Question #219171
A typical copper might have 2*10^21 free electrons in 1cm of it's length. Suppose that the drift speed of the electrons along the wire is 0.05cm/s
a. How many electrons will pass through a cross section wire each second
b. How large a current would be flowing in the wire
1
Expert's answer
2021-07-21T09:51:43-0400

Gives

Number of free electrons (n)=2×10212\times10^{21}

Length(l)=1Cm

V=0.05cm/sec

Number per second

=numberlength×velocity\frac{number}{length}\times velocity


=2×10211×0.05=1×1020electron/sec=\frac{2\times10^{21}}{1}\times0.05=1\times10^{20} electron/sec

(b)

I=Qt=1×1020×1.6×10191=16AI=\frac{Q}{t}=\frac{1\times10^{20}\times1.6\times10^{-19}}{1}=16A


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