Question

Calculate the drift velocity of the free electrons with mobility of (3)5\(\times\)10\(^{-3}\) m\(^2\)/Vs in copper for an electric field strength of 0.5 V/m.

• A. (3)5 m/s
• B. (1)75\(\times\)10\(^{3}\) m/s
• C. 1(1)5 m/s
• D. (1)75\(\times\)10\(^{-3}\) m/s

Hint:

Drift Velocity and Mobility. Drift velocity \(v_d\) is the average velocity of charge carriers under an electric field E. \(v_d = \mu E\), where \(\mu\) is the mobility. Ensure units are consistent (m, V, s).

Answer 1

The Correct Option is D

Drift velocity (\(v_d\)) is the average velocity attained by charged particles (like electrons) in a material due to an electric field (\(E\)).
It is related to the electric field and the charge carrier mobility (\(\mu\)) by the formula: $$ v_d = \mu E $$ Given: Electron mobility \(\mu = (3)5 \times 10^{-3}\) m\(^2\)/Vs Electric field strength \(E = 0.
5\) V/m Substitute the values: $$ v_d = ((3)5 \times 10^{-3} \, \text{m}^2/\text{Vs}) \times (0.
5 \, \text{V/m}) $$ The units (m\(^2\)/Vs) \(\times\) (V/m) = m/s, which is correct for velocity.
$$ v_d = ((3)5 \times 0.
5) \times 10^{-3} \, \text{m/s} $$ $$ v_d = (1)75 \times 10^{-3} \, \text{m/s} $$ The drift velocity is \((1)75 \times 10^{-3}\) m/s.