Q88 / 120 · VITEEE 2014 · Physics

An electron moves at right angle to a magnetic field of $5 \times 10^{-2} \mathrm{~T}$ with a speed of $6 \times 10^{7} \mathrm{~m} / \mathrm{s}$. If the specific charge of the electron is $1.7 \times 10^{11} \mathrm{C} / \mathrm{kg}$. The radius of the circular path will be

  • A
    $2.9 \mathrm{~cm}$
  • B
    $3.9 \mathrm{~cm}$
  • Correct answer
    C
    $2.35 \mathrm{~cm}$
  • D
    $2 \mathrm{~cm}$

Solution

Radius of circular path $\begin{aligned} \mathrm{r} &=\frac{\mathrm{mV}}{\mathrm{B}}=\frac{\mathrm{V}}{\left(\frac{\mathrm{e}}{\mathrm{m}}\right) \mathrm{B}} \\ \mathrm{r} &=\frac{6 \times 10^{7}}{1.7 \times 10^{11} \times 15 \times 10^{-2}} \\ &=2.35 \times 10^{-2} \mathrm{~m}=2.35 \mathrm{~m} \end{aligned}$

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

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