JEE Main 2024 — Current Electricity Question with Solution

To calculate the amount of electric charge $Q$ that crosses through a section of the wire over a period of $20$ seconds, given the current varies with time as $I=I_0+\beta t$, where $I_0=20$ A (initial current) and $\beta =3$ A/s (rate of change of current with time), we use the concept of integration from calculus because the current is not constant but changes linearly with time.

The electric charge $Q$ is the integral of current $I$ over the time interval from $0$ to $20$ seconds. The formula for $Q$ is given by:

$Q={\int }_0^{T}I(t)dt$

Where $I(t)=I_0+\beta t$ and $T=20$ s. Substituting the given values:

$Q={\int }_0^{20}(20+3t)dt$

This integral can be solved in two parts:

1. The integral of the constant term $20$:

$\int 20dt=20t$

2. The integral of the linear term $3t$:

$\int 3tdt=\frac{3}{2}{t}^2$

Thus, combining these and evaluating from $0$ to $20$ seconds:

$Q={\left[20t+\frac{3}{2}{t}^2\right]}_0^{20}$

$Q=\left[20(20)+\frac{3}{2}(20{)}^2\right]-\left[20(0)+\frac{3}{2}(0{)}^2\right]$

$Q=400+600=1000$ C

Therefore, the amount of electric charge that crosses through a section of the wire in $20$ seconds is $1000$ C.