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Torque

A magnetic ne...

Question

A magnetic needle of magnetic moment $6.7 \times 10^{- 2} {Am}^{2}$ and moment of inertia $7.5 \times 10^{- 6} {kg m}^{2}$ is performing simple harmonic oscillations in a magnetic field of $0.01 T$ . Time taken for $10$ complete oscillations is - (in second)

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Solution

Given:
$M = 6.7 \times 10^{- 2} {Am}^{2}$
$I = 7.5 \times 10^{- 6} {kg m}^{2}$
$B = 0.01 T$
Time period of oscillation of needle is given by
$T = 2 π \sqrt{\frac{I}{M B}}$
$T = 2 π \sqrt{\frac{7.5 \times 10^{- 6}}{6.7 \times 10^{- 2} \times 0.01}}$
$T = 2 π \sqrt{\frac{75}{67 \times 100}}$
$T = \frac{2 π}{10} \times \sqrt{\frac{75}{67}} \times 10$
Time taken for for $10$ oscillation is
$t = 10 T = 2 π \sqrt{\frac{75}{67}}$
$T = 6.28 \times \sqrt{\frac{75}{67}} = 6.65 s$

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