Question
Consider circuit in Fig. How much energy is absorbed by electrons from the initial state of no current $($ignore thermal motion$)$ to the state of drift velocity?
✓
Answer
Key concept : Relation between current and drift velocity is given by $I =$ ne $Av_d,$
where vd is the drift speed of electrons and n is the number density of electrons.
According to the $Ohm'$ s law current in the circuit
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{6\text{V}}{6\Omega}=1\text{A}$
But, $\text{I}=\text{neAv}_\text{d}$
or $\text{v}_\text{d}=\frac{\text{I}}{\text{neA}}$
On substituting the value,
For $n =$ number of electrons/volume $= 10^{29}/m^3$
length of circuit $= 10\ cm$, cross $-$ section $= A = (1mm)^2$
$\text{v}_\text{d}=\frac{1}{10^{29}\times16\times10^{-19}\times10^{-6}}$
$=\frac{1}{1.6}\times10^{-4}\text{m/s}$
Therefore, the energy absorbed in the form of $KE$ is given by
Total $KE = KE$ of $1$ electron $\times$ on. of electrons
$\text{KE}=\frac{1}{2}\text{m}_\text{e}\text{v}_\text{d}^2\times\text{nAl}$
$=\frac{1}{2}\times9.1\times10^{31}\times\frac{1}{2.56}\times10^{-8}\times{10}^{29}\times{10}^{-6}\times10^{-1}$
$=2\times10^{-17}\text{J}$
where vd is the drift speed of electrons and n is the number density of electrons.
According to the $Ohm'$ s law current in the circuit
$\text{I}=\frac{\text{V}}{\text{R}}$
$\text{I}=\frac{6\text{V}}{6\Omega}=1\text{A}$
But, $\text{I}=\text{neAv}_\text{d}$
or $\text{v}_\text{d}=\frac{\text{I}}{\text{neA}}$
On substituting the value,
For $n =$ number of electrons/volume $= 10^{29}/m^3$
length of circuit $= 10\ cm$, cross $-$ section $= A = (1mm)^2$
$\text{v}_\text{d}=\frac{1}{10^{29}\times16\times10^{-19}\times10^{-6}}$
$=\frac{1}{1.6}\times10^{-4}\text{m/s}$
Therefore, the energy absorbed in the form of $KE$ is given by
Total $KE = KE$ of $1$ electron $\times$ on. of electrons
$\text{KE}=\frac{1}{2}\text{m}_\text{e}\text{v}_\text{d}^2\times\text{nAl}$
$=\frac{1}{2}\times9.1\times10^{31}\times\frac{1}{2.56}\times10^{-8}\times{10}^{29}\times{10}^{-6}\times10^{-1}$
$=2\times10^{-17}\text{J}$
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