	imes 0.5 	imes 10^{-4}}$$M = \frac{4 	imes 9.8596 	imes 49 	imes 10^{-2}}{77.44 	imes 0.5 	imes 10^{-4}}$$M = \frac{1932.48 	imes 10^{-2}}{38.72 	imes 10^{-4}}$$M = 49.909 	imes 10^2 \approx 5000 \,A-m^2$

$A$ bar-magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)

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