Electromagnetism question

Question

This problem is in part an exercise in unit conversion. Silver is the element with the highest electrical conductivity. Wikipedia’s silver entry gives the density of silver to be 10.49 g/cm3 and its atomic weight (or relative atomic mass) as 107.86 in atomic units. Only one of the 47 electrons attached to each silver atom is mobile.

a. In the lab you are given a silver wire that is 102 μm in diameter and 12.0 m long. You measure its resistance and find it is R = 23.9 Ω. What is the electrical conductivity σ of silver? (Use the appropriate precision given the precision of the numbers you use!)

b. Use equation 7.6 in the text to estimate the mean free path λ of electrons in silver. Look up the charge and mass of the electron and use the information above to find n. (f = 1.) Instead of some thermal velocity, use the quantum mechanical Fermi velocity, which is c/200, with c the speed of light.

c. Suppose you now are given a cube of silver, 1 m on each edge. You apply 1 V between two of the faces. What current will flow?

Answer 1

I will help you answer part a) and c) of your question because the equation referenced in part b) is not included.

a) conductivity σ is the inverse of resistivity ρ. So first let's find ρ and then invert it.

where R is the resistance, A is the cross-sectional area and l is the length

Assuming cylindrical wire, the cross-sectional area A = π r², and r = 1/2 x 102μm = 51.0 μm

A = π (51.0 μm)²

= π (51.0 x 10^-6m)²

A = 8.17 x 10^-9 m²

Now we can plug in our numbers into the resistivity equation:

ρ = 23.9 Ω x 8.17 x 10^-9 m² / 12.0 m

= 1.63 x 10^-8 Ωm

Now we can calculate for conductivity σ

σ = 1/1.63 x 10^-8 Ωm

= 6.13 x 10⁷ Ω^-1m^-1

c)

To find current flow for a 1V potential difference across 1m³ of silver we will be combining Ohm's law V= IR with the above resistivity equation we already used.

The idea is to solve for resistance R in this cube of silver and then once we have R, we can solve for I=V/R

Rearranging the resistivity equation to solve for

R = ρl/A

= 1.63 x 10^-8 Ωm x 1m / 1m²

= 1.63 x 10^-8 Ω

Now using Ohm's law to solve for current:

I = V/R

= 1V / 1.63 x 10^-8 Ω

= 6.13 x 10⁷ A

Which is an absolutely huge amount of current. In real life you likely would not be able to find a power supply that would be able to provide that much current, so likely the actual current would be limited by the internal resistance of the power supply.