QJEEM24Phb20f: JEE Main 2024 Physics - Class 12 JEEM24Phb20f Solution: The current in a conductor is expressed as , where is in Ampere …

To find the amount of electric charge that flows through a section of the conductor, we have to integrate the current over the given time interval. The current $I(t)$ as a function of time $t$ is given by:

I = 3 t^{2} + 4 t^{3}

The electric charge $Q$ that flows through the conductor from time $t = 1$ s to $t = 2$ s is calculated by integrating the current $I(t)$ with respect to time over this interval:

Q = \int_{t_{1}}^{t_{2}} I (t) d t

Substituting the given limits ( $t_1=1$ and $t_2=2$ ) and the expression for $I(t)$ , we get:

Q = \int_{1}^{2} (3 t^{2} + 4 t^{3}) d t

Now we'll integrate the function with respect to $t$ :

Q = {[\frac{3}{3} t^{3} + \frac{4}{4} t^{4}]}_{1}^{2}

Simplifying the integrated function:

Q = {[t^{3} + t^{4}]}_{1}^{2}

Substitute the upper and lower limits of the integration:

Q = [(2)^{3} + (2)^{4}] - [(1)^{3} + (1)^{4}]

Q = [8 + 16] - [1 + 1]

Q = 24 - 2

Q = 22 C

Therefore, the amount of electric charge that flows through the section of the conductor from $t=1$ s to $t=2$ s is 22 Coulombs.

QJEEM24Phb20f: JEE Main 2024 Physics - Class 12 JEEM24Phb20f Solution: The current in a conductor is expressed as , where is in Ampere …

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