Question

Mobility of charged particles is $6\times {10}^{-2}{\text{m}}^2{\text{V}}^{-1}{\text{s}}^{-1}$. If the electric field applied is $9\times {10}^{-3}\text{V/m}$, then the drift speed would be

• A. $2\text{m/s}$
• B. $2\times {10}^{-4}\text{mm/s}$
• C. $0.69\times {10}^{-4}\text{m/s}$
• D. $0.54\text{mm/s}$

Answer 1

The correct option is D $0.54\text{mm/s}$

Mobility $\left(\mu \right)$ is defined as,

$\mu =\frac{v_d}{E}$

So,
$v_d=\mu E=6\times {10}^{-2}\times 9\times {10}^{-3}=0.54\text{mm/s}$