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JEE Main 2019 (10 Jan, Shift 2) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from JEE Main 2019 (10 Jan, Shift 2)

Q75JEE Main 2019 · 10 JAN · SHIFT 2Difficult

Current Loop as a Magnetic Dipole

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and Tc respectively, then -

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Answer: (b)

  1. For small oscillations of a dipole: T=2πIμBT = 2\pi\sqrt{\frac{I}{\mu B}}.
  2. Hoop: Ih=MR2I_h = MR^2, μh=2μ\mu_h = 2\mu, so Ihμh=MR22μ\frac{I_h}{\mu_h} = \frac{MR^2}{2\mu}.
  3. Cylinder: Ic=MR22I_c = \frac{MR^2}{2}, μc=μ\mu_c = \mu, so Icμc=MR22μ\frac{I_c}{\mu_c} = \frac{MR^2}{2\mu}.
  4. The ratios are equal, so Th=TcT_h = T_c. The answer is (b).
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