Question

Calculate the drift velocity of electrons in copper and current density in wire of diameter 0.16 cm which carries a steady current of 10 A. Given n = 8.46 × 10^28 m^–3.​

Answer 1

Answer:

1.088*10^-27

Explanation:

$v = \frac{ne \times area}{i}$

Answer 2

Drift Velocity $v_d = 3.676 \times 10^{-4} \ m/s$ and Current Density $j = 4.976 \times 10^{-4} \ A/m^{2}$

Step-by-step Explanation:

Given: Diameter of the wire D: 0.1 6 cm = 0.16 × 10⁻² m

Current I = 10A

Number density n = 8.46 × 10²⁸ m⁻³

To Find: The Drift Velocity and Current Density of electrons

Solution:

• Calculating the drift velocity ($v_d$) of the electrons

The Expression for finding the drift velocity is given by:-

$v_d = \frac{I}{nAQ} = \frac{I}{n(\pi D^{2}/4 )Q}$

Where $Q$ is the charge of the electron which is $Q = 1.6 \times 10^{-19} \ C$

Substituting the given values in the above formula, we get;

$\Rightarrow v_d = \frac{10}{8.46 \times 10^{28} \times (3.14 \times((0.16)^2/4)) \times 1.6 \times 10^{-19} } = 3.676 \times 10^{-4} \ m/s$

• Calculating the current density ($j$)

The relation between drift velocity and the current density is given by:-

$j = nQv_d$

Substituting the given values and calculating, we get,

$\Rightarrow j = 8.46 \times 10^{28} \times 1.6 \times 10^{-19} \times 3.676 \times 10^{-4} = 4.976 \times 10^{-4} \ A/m^{2}$

Hence, the drift velocity is $v_d = 3.676 \times 10^{-4} \ m/s$ and the current density is $j = 4.976 \times 10^{-4} \ A/m^{2}$