Relation between electric current and drift velocity

Derivation→ Let us consider

  • The length of the conductor = $l$

  • The cross-section area of the conductor = $A$

  • The total number of free electrons inside the conductor = $N$

  • The current flow in the conductor = $i$

  • Flow of electrons in conductor
    The flow of Electron in Conductor
  • The Relaxation time between the two successive collisions =$\tau$
  • According to the law of current density

    $J=\frac{i} {A} $

    $J=\frac{q} {A \tau} \qquad \left( \because i=\frac{q}{\tau} \right)$

    $J=\frac{N \: e }{A \tau} \qquad \left( \because q=Ne \right)$

    Now multiply by length of conductor $l$ in above equation. Therefore we get

    $J=\frac{N \: e \: l}{A \tau\: l} $

    $J=\frac{N \: e \: l}{V \tau\: }\qquad \left(\because V=A.l \right)$

    Where $V$ is volume of the conductor.

    $J=\frac{n \: e \: l}{ \tau }\qquad \left( \because n=\frac{N}{V} \right)$

    Where $n$ is total number of electrons per unit .

    $J=n \: e \: v_{d}\qquad \left( \because v_{d}=\frac{l}{\tau}\right)$

    Where $v_{d}$ are known as drift velocity of charged particles.

    Now substitute the value of current density $J$ from equation $(1)$ to above equation then above equation can be written as


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