Question

A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the density of free electrons in the metal?Express your answer numerically in m−3 to two significant figures.

Answer 1

Answer:

$6.9\times 10^{28}m^{-3}$

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=r$r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m$

$1mm=10^{-3} m$

Current=I=8 A

Drift velocity=$v_d=5.4\times 10^{-5} m/s$

We have to find the density of free electrons in the metal

We know that

Density of electron=$n=\frac{I}{v_deA}$

Using the formula

Density of free electrons=$\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}$

By using Area of wire=$\pi r^2$

$\pi=3.14\\e=1.6\times 10^{-19} C$

Density of free electrons=$6.9\times 10^{28}m^{-3}$

Answer 2

Explanation:

We will calculate the density as follows.

               n = $\frac{i}{v_{d}eA}$

As the metallic wire is in the shape of a circle. So, its area will be calculated as follows.

                 Area = $\pi \times r^{2}$

                          = $3.14 \times (2.06 \times 10^{-3})^{2}$

                $v_{d}$ = $5.40 \times 10^{-5}$

                        i = 8 A

Hence, we will calculate the density of free electrons as follows.

              n = $\frac{i}{v_{d}eA}$

                 = $\frac{8}{5.40 \times 10^{-5} \times 3.14 \times (2.06 \times 10^{-3})^{2} \times 1.60 \times 10^{-19}}$

                  = $6.950 \times 10^{28} m^{-3}$

Thus, we can conclude that the density of free electrons in the metal is $6.950 \times 10^{28} m^{-3}$.