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JEE Main 2021 (27 Jul, Shift 1) · Vibration Magnetometer

Vibration Magnetometer question from JEE Main 2021 (27 Jul, Shift 1)

Q12JEE Main 2021 · 27 JUL · SHIFT 1Medium

Vibration Magnetometer

In a uniform magnetic field, the magnetic needle has a magnetic moment 9.85 ×\times 10−-2 A m2 and moment of inertia 5 ×\times 10−-6 kg m2. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is _______________ mT. [Take π\pi2 as 9.85]

Numerical answer type. Work it out, then open the answer.

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Answer: 8

  1. 10 oscillations in 5 s: T=0.5T = 0.5 s.
  2. B=4π2IMT2=4×9.85×5×10−69.85×10−2×0.25B = \frac{4\pi^2 I}{MT^2} = \frac{4 \times 9.85 \times 5 \times 10^{-6}}{9.85 \times 10^{-2} \times 0.25}
  3. B=20×10−62.5×10−3=8×10−3 TB = \frac{20 \times 10^{-6}}{2.5 \times 10^{-3}} = 8 \times 10^{-3}\ \text{T}
  4. So the answer is 8 mT.
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