Question

An electron travels at a distance of 0.10m in an electric field if intensity 3200V/m, enters perpendicular to the field with a velocity $4 \times {10^7}m/s$, what is its deviation in its path:
A) 1.76mm.
B) 17.6mm.
C) 176mm.
D) 0.176mm.

Answer 1

Hint: Electric field intensity at any given point is the strength of electric field at that given point. It is also defined as the force experienced by unit positive charge placed at a particular distance.

Complete step by step solution:
Given that,
Electron travels at a distance $(d) = 0.10m$
Electric field intensity $E = 3200V/m$
And the velocity of an electron is $(v) = 4 \times {10^7}m/s$
Charge on an electron is $(q) = 1.6 \times {10^{ - 19}}C$
Mass of an electron is $(m) = 9.1 \times {10^{ - 31}}kg$
The velocity, electric and magnetic field vectors are in the same direction and the magnetic field and velocity vector are parallel so there is no magnetic force.
Thus, ${F_m} = vqB\sin {0^ \circ } = 0$
The electric force on the charge is given as ${F_e} = qE$
And the displacement along x-axis after time $t$ is given by,
$x = vt + \dfrac{1}{2}{a_y}t$
$x = vt + \dfrac{{qE{t^2}}}{{2m}}$
Now come to the question,
Let us consider that the acceleration along $x$- axis,
${a _x} = 0$
And the acceleration along $y$- axis,
${a_y} = \dfrac{{qE}}{m}$
\[ = \dfrac{{1.6 \times {{10}^{ - 19}} \times 3200}}{{9.1 \times {{10}^{ - 31}}}}\]
$ = \dfrac{{5120 \times {{10}^{12}}}}{{9.1}} = 562.63 \times {10^{12}}m/{s^2}$ ……(1)
Now, for x- axis and calculate time from here,
${S_x} = {u_x} \times t$
$0.1 = 4 \times {10^7} \times t$
The time is
$t = \dfrac{{0.1}}{4} \times {10^{ - 7}}$ …………….( 2)
For Y- axis,
${s_y} = ({u_y}t + \dfrac{1}{2}{a_y}{t^2})$ ……………..(3)
Putting the values of equation (1)and (2) into equation (3), we get the value of
${s_y} = 0 + \dfrac{1}{2} \times 562.63 \times {10^{12}} \times \dfrac{{{{\left( {0.1} \right)}^2}}}{{{{\left( 4 \right)}^2}}} \times {10^{ - 14}}$
${s_y} = 0 + \dfrac{1}{2} \times 562.63 \times {10^{12}} \times \dfrac{{{{\left( {0.1} \right)}^2}}}{{16}} \times {10^{ - 14}}$
${s_y} = \dfrac{{281.315 \times {{10}^{ - 2}} \times {{10}^{ - 14}} \times {{10}^{12}}}}{{16}}$
${s_y} = 17.58 \times {10^{ - 4}}m$
Conversion into $mm({10^{ - 3}})$
$ = 17.58 \times {10^{ - 1}}mm$
$ = 1.758mm$
${s_y} = 1.76mm$

Hence, The correct option of this question is (B)


Note: If electric field is not present, then the particle revolves along a circle in the $XY$ plane. Accelerated or decelerated depending on the polarity of charge and direction of electric field.