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AP EAMCET · PHYSICS · Magnetic Effects of Current

The distance moved by a charged particle along the magnetic field (the component of velocity is parallel to the magnetic field) in one rotation is given by ( \(\mathrm{m}\) - mass of the particle, \(\mathrm{v}\) - velocity of the particle, \(\mathrm{q}\) charge of the particle, B - magnetic field)

  1. A \(\frac{2 \pi m v}{q B}\)
  2. B \(\frac{\pi \mathrm{mv}}{\mathrm{qB}}\)
  3. C \(\frac{4 \pi \mathrm{mv}}{\mathrm{qB}}\)
  4. D \(\frac{2 \pi m v}{q B^2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{2 \pi m v}{q B}\)

Step-by-step Solution

Detailed explanation

If the velocity component \(\mathrm{v}_{\mathrm{y}}\) is perpendicular to the magnetic field \(\mathrm{B}\), the magnetic force acts like a centripetal force \(\mathrm{qv}_{\mathrm{y}} \mathrm{B}\).…