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Question: The current in a conductor varies with time as \[I = 2t + 3{t^2}\] \(amp\). Where, \[I\] is in amper...

The current in a conductor varies with time as I=2t+3t2I = 2t + 3{t^2} ampamp. Where, II is in ampere and tt is in second. Electric charge flowing through a section of the conductor during t=2sect = 2\sec to t=3sect = 3\sec is:
(A) 10C10C
(B) 24C24C
(C ) 33C33C
(D) 44C44C

Explanation

Solution

The rate of flow of charge is called current. When a current flows through a cross section the electrons of the conductor atoms flow through the same cross section in the opposite direction. So, the rate of flow of charge with respect to time is current and hence charge flowing through a particular cross section for a particular interval of time is the area of the current-time curve for that interval.

Complete step-by-step answer:
In the question, the expression of current is given with respect to time, where II is in ampere and tt is in second.
We know the rate of change of charge is known as current. Thus, if we differentiate, charge flowing through a conductor with respect to time, we obtain the amount of current flowing through it.
In other words:
I=dqdtI = \dfrac{{dq}}{{dt}}
Here, qq is the amount of charge flowing.
This is termed as instantaneous current.
In the question, we need to find the amount of charge flowing from the expression of current, thus we integrate the expression of current and obtain the charge flowing.
SI unit of current is Ampere, and that of charge is Coulomb.
Thus,
q=Idtq = \smallint Idt
We need to find the charge flowing through the conductor form 2sec2\sec to 3sec3\sec , thus we integrate within that interval:
The amount of charge flowing through the conductor during 2sec2\sec to 3sec3\sec
q=23Idtq = \smallint _2^3Idt
Putting the expression forII, we get:
q=232t+3t2dtq = \smallint _2^32t + 3{t^2}dt
On integrating, we get:
q=[t2+t3]23q = \left[ {{t^2} + {t^3}} \right]_2^3
Putting the values of interval, we get:
q=(9+27)(4+8)q = (9 + 27) - (4 + 8)
Therefore, we obtain:
q=24Cq = 24C

Thus, option (B) is correct.

Note: The product of current and corresponding time will give us the instantaneous charge for that particular moment. However, for finding the total charge flowing for a particular interval of time, the area of the current-time curve will be equal to the total charge flowing for that particular interval of time