Question

A stream of electrons is moving with a drift velocity of 5mm/sec. If number of electrons be 6*10^23/cm^3,calculate the current density

Answer 1

$\large\underline{\bigstar \: \: {\sf Given-}}$

• Stream of Electron is moving with a drift velocity ${\sf v_d=5\:mm/s}$
• Electron density (n)= 6×10²³ /cm³

$\large\underline{\bigstar \: \: {\sf To \: Find -}}$

• Current Density ( J )

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

$\implies\underline{\boxed{\sf Current (I)=neAv_d}}$

$\implies\underline{\boxed{\sf Current \: Density (J)=\dfrac{I}{A}}}$

$\large\underline{\bigstar \: \: {\sf Solution-}}$

Convert ${\bf v_d}$ from mm/s to m/s -

$\implies{\sf v_d=5\:mm/s }$

$\implies{\bf v_d=5 \times 10^{-3}\:m/s}$

Convert electron density (n) from /cm³ to /m³ -

$\implies{\sf n=6 \times 10^{23}\:/cm^3}$

$\implies{\sf n =6\times 10^{23} \times 10^{-6}}$

$\implies{\bf n = 6 \times 10^{17}}$

Now , On putting value

$\implies{\sf I = neAv_d}$

e = ${\sf 1.6 \times 10^{-19}\:C}$

A = Cross-section Area

$\implies{\sf I =6\times 10^{17}×1.6\times 10^{-19} \times A \times 5×10^{-3} }$

$\implies{\sf 48\times 10^{-5}\times A}$

$\large\implies{\sf Current \: Density (J)=\dfrac{I}{A} }$

$\implies{\sf \dfrac{48×10^{-5}\times A}{A}}$

$\implies{\sf 48 \times 10^{-5}\:A }$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Current Density is ${\bf 48\times 10^{-5}\:A}$