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Vibration Magnetometer question from JEE Main 2017

Q14JEE Main 2017Medium

Vibration Magnetometer

A magnetic needle of magnetic moment 6.7 ×\times 10−2 A m2 and moment of inertia 7.5 ×\times 10−6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:

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Answer: (b)

  1. T=2πIMB=2π7.5×10−66.7×10−2×0.01T = 2\pi\sqrt{\frac{I}{MB}} = 2\pi\sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times 0.01}}
  2. IMB=0.0112⇒T=2π×0.1058≈0.665 s\frac{I}{MB} = 0.0112 \Rightarrow T = 2\pi \times 0.1058 \approx 0.665\ \text{s}
  3. 10 oscillations: 10T≈6.65 s10T \approx 6.65\ \text{s}
  4. So the answer is (b) 6.65 s.
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