Question

An electron beam has an aperture $1.00{\text{mm}}^2$. A total of $6.0\times {10}^{16}$ electrons go through any perpendicular cross-section per second. Find the current$\left(i\right)$ and the current density$\left(j\right)$ of the beam.

• A. $i=6.3\times {10}^{-3}\text{A},j=9.6\times {10}^3{\text{A/m}}^2$
• B. $i=6.3\times {10}^{-3}\text{A},j=6.3\times {10}^3{\text{A/m}}^2$
• C. $i=9.6\times {10}^{-3}\text{A},j=6.3\times {10}^3{\text{A/m}}^2$
• D. $i=9.6\times {10}^{-3}\text{A},j=9.6\times {10}^3{\text{A/m}}^2$

Answer 1

The correct option is D $i=9.6\times {10}^{-3}\text{A},j=9.6\times {10}^3{\text{A/m}}^2$

Current,
$i=\frac{q}{t}=\frac{ne}{t}$
$\Rightarrow i=\frac{6.0\times {10}^{16}\times 1.6\times {10}^{-19}}{1}=9.6\times {10}^{-3}\text{A}$

Also, current density,
$j=\frac{i}{A}=\frac{9.6\times {10}^{-3}}{1\times {10}^{-6}}=9.6\times {10}^3{\text{A/m}}^2$

Key Learning: $i=\frac{q}{t}$ and $j=\frac{i}{A}$