Chapter 14: Q7CP (page 546)

The mobility of the mobile electrons in copper is $4.5 \times 10^{- 3} (m / s) / (N / C)$ . How large an electric field would be required to give the mobile electrons in a block of copper a drift speed of $1 \times 10^{- 3} m / s$ ?

Short Answer

Expert verified

$0.22 N / C$

Step by step solution

Identification of given data

The given data is listed below as,

The drift speed of the electrons is, $ν = 1 \times 10^{- 3} m / s$
The mobility of mobile electrons in copper $μ = 4.5 \times 10^{- 3} (m / s) / (N / C)$

Explanation of Drift speed

The drift speed can be determined by taking the product of the electric field and the mobility of ions. It is the average velocity attained by the charged particles in the presence of the electric field.

Calculation of the electric field for the mobile electrons

The expression for the drift speed of ions is as follows,

$ν = E \times μ$

Here, $E$ is magnitude of the electric field,and $μ$ is the mobility of chloride ions.

For $ν = 1 \times 10^{- 3} m / s$ and $μ = 4.5 \times 10^{- 3} (m / s) / (N / C)$ .

role="math" localid="1653918994881" $(1 \times 10^{- 3} m / s) = E \times (4.5 \times 10^{- 3} (m / s) / (N / C)) E = \frac{(1 \times 10^{- 3} m / s)}{(4.5 \times 10^{- 3} (m / s) / (N / C))} = 0.22 N / C$