current density J = I/A
in region 1
J1 = I/A1
A1 = I/J1
dirft speed v1 = I/(n*e*A1)
v1 = I*J1/(n*e*I) = J1/(n*e)
I = current in the conductor
n = electron density = 6*10^28
v1 = 3.8*10^3/(6*10^28*1.6*10^-19)
v1 = 3.95*10^-7 m/s
in region 2
A2 = A1/2 = I/(2*J1)
v2 = 2*J2/(n*e) = 2*3.8*10^3/(6*10^28*1.6*10^-19)
v2 = 7.92*10^-7 m/s
A current-carrying conductor made of aluminum gradually narrows as shown in the figure below. The cross-sectional...
An aluminum wire carrying a current of 1.80 A has a cross-sectional area of 2.5x10−6 m2. The density of aluminum is 2.7 g/cm3. (Assume three electrons are supplied by each atom.) (a) Find the charge density, n, for aluminum?
An aluminum wire having a cross-sectional are equal to 530 X 10m carries a current of 7.00 A. The density of aluminum is 2.70 /em! Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wre. 24.4719 X The equation for the rift velocity includes the number of charge carriers per volume, which in the cases qual to the number of atoms per volume. How do you calculate that if you...
1. An aluminum wire having a cross-sectional area equal to 3.50 10-6 m2 carries a current of 6.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire. 0.000193452 Incorrect: Your answer is incorrect. The equation for the drift velocity includes the number of charge carriers per volume, which in this case is equal to the number of atoms per volume. How...
Please answer this question:
An aluminum wire having a cross-sectional area of 2.10 ✕
10−6 m2 carries a current of 6.00 A. The
density of aluminum is 2.70 g/cm3. Assume each aluminum
atom supplies one conduction electron per atom. Find the drift
speed of the electrons in the wire.
Please answer correctly!!!! Thank you so much in advance!
Safari File Edit View History Bookmarks Window Help 恸 令49%DEE1 Mon 3:12:37 PM 6.585*10*-5 X Your response is off by a multiple...
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all of 8 please.
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