An 18 gauge copper wire (diameter 1.02 mm) carries a current with a current density of 2.90×10^6 A/m2. The density of free electrons for copper is 8.5×10^28 electrons per cubic meter.Calculate the current in the wire.Express your answer with the appropriate units.Calculate the magnitude of the drift velocity of electrons in the wire.Express your answer with the appropriate units.

Question
Asked Nov 12, 2019
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An 18 gauge copper wire (diameter 1.02 mm) carries a current with a current density of 2.90×10^6 A/m2. The density of free electrons for copper is 8.5×10^28 electrons per cubic meter.

Calculate the current in the wire.
Express your answer with the appropriate units.
Calculate the magnitude of the drift velocity of electrons in the wire.
Express your answer with the appropriate units.
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Expert Answer

Step 1

The equation for the current in the wire is given by

I JA
Here, I is the current, J is the current density and A is the area of cross section
The equation for the area of cross section of the wire is given by
ла?
4
Here, d is the diameter of the wire.
Thus
ла'л
I
4
Substitute the numerical values in the above equation
T(1.02x103 m(2.90 x 10 A/m2)
I =
4
= 2.37 A
The equation for the drift velocity of the electrons in the wire is given by
I
neA
is the drift velocity, n is the free electron density and e is the magnitude of charge of the
Here,
electron
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I JA Here, I is the current, J is the current density and A is the area of cross section The equation for the area of cross section of the wire is given by ла? 4 Here, d is the diameter of the wire. Thus ла'л I 4 Substitute the numerical values in the above equation T(1.02x103 m(2.90 x 10 A/m2) I = 4 = 2.37 A The equation for the drift velocity of the electrons in the wire is given by I neA is the drift velocity, n is the free electron density and e is the magnitude of charge of the Here, electron

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Step 2

Putting the equatio...

41
nd'ne
Substitute the numerical values in the above equation.
4(2.37 A)
(1.02x10 m) (8.5x102) (1.60 x10"
C)
2.13x10 m/s
help_outline

Image Transcriptionclose

41 nd'ne Substitute the numerical values in the above equation. 4(2.37 A) (1.02x10 m) (8.5x102) (1.60 x10" C) 2.13x10 m/s

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