Q89 / 160 · JIPMER 2009 · Physics

Under the influence of a uniform magnetic field a charged particle is moving in a circle of radius $R$ with constant speed $v$. The time period of the motion

  • A
    depends on $v$ and not on $R$
  • B
    depends on both $R$ and $v$
  • Correct answer
    C
    is independent of both $R$ and $v$
  • D
    depends on $R$ and not on $v$

Solution

When magnetic field is perpendicular to motion of charged particle, then Centripetal force $=$ magnetic force ie, $\frac{m v^2}{R}=B q v$ or $R=\frac{m v}{B q}$ Further, time period of the motion or $T=\frac{2 \pi R}{v}=\frac{2 \pi\left(\frac{m v}{B q}\right)}{v}$ or $\quad T=\frac{2 \pi m}{B q}$ It is independent of both $R$ and $v$.

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

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