Question

An electron beam has an aperture of 1.0 mm^(2). A total of 6xx10^(16) electrons flow through any perpendicular cross-section per second. Calculate (i) the current (ii) the current density in the electron beam.

Answer 1

Answer:

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}