Consider an electrophysiology experiment in which a 0.200 ~mm0.200 mm diameter round silver wire is used as a stimulation electr
ode. In this experiment, a current of 4.69~ mA4.69 mA is applied through the wire. The density of free electrons in silver is 5.86 \times 10^{28}~ \frac{\text{electrons}}{m^3}5.86×10 28 m 3 electrons . Calculate the magnitude of the drift velocity of the electrons in the silver wire under these conditions. Enter your answer in units of \frac{\text{meters}}{\text{second}} second meters using calculator scientific notation (e.g., 1.23e-41.23e−4 for 0.0001230.000123).
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1. Amperes, is the SI unit (also a fundamental unit) responsible for current.
2. Δq over Δt technically
Rearrange for Δq
I x Δt = Δq
1.5mA x 5 = Δq
Δq = 0.0075
Divide this by the fundamental charge "e"
Electrons: 0.0075 / 1.60 x 10^-19
Electrons: 4.6875 x 10^16 or 4.7 x 10^16
3. So we know that the end resistances will be equal so:
ρ = RA/L
ρL = RA
ρL/A = R
Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper
We are looking for L2 so we can isolate using algebra to get:
If you fill in those values you get 0.0205
or 2.05 cm