A magnetic needle of magnetic moment 6 times | vedclass

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(C) The time period $T$ of a magnetic needle performing simple harmonic motion in a magnetic field is given by the formula:$T = 2 \pi \sqrt{\frac{I}{M

Vedclass · Accepted Answer

$25.12$

Vedclass · Answer

(C) The time period $T$ of a magnetic needle performing simple harmonic motion in a magnetic field is given by the formula:$T = 2 \pi \sqrt{\frac{I}{MB}}$Given:Magnetic moment $M = 6 	imes 10^{-2} 	ext{ A m}^2$Moment of inertia $I = 9.6 	imes 10^{-5} 	ext{ kg m}^2$Magnetic field $B = 0.01 	ext{ T}$Substituting the values:$T = 2 	imes 3.14 	imes \sqrt{\frac{9.6 	imes 10^{-5}}{6 	imes 10^{-2} 	imes 0.01}}$$T = 6.28 	imes \sqrt{\frac{9.6 	imes 10^{-5}}{6 	imes 10^{-4}}}$$T = 6.28 	imes \sqrt{0.16}$$T = 6.28 	imes 0.4 = 2.512 	ext{ s}$The time taken for $10$ oscillations is $10 	imes T = 10 	imes 2.512 = 25.12 	ext{ s}$.

$A$ magnetic needle of magnetic moment $6 \times 10^{-2} \text{ A m}^2$ and moment of inertia $9.6 \times 10^{-5} \text{ kg m}^2$ performs simple harmonic motion in a magnetic field of $0.01 \text{ T}$. The time taken to complete $10$ oscillations is (Take $\pi = 3.14$): (in $\text{ s}$)

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