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

Step-by-step explanation:

We will calculate the density as follows.

n =
`(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 * r^(2)`

=
`3.14 * (2.06 * 10^(-3))^(2)`

`v_(d)` =
`5.40 * 10^(-5)`

i = 8 A

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

n =
`(i)/(v_(d)eA)`

=
`(8)/(5.40 * 10^(-5) * 3.14 * (2.06 * 10^(-3))^(2) * 1.60 * 10^(-19))`

=
`6.950 * 10^(28) m^(-3)`

Thus, we can conclude that the density of free electrons in the metal is
`6.950 * 10^(28) m^(-3)`.

Answer 2

`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
`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)`