Question

answer it
no illogical answers​

Answer 1

Solution;-

→ The e⁻   Moves random at high thermal Velocities ( ≈ 10⁵ m/s ) between two collision But also keeps an drifting opposite to electric field at small velocity

Calculation of drift velocity

→ if an e⁻ moving with random thermal velocity μ₁ Accelerate for time τ              ( relaxation time between two Successive Collision ) and attains Velocity V

Drift Velocity Vd ⇒ Avg Velocity

$\to\sf \overrightarrow{\sf{v}}d=\dfrac{-e\overrightarrow{\sf{E}}}{m} \tau$

where

→ v = El ⇒ E= V/l

We get

$\to\sf \overrightarrow{\sf{v}}d=\dfrac{-eV}{ml} \tau$

We know that

→ i = vdenA

On substituting in Vd we get

$\to\sf I=neA\bigg(\dfrac{eV}{ml}\bigg) \tau$

$\to\sf I=\dfrac{ne^{2} }{m} \tau\dfrac{A}{I} .V$

$\to\sf V=\dfrac{m}{ne^{2} } \tau\dfrac{I}{A} .I$

We know that ohms law v = IR  where R = V/I

$\to\sf \dfrac{V}{I} =\dfrac{m}{ne^{2} } \tau\dfrac{I}{A}$

$\to\sf R=\dfrac{m}{ne^{2} } \tau\dfrac{I}{A}$

Hence we derive