Question

The current density of a silver wire is 3.20x106 A/m² . If the current and the drift velocity of the electrons through the wire is 2.00 A and 3.20x10-4 m/s respectively. . Part A - Find the diameter of the wire and the density of free electrons. 8.92x10-4 m 6.25x1028 8.92x10-4 m , 1.25x1027 6.31*10-4 m 6.25x1028 6.31*10-4 m , 1.25x1027 7.96x10-7 m , 1.25x1027

Answer 1

Current density j is related to current I by the formula

j = I/A

Where A is the area of cross section of the wire.

Now, given I = 2.00 A

j = 3.20*10⁶ A/m^{​​​​​​2}

Using these values in expression for j, we get

3.20*10^{6 ₌} 2/A

Or A = 2/3.20*10⁶

A = 0.625*10^-6 m^{​​​​​​2}

Now, area of cross section A is related to diameter D as

A = πD^{​​​​​2}/4

Or D^{​​​​​​2} = A*4/π

= 0.625*10^-6*4/3.14

D^{​​​​​​2} =( 0.796*10^-6)

Taking square root of both sides

D = 0.892*10^-3 m

= 8.92*10^-4  m

Further, current and number density n are related by the formula

I = nAev

Where e is the charge over the electrons.

Or we can write

n = l/Aev

= 2.00/0.625*10^-6*1.6*10^-19*3.20*10^-4

= 2.00/3.20*10^-29

= 0.625*10²⁹

n = 6.25*10²⁸ m^{​​​​​​-3}​​​​m-3

Hence, option 1st is the correct answer.