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Question

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

Vedclass · Accepted Answer

$1.6 	imes 10^{-4} 	ext{ T}$

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(B) The time period of oscillation for a magnetic needle in a magnetic field is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB}}$.Given: $T = 2 	ext{ s}$,$I = 8 	imes 10^{-6} 	ext{ kg m}^2$,$M = 5 	imes 10^{-2} 	ext{ A m}^2$.Substituting the values into the formula: $2 = 2\pi \sqrt{\frac{8 	imes 10^{-6}}{5 	imes 10^{-2} 	imes B}}$.Dividing by $2$: $1 = \pi \sqrt{\frac{8 	imes 10^{-6}}{5 	imes 10^{-2} 	imes B}}$.Squaring both sides: $1 = \pi^2 \left( \frac{8 	imes 10^{-6}}{5 	imes 10^{-2} 	imes B} ight)$.Rearranging for $B$: $B = \frac{\pi^2 	imes 8 	imes 10^{-6}}{5 	imes 10^{-2}}$.Using $\pi^2 \approx 9.86$ (or approximately $10$): $B = \frac{9.86 	imes 8 	imes 10^{-6}}{5 	imes 10^{-2}} \approx 1.577 	imes 10^{-4} 	ext{ T}$.Rounding to the nearest option,$B \approx 1.6 	imes 10^{-4} 	ext{ T}$.

$A$ magnetic needle has a magnetic moment of $5 \times 10^{-2} \text{ A m}^2$ and a moment of inertia of $8 \times 10^{-6} \text{ kg m}^2$. It has a period of oscillation of $2 \text{ s}$ in a magnetic field $\vec{B}$. The magnitude of the magnetic field is approximately:

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