You have a rectangular solid. You would like to pass current through it so that it has the least resistance. Should you direct the current perpendicular to side A, B, or C? Explain. b: In a different experiment, current is passed through side A (and out the opposite side). Use 0.2m×0.1m×0.02m for the dimensions of the solid and assume that the solid is made of copper so its resistivity is rho = 1.69×10−8 Ωm and ne− = 8.49×1028 e−/m3. The potential difference across the solid is maintained at 0.001 V. Determine the drift velocity of the electrons and then how long it would take a single electron to pass completely through the copper.

Respuesta :

Answer:

a) we know that resistance is directly proportional to the length of the conductor as length increases the resistance also increases so to direct the current with least resistance, then the direction is perpendicular to side B only other sides have a larger length than B.

b) [tex]V_d[/tex] = 4.35 × 10⁻⁴ m/s

time = 229.56 s

Explanation:

Given:

a) we know that resistance is directly proportional to the length of the conductor as length increases the resistance also increases so to direct the current with least resistance, then the direction is perpendicular to side B only other sides have a larger length than B.

Dimension =  0.2m × 0.1m × 0.02m

ρ = 1.69×10⁻⁸ Ωm

[tex]n_{e^-}[/tex] = 8.49 × 10²⁸ e−/m³

Potential difference across the solid = 0.001 V

Now,

from Ohm's law V = I × R

here

V is potential difference

I is the current

R is the resistance

or

R = [tex]\frac{\textup{I}}{\textup{A}}[/tex]

also,

R = [tex]\frac{\rho\times l}{A}[/tex]

Here, l is the length

A is the area

and,

the drift speed, [tex]V_d[/tex] = [tex]\frac{J}{ne} = \frac{I}{ne\times A}[/tex]

substituting value of I in the above equation, we get

[tex]V_d[/tex]= [tex]\frac{V}{\rho\times l\times ne}[/tex]

on substituting the values, we get

[tex]V_d = \frac{0.001}{1.69\times10^{-8}\times0.1\times8.49\times10^{28}\times1.6\times10^{-19}} \ \textup{m/s}[/tex]

or

[tex]V_d[/tex] = 4.35 × 10⁻⁴ m/s

also,

the time taken, t = [tex]\frac{\textup{Distance}}{\textup{speed}}[/tex]

= [tex]\frac{\textup{0.1}}{\textup{4.35\times10^{-4}}}[/tex]

= 229.56 s

ACCESS MORE