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PhysicsNEET 2024Bar magnet and magnetic field lines

Magnetism and Matter

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds. The moment of inertia of the needle is 9.8×10−6 kg m29.8 \times 10^{-6}\ \mathrm{kg\,m^2}. If the magnitude of the magnetic moment of the needle is x×10−5 A m2x \times 10^{-5}\ \mathrm{A\,m^2}, then the value of xx is:

  1. A.128π2128\pi^2
  2. B.50π250\pi^2
  3. C.1280π21280\pi^2✓
  4. D.5π25\pi^2

Solution

Oscillation period T=520=0.25T = \frac{5}{20} = 0.25 s. For a magnetic needle, T=2πIMBT = 2\pi\sqrt{\frac{I}{MB}}, so M=4π2IT2B=4π2×9.8×10−60.0625×0.049=1280π2×10−5M = \frac{4\pi^2 I}{T^2 B} = \frac{4\pi^2 \times 9.8\times10^{-6}}{0.0625 \times 0.049} = 1280\pi^2 \times 10^{-5} A m². Hence x=1280π2x = 1280\pi^2.

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