Question

An electron beam has an aperture $1m{m}^2$. A total of $6.0\times {10}^{16}$ electrons go through any perpendicular cross-section per second. Find the current density in the beam.

• A. $7.6\times {10}^{3}A{m}^{-2}$
• B. $9.6\times {10}^{3}A{m}^{-2}$
• C. $8.6\times {10}^{3}A{m}^{-2}$
• D. $6.6\times {10}^{3}A{m}^{-2}$

Answer 1

The correct option is B $9.6\times {10}^{3}A{m}^{-2}$

The total charge crossing a perpendicular cross-section in 1 second is
$q=ne$
$=6.0\times {10}^{16}\times 1.6\times {10}^{-19}C$
$=9.6\times {10}^{-3}C$

The current is
$i=\frac{q}{t}$
$=\frac{9.6\times {10}^{-3}C}{1s}=9.6\times {10}^{-3}A$

As the charge is negative, the current is opposite to direction of motion of the beam.

The current density is
$j=\frac{i}{S}=\frac{9.6\times {10}^{-3}A}{1.0m{m}^2}=9.6\times {10}^{3}A{m}^{-2}$