Q129 / 150 · MHT CET 2022 (10 Aug Shift 1) · Physics

A charge moves in a circular path perpendicular to a magnetic field. The time period of the revolution is independent of

  • A
    strength of charge.
  • B
    magnetic field.
  • C
    mass of the charge.
  • Correct answer
    D
    velocity of the charge.

Solution

The time period of a charged particle undergoing circular motion under uniform perpendicular magnetic field is given by $T=\frac{2 \pi m}{q B}$ So, it is independent of velocity.

Hence, the correct answer is option: 'D'.

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