Question

An electron beam has an aperture $1.0{\text{mm}}^2$. A total of $6.0\times {10}^{16}$ electrons go through any perpendicular cross-section per second. Find $\left(a\right)$ the current and $\left(b\right)$ the current density in the beam.

• A. $6.3\times {10}^{-3}\text{A}\&6.3\times {10}^3{\text{A/m}}^2$
• B. $3.2\times {10}^{-3}\text{A}\&3.2\times {10}^3{\text{A/m}}^2$
• C. $9.6\times {10}^3\text{A}\&9.6\times {10}^{-3}{\text{A/m}}^2$
• D. $9.6\times {10}^{-3}\text{A}\&9.6\times {10}^3{\text{A/m}}^2$

Answer 1

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

$\left(a\right).$ The total charge crossing a perpendicular cross-section in one second is

$q=ne=6.0\times {10}^{16}\times 1.6\times {10}^{-19}\text{C}$

$\Rightarrow q=9.6\times {10}^{-3}\text{C}$

So, current will be

$I=\frac{q}{t}=\frac{9.6\times {10}^{-3}\text{C}}{1\text{s}}=9.6\times {10}^{-3}\text{A}$

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

$\left(b\right).$ The current density is defined as

$J=\frac{I}{A}=\frac{9.6\times {10}^{-3}\text{A}}{(1.0\text{mm}{)}^2}=\frac{9.6\times {10}^{-3}A}{1.0\times {10}^{-6}{m}^2}$

$\Rightarrow J=9.6\times {10}^3{\text{A/m}}^2$

Hence, option $\left(a\right)$ is correct.