Question

Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carrying a current of 30 A (values similar to those for the electric wire to your study lamp). Assume one electron per atom of copper contributes to the current. The atomic mass of copper is 63.5 g/mol and its density is 8.93 g/cm3 . Avogadro’s number is 6.022 × 1023 and the fundamental charge is 1.602 × 10−19 C. Answer in units of m/s.

Answer 1

Answer:

The correct answer is $2.8*10^{-5}ms^{-1}$

Explanation:

The formula for the electron drift speed is given as follows,

$u=I/nAq$

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is $8.93/63.5 molcm^{-3}$. Converting this number to m³ using very elementary unit conversion we get $140384molm^{-3}$. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

$n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}$

if we convert the area from mm³ to m³ we get $A=80*10^{-6}m^{2}$.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

$u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}$

which is our final answer.

Answer 2

The average drift speed of the electrons is 2.8 x 10^-5ms^-1

Calculations and Parameters:

To find the electron drift speed, we use the formula:

u= I/aAq

Where:

• n is the number of electrons per unit m³,
• q is the charge on an electron
• A is the cross-sectional area of the copper wire
• I is the current.

Hence, we can see that the number of moles of copper atoms in a cm³ of the volume of the conductor is 8.93/ 63.5molcm^-3.

n= 140384 x 6.02 x 10^23

= 8.45 x 10^28 electrons.m^-3

After the conversion of units and putting the values together we would get:

u= 2.8 x 10^-5ms^-1

Read more about average speed here:
https://brainly.com/question/6504879