An electron beam has a aperture of `1.0mm^2`. A total of `6.0xx10^16` electrons go throgh any perpendicular cross section per second.Find a. the current and b. the current density in the beam.

Explanation

a. The current is given by `i=q/t=("ne")/t`
Susitituting the values we have
`i=((6.0xx10^16)(1.6xx10^19))/1`
`=9.6xx10^-3A`
b. The current density is
`j=i/A=(9.6xx10^-3)/10^-6`
`=9.6x10^3A//m^2`