Question

A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

Answer 1

Answer:

a)n= 3.125 x $10^{19$ electrons.

b)J= 1.515 x $10^{6$ A/m²

c)$V_{d$ =1.114 x $10^{4$m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x $10^{-3$ m

radius 'r' = d/2 => 1.025 x $10^{-3$ m

no. of electrons 'n'= 8.5 x $10^{28}$

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x $10^{-19$C

n= Q/e => 5/ 1.6 x $10^{-19$

n= 3.125 x $10^{19$ electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x $10^{-3$)²)

J= 1.515 x $10^{6$ A/m²

c) The typical speed'$V_{d$' of an electron is given by:

$V_{d$ = $\frac{J}{n|q|}$

    =1.515 x $10^{6$ / 8.5 x $10^{28}$ x |-1.6 x $10^{-19$|

$V_{d$ =1.114 x $10^{4$m/s

d) According to these equations,

J= I/A

$V_{d$ = $\frac{J}{n|q|}$ =$\frac{I}{nA|q|}$

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

Answer 2

Answer:

a) 3.1205*10^19 electron/s

b) 1.51*10^6 A/m^2

c) 1.11*10^-4 m/s

Explanation:

a) to find the number of electrons you use the current in the wire, and the following formula:

$I=5.00A=5.00\frac{C}{s}\\\\1C=6.241*10^{18}e\\\\I=5.00(6.241*10^{18}e)/s=3.1205*10^{19}\frac{e}{s}$

3.1205*10^19 electrons per second

b) To find the current density you use the formula:

$J=\frac{I}{A}$

I: current in the wire

A: cross area of the wire

$J=\frac{I}{\pi r^2}=\frac{5.00A}{\pi(\frac{2.05}{2}*10^{-3})^2}=1.51*10^6\frac{A}{m^2}$

c) To find the speed you use the formula for the drift speed of electrons in the wire:

$I=nqv_dA\\\\v_d=\frac{I}{nqA}$

n: number of free electrons

q: charge of the electron = 1.6*10^{-19}C

$v_d=\frac{5.00C/s}{(8.5*10^{28}m^{-3})(1.6*10^{-19}C)(3.30*10^{-6}m^2)}=1.11*10^{-4}\frac{m}{s}$

d) if the diameter of the wire is increased, the number of electron that pass trough the lighy bulb each second is the same.

The current density decreases because J=I/A. If A increases J decreases.

The drift vellocity of the electrons decreases, again, because in the formula for vd the Area is in the denominator.a