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

`6.9* 10^(28)m^(-3)`

Step-by-step explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=
`r=(d)/(2)=(4.12)/(2)=2.06mm=2.06* 10^(-3) m`

`1mm=10^(-3) m`

Current=I=
`8 A`

Drift velocity=
`v_d=5.4* 10^(-5) m/s`

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

We know that

Density of electron=
`n=(I)/(v_deA)`

Using the formula

Density of free electrons=
`(8)/(5.4* 10^(-5)* 1.6* 10^(-19)* 3.14* (2.06* 10^(-3))^2)`

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

`\pi=3.14`

`e=1.6* 10^(-19) C`

Density of free electrons=
`6.9* 10^(28)m^(-3)`