Question

An electron beam has an aperture of 1.0mm square. A total of 6 * 10 ki power 16 electrons flows through any perpendicular cross section per second. Calculate :
(1) The current
(2) The current density in electron beam.

Answer 1

Given,

Aperture = A = 1.0 $mm^{2}$  = $10^{-6} m^{2}$

Number of electrons = n =$6*10^{16}$

Time = t = 1 s

 

To find,

1) The current = I = $\frac{q}{t}=\frac{ne}{t}$

Where, e =$1.6*10^{-19}$

On putting the values,

I = $\frac{6*10^{16} *1.6*10^{-19}}{1}$

I = $9.6*10^{-3}A$

2) Current density  

Current density = j = $\frac{I}{A}$

On putting the values we get,

j = $\frac{9.6*10^{-3}}{10^{-6}}$

j = $9.6*10^{3} A m^{-2}$