In a uniform magnetic field of 0.049 T, a ma | vedclass

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(C) The time period of oscillation for a magnetic needle in a uniform magnetic field is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$, where $I$ is the mom

Vedclass · Accepted Answer

$1280$

Vedclass · Answer

(C) The time period of oscillation for a magnetic needle in a uniform magnetic field is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$, where $I$ is the moment of inertia, $M$ is the magnetic moment, and $B$ is the magnetic field strength.Given: $B = 0.049 \ T$, $I = 9.8 	imes 10^{-5} \ kg \ m^2$, and the needle performs $20$ oscillations in $5 \ s$.The time period $T = \frac{5 \ s}{20} = 0.25 \ s = \frac{1}{4} \ s$.Substituting the values into the formula: $\frac{1}{4} = 2 \pi \sqrt{\frac{9.8 	imes 10^{-5}}{M 	imes 0.049}}$.Squaring both sides: $\frac{1}{16} = 4 \pi^2 	imes \frac{9.8 	imes 10^{-5}}{M 	imes 0.049}$.Rearranging for $M$: $M = \frac{4 \pi^2 	imes 9.8 	imes 10^{-5} 	imes 16}{0.049} = \frac{4 \pi^2 	imes 9.8 	imes 10^{-5} 	imes 16}{0.049}$.Since $9.8 / 0.049 = 200$, we get $M = 4 \pi^2 	imes 200 	imes 16 	imes 10^{-5} = 12800 \pi^2 	imes 10^{-5} \ A \ m^2$.Wait, re-calculating: $M = \frac{4 \pi^2 	imes 9.8 	imes 10^{-5} 	imes 16}{49 	imes 10^{-3}} = 4 \pi^2 	imes 0.2 	imes 16 	imes 10^{-2} = 12.8 \pi^2 	imes 10^{-2} = 1280 \pi^2 	imes 10^{-5} \ A \ m^2$.Thus, $x = 1280 \pi^2$.

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