Q158 / 200 · Manipal (MET) 2011 · Physics

An electron moving with velocity $2 \times 10^{-7} \mathrm{~m} / \mathrm{s}$, describes a circle in a magnetic field of strength $2 \times 10^{-2} \mathrm{~T}$. If $\frac{e}{m}$ of electron is $1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$. Then the diameter of the circle will be

  • A
    11 cm
  • Correct answer
    B
    1.1 cm
  • C
    1.1 mm
  • D
    1.1 m

Solution

The force exerted on electron moving with velocity $v$ is $F=e v B \sin \theta$ where $\theta=90^{\circ}$ $\therefore \quad F_B=e v B$...(i) Centripetal force is given by $F_C=\frac{m v^2}{r}$...(ii) Equating Eqs. (i) and (ii), we get $r=\frac{m v^2}{e B}$ Diameter $d=2 r=\frac{2 v}{(e / m) B}$ $\begin{aligned} \Rightarrow \quad d^{\prime} & =\frac{2 \times 2 \times 10^7}{1.76 \times 10^{11} \times 2 \times 10^{-2}} \\ & =1.1 \mathrm{~cm}\end{aligned}$

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

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