In a closed circuit, the current / (in ampere) at an instant of time t (in second) is given by l = `4-0.08t`. The number of electrons flowing in 50 s through the cross-section of the conductor is

Options

A. `1.25xx10^(19)`
B. `6.25xx10^(20)`
C. `5.25xx10^(19)`
D. `2.55xx10^(20)`

Answer

B

Explanation

Let the number of electrons = N
`I=4-0.08tA`
or `(dQ)/(dt)=4-0.08tAorQ=underset(0)overset(50)int(4-0.08t)dtC`
or `Ne=[4t-(0.08t^(2))/(2)]_(0)^(50)=100C" "(because Q=Ne)`
or `N=(100)/(e)=(100)/(1.6xx10^(-19))=6.25xx10^(20)`