Please Solve this, especially the area of cross section part
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Answer:
2.5×10-4 ms-1
in a small time interval dt, the number of electrons leaving a cross section is (4×1028)(10−5)vdt.
Therefore, charge crossing in time dt is :
dq=(1.6×10−19)(4×1028)(10−5)vdt
Hence, I=dtdq=6.4×104v=16
This gives: v=2.5×10−4 m/s
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