the diameter of copper wire is 2 mm , a steady current of 6.25 A is generated by 8.5×10^28/m^3 electrons flowing through it . calculate the drift velocity of conduction electrons
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Final. Answer :0.15 mm / s
Understanding :
1) CURRENT in a wire is given by
I. = neAV
where
I. = current
n = electron density
A = Area of cross section
V = Drift Velocity.
Steps
1) I = 6 .25 A
n = 8.25 * 10^(28)
radius = 1mm = 10^(-3) m
A = π * (10^-3)^2
= π *10^(-6) m^2
Th^n,
I = neAV
=> 6.25 = 8.5 * 10^(28) * 1.6 * 10(-19) * π * 10^(-6)
* V
=> V =
\frac{6.25 \times {10}^{ - 3} }{8.5 \times 1.6 \times \pi} = 0.15 \times {10}^{ - 3} = 0.15mm \: per \: s8.5×1.6×π6.25×10−3=0.15×10−3=0.15mmpers
ℍᝪℙℰ ⅈᝨ ℍℰℒℙՏ ℽᝪႮ
Understanding :
1) CURRENT in a wire is given by
I. = neAV
where
I. = current
n = electron density
A = Area of cross section
V = Drift Velocity.
Steps
1) I = 6 .25 A
n = 8.25 * 10^(28)
radius = 1mm = 10^(-3) m
A = π * (10^-3)^2
= π *10^(-6) m^2
Th^n,
I = neAV
=> 6.25 = 8.5 * 10^(28) * 1.6 * 10(-19) * π * 10^(-6)
* V
=> V =
\frac{6.25 \times {10}^{ - 3} }{8.5 \times 1.6 \times \pi} = 0.15 \times {10}^{ - 3} = 0.15mm \: per \: s8.5×1.6×π6.25×10−3=0.15×10−3=0.15mmpers
ℍᝪℙℰ ⅈᝨ ℍℰℒℙՏ ℽᝪႮ
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