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In a closed circuit, the current I (in a...

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

A

`1.25xx10^(19)`

B

`6.25xx10^(20)`

C

`5.25xx10^(19)`

D

`2.55xx10^(20)`

Text Solution

Verified by Experts

The correct Answer is:
B
`I=4-0.08tA`
or `(dq)/(dt)=4-0.08tA`
or `q=int_(0)^(50)(4-0.08t)dtC`
or `Ne=[4t-(0.08t^(2))/(2)]_(0)^(50)=100C`
where N is number of electrons.
or `N=(100)/(e)=(100)/(1.6xx10^(-19))`
`=6.25xx10^(20)`
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