What are (a) the current density and (b) the electron drift speed? 1. 2. The cur
ID: 1280628 • Letter: W
Question
What are (a) the current density and (b) the electron drift speed?
1.
2.
The current in a 1.10mm x 1.10mm square aluminum wire is 2.80A . What are (a) the current density and (b) the electron drift speed? 2.31 MA/m2 Submit My Answers Give Up Correct m / s Submit My Answers Give Up The wires leading to and from a 0.12-mm-diameter lightbulb filament are 1.5 mm in diameter. The wire to the filament carries a current with a current density of 4.3x105 A/m2 . What is the current in the filament? Express your answer to two significant figures and include the appropriate units. I = Value Units Submit My Answers Give Up What is the current density in the filament? Express your answer to two significant figures and include the appropriate units. J = Value UnitsExplanation / Answer
(A)
(a) To find current density (J), the equation is just J = I / A, where I = the current and A is the area of the surface.
In this case
I = 2.8 Amp
and the cross sectional area A isn't given, but you can find it using 1.10mm x 1.10mm.
Don't forget though to convert mm into m before dividing the current by area, to get m^2. (1000mm = 1m -> 1.10mm = 0.0011m)
A = 0.0011m(0.0011m) = 1.21E-6
Thus, you have J = (2.80A) / (1.21E-6) = 2314049.58 A/m^2.
(b) The electron drift speed can be found using this equation:
v = J/nq, where v is the drift speed you are trying to find, J is the density you just found, q = the charge of an electron, and n = the number of charges carried per unit volume (the electron density).
For aluminum n = 6.02E28 electrons/m^3
So
v = 2314049.58 / ((6.02E28)(1.6E-19) in m/s
v = 0.000240 m/s
(B)
Area of wire to filament = pi*r^2
= pi*(d/2)^2
d = 1.5mm = 1.5/1000m
Area of wire = pi*(1.5/2000)^2
= 1.767e-6 m^2
current density = 4.3e5 (I assume when you wrote 4.3x105 you mean 4.3 x 10^5 ?)
current = 0.76A
current density in the filament = current / area
= 0.76 / ( pi*(0.12/2000)^2 )
67198754 A/m^2
= 6.72 e7 A/m^2
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