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MHT CET 2021 (24 Sep, Shift 2) · Current Loop as a Magnetic Dipole

Current Loop as a Magnetic Dipole question from MHT CET 2021 (24 Sep, Shift 2)

Q49MHT CET 2021 · 24 SEP · SHIFT 2Easy

Current Loop as a Magnetic Dipole

A particle of charge 'q' and mass 'm' moves in a circular orbit of radius 'r' with angular speed 'ω\omega'. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

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Answer: (c)

  1. μ=qωr22\mu = \frac{q\omega r^2}{2} and L=mωr2L = m\omega r^2.
  2. μL=q2m\frac{\mu}{L} = \frac{q}{2m}: depends only on qq and mm, so the answer is (c).
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