Question

The current density at a point is $\overrightarrow{J}=(2\times {10}^4\hat{j}){\text{Am}}^{-2}$. Find the rate of charge flow through a cross-sectional area $\overrightarrow{S}=(2\hat{i}+3\hat{j}){\text{cm}}^2$.

• A. $15$
• B. $20$
• C. $7$
• D. $6$

Answer 1

The correct option is D $6$

As we know that rate of charge flow is current.
So.
$\int dI=\int \overrightarrow{J}\cdot \overrightarrow{dS}$

$\Rightarrow I=\overrightarrow{J}\cdot \overrightarrow{S}$

putting the given value

$\Rightarrow I=\left[\right(2\times {10}^4\left)\hat{j}\right]\cdot \left[\right(2\hat{i}+3\hat{j})\times {10}^{-4}]$

$\therefore I=6\text{A}$

Hence, option $\left(a\right)$ is correct.

Why this question ? For a finite area, $I=\int \overrightarrow{J}\cdot \overrightarrow{dS}$ The direction of the current density is the same as the direction of the current. Thus, it is along the motion of the moving charges. If a current $I$ is uniformly distributed over an area, then $I=\overrightarrow{J}\cdot \overrightarrow{S}$