Q134 / 150 · MHT CET 2023 (10 May Shift 1) · Physics

Electron of mass ' $\mathrm{m}$ ' and charge ' $\mathrm{q}$ ' is travelling with speed ' $v$ ' along a circular path of radius ' $R$ ', at right angles to a uniform magnetic field of intensity ' $\mathrm{B}$ '. If the speed of the electron is halved and the magnetic field is doubled, the resulting path would have radius

  • A
    $4 \mathrm{R}$
  • B
    $2 \mathrm{R}$
  • C
    $\frac{\mathrm{R}}{2}$
  • Correct answer
    D
    $\frac{\mathrm{R}}{4}$

Solution

From cyclotron motion and uniform circular motion, i.e., $\mathrm{q} v \mathrm{vB} \sin 90^{\circ}=\frac{m v^2}{\mathrm{R}}$ $\therefore \quad \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$ Given: $\mathrm{v}^{\prime}=\frac{\mathrm{v}}{2}$ and $\mathrm{B}^{\prime}=2 \mathrm{~B}$ $\begin{aligned} \therefore \quad \mathrm{R}^{\prime} & =\frac{\mathrm{mv}}{\mathrm{qB}^{\prime}} \\ & =\frac{\mathrm{m} \frac{\mathrm{v}}{2}}{\mathrm{q} 2 \mathrm{~B}}=\frac{1}{4} \frac{\mathrm{mv}}{\mathrm{qB}} \\ \Rightarrow \mathrm{R}^{\prime} & =\frac{1}{4} \mathrm{R} \end{aligned}$

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

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