Question

An electron is projected horizontally with a kinetic energy of 10 keV. A magnetic field of strength 1.0 × 10⁻⁷ T exists in the vertically upward direction. (a) Will the electron deflect towards the right or left of its motion? (b) Calculate the sideways deflection of the electron while travelling through 1 m. Make appropriate approximations.

Answer 1

Given:
The kinetic energy of the electron projected in the horizontal direction, K.E = 10 keV = 1.6 × 10⁻¹⁵ J
Magnetic field, B = 1 × 10⁻⁷ T
The direction of magnetic field is vertically upward.
(a) The direction can be found by the right-hand screw rule. So, the electron will be deflected towards left.
(b) Kinetic energy,
$$\begin{gathered} \mathrm{K}.\mathrm{E}.=\frac{1}{2}m{v}^2 \\ v=\sqrt{\frac{K.E.\times 2}{m}} \end{gathered}$$
Magnetic force,
$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$ ....(i)
and F = ma
Therefore, equation (i) will be
$a=\frac{q\overrightarrow{v}\times \overrightarrow{B}}{m}$
Applying equation of motion
s = ut + $\frac{1}{2}$ at²,

t= Time taken to cross the magnetic field
As there is no force acting on the electron in the horizontal direction, the velocity of the electron remain constant in this direction.
So, the time taken to cross a distance of 1m in the horizontal direction in the magnetic field,
$$\begin{gathered} t=\frac{d}{v_{\mathrm{horizontal}}} \\ =\frac{1}{\sqrt{{\displaystyle \frac{2\times K.E}{m}}}} \end{gathered}$$
Putting the respective value of the variables, we get:
$$\begin{gathered} y=\frac{1}{2}\frac{qvB}{m}{t}^2 \\ =\frac{1}{2}\frac{qvB}{m}\frac{1}{{\displaystyle \frac{2\times K.E}{m}}} \\ =\frac{qvB}{4\times K.E} \\ =\frac{1}{4}\frac{1.6\times {10}^{-19}\times 1\times {10}^{-7}\times \sqrt{{\displaystyle \frac{2\times \mathrm{K}.\mathrm{E}}{\mathrm{m}}}}}{10\times {10}^{-16}} \\ =0.148\times {10}^{-3}m \\ =0.01482\mathrm{cm} \end{gathered}$$

s = 0.0148 = 1.5 × 10⁻² cm