Question

Write Ohm's law. Deduce Ohm's law on the basis of drift velocity of electron.

Hint:

Use \(v_d = eE\tau/m\) and \(I = neAv_d\) with \(E = V/L\); collecting constants gives \(V = IR\).

Answer 1

Step 1: Statement of Ohm's law.
At constant temperature (and other physical conditions), the current \(I\) flowing through a conductor is directly proportional to the potential difference \(V\) applied across its ends:
\[ V \propto I \quad\Rightarrow\quad V = IR \]
where \(R\) is the resistance of the conductor.

Step 2: Drift velocity of electrons.
When a potential difference \(V\) is applied across a conductor of length \(L\), an electric field is set up:
\[ E = \frac{V}{L} \]
Each free electron (charge \(e\), mass \(m\)) is accelerated but, due to collisions, acquires a small average drift velocity
\[ v_d = \frac{eE\tau}{m} \]
where \(\tau\) is the average relaxation time between collisions.

Step 3: Relate current to drift velocity.
If \(n\) is the number of free electrons per unit volume and \(A\) the area of cross-section, the current is
\[ I = neAv_d \]

Step 4: Combine the equations.
Substitute \(v_d = \dfrac{eE\tau}{m}\) and \(E = \dfrac{V}{L}\):
\[ I = neA\cdot\frac{eE\tau}{m} = \frac{ne^2A\tau}{m}\cdot\frac{V}{L} \]
\[ I = \left(\frac{ne^2A\tau}{mL}\right)V \]

Step 5: Identify Ohm's law.
Rearranging,
\[ V = I\cdot\frac{mL}{ne^2A\tau} = IR \]
where the resistance is
\[ R = \frac{mL}{ne^2A\tau} \]
Since \(m, n, e, \tau, A, L\) are constants for a given conductor at a fixed temperature, \(R\) is constant, so \(V \propto I\). This is Ohm's law.
\[\boxed{V = IR,\qquad R = \frac{mL}{ne^2A\tau}}\]