In a uniform magnetic field,a magnetic needle | vedclass

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(A) The time period $T$ of a magnetic needle oscillating in a uniform magnetic field $B$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$.Given

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(A) The time period $T$ of a magnetic needle oscillating in a uniform magnetic field $B$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$.Given: Magnetic moment $M = 9.85 	imes 10^{-2} \, A \cdot m^{2}$,Moment of inertia $I = 5 	imes 10^{-6} \, kg \cdot m^{2}$.The needle performs $10$ oscillations in $5 \, s$,so the time period $T = \frac{5}{10} = 0.5 \, s$.Substituting the values into the formula: $0.5 = 2\pi \sqrt{\frac{5 	imes 10^{-6}}{9.85 	imes 10^{-2} 	imes B}}$.Squaring both sides: $0.25 = 4\pi^{2} \left( \frac{5 	imes 10^{-6}}{9.85 	imes 10^{-2} 	imes B} ight)$.Using $\pi^{2} = 9.85$: $0.25 = 4 	imes 9.85 	imes \frac{5 	imes 10^{-6}}{9.85 	imes 10^{-2} 	imes B}$.$0.25 = 4 	imes \frac{5 	imes 10^{-6}}{10^{-2} 	imes B} = \frac{20 	imes 10^{-6}}{10^{-2} 	imes B} = \frac{20 	imes 10^{-4}}{B}$.$B = \frac{20 	imes 10^{-4}}{0.25} = 80 	imes 10^{-4} = 8 	imes 10^{-3} \, T$.Since $1 \, T = 1000 \, mT$,$B = 8 \, mT$.

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