Question

An electron travels at a distance of 0.10 m in an electric field if intensity 3200 V/m, enters perpendicular to the field with a velocity $4\times {10}^{7}m/s,$ what is its deviation in its path:

• A. 1.76 mm.
• B. 17.6 mm.
• C. 176 mm.
• D. 0.176 mm.

Answer 1

The correct option is A 1.76 mm.

Distance $=0.10m$

electric field if intensity $=3200V/m$

velocity $=4\times {10}^{7}m/s$

Now, on $y$ direction

$y=\frac{1}{2}\frac{eE}{m}{t}^2$

Putting the value of charge mass and given electric field intensity and get the relation in $y$ and $t$.

Calculation $t$ from :

$x=Vt$

or, $10=4\times {10}^7\times t$

Now,

$E=9\times {10}^9\times \frac{q}{r^2}$

or, $4\times {10}^7=9\times {10}^9\times \frac{q}{{\left(0.10\right)}^2}$

or, $q=\frac{4\times {10}^7\times {\left(0.10\right)}^2}{9\times {10}^9}$

$={\left(\frac{1}{10}\right)}^2\times \frac{4}{9\times {10}^2}$

$=\frac{1}{100}\times \frac{4}{9}\times \frac{1}{100}$

$=\frac{0.44}{10000}=0.000044C$

$=\frac{0.000044\times {10}^6}{{10}^6}$

$=44\mu C$

Velocity $=\frac{qB}{mr}=\frac{qE}{r}$

or, $r=\frac{qE}{V}=\frac{44\times {10}^{-6}\times 4\times {10}^7}{4\times {10}^7}=1.76mm$