An electron is projected horizontally with a kinetic energy of $10 k e V$ . A magnetic field of strength $1.0 \times 10^{- 7} T$ exists in the vertically upward direction. (a) Will the electron deflect towards right or towards left of its motion ? (b) Calculate the sideways deflection of the electron in travelling through $1 m$ . Make appropriate approximations.

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Solution

$\begin{matrix} K . E = 10 K e V = 1.6 \times 10^{- 15} J, \to B = 10^{- 7} T B y r i g h t - h a n d r u l e, t h e e l e c t r o n w i l l b e d e f l e c t e d t o w a r d s l e f t . \frac{1}{2} m v^{2} = K . E v = \sqrt{\frac{K E \times 2}{m}} F = q v B a = \frac{q v B}{m_{e}} A p p l y i n g s = u t + \frac{1}{2} a t^{2} = \frac{1}{2} \frac{q v B}{m_{e}} \times \frac{x^{2}}{v^{2}} = \frac{q B x^{2}}{2 m_{e} v} = \frac{q B x^{2}}{2 m_{e} \sqrt{\frac{K E \times 2}{m}}} = \frac{1}{2} \times \frac{1.6 \times 10^{- 19} \times 10^{- 7} \times 1^{2}}{9.1 \times 10^{- 31} \sqrt{\frac{1.6 \times 10^{- 15} \times 2}{9.1 \times 10^{- 31}}}} s = 0.0148 ≃ 1.5 \times 10^{- 2} c m \end{matrix}$

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