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**Hint:**The amount of charge per unit that flows through the unit of area of chosen cross section. The total amount of current flowing through the one unit of cross section. The S.I unit of current density is ampere per meter square, whereas the ampere is the unit of current and meter square is the unit of cross sectional area.

**Complete step by step answer:**

Given that, current density at certain point is $\overrightarrow{J} = \left( {2 \times {{10}^4}\hat{j}} \right)J{m^{ - 2}}$. An area of cross sectional is $\overrightarrow{S}= \left( {2\hat{i} + 3\hat{j}} \right)c{m^2}$. From the given data, we have to find out the rate of charge flow that means current. From the current relation, $I = \overrightarrow{J} \times \overrightarrow{A}$.

From the given data, substituting the value of $\mathop j\limits^ \to $ and area of cross sectional is nothing but $ \overrightarrow{A}$ and we are converting $c{m^2}$ into ${m^2}$,

By substituting the values we get,

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

In multiplication products $\hat{i} $ becomes $0$, and $\hat{j}$ becomes $1$. Then we get,

$I = 3 \times 2 \times {10^4} \times {10^{ - 4}} \\

\therefore I = 6\,A$

Here as the bases are equal the exponents must be added, by adding those exponents we get zero.

**Hence, the rate of charge flow that means the current is, $I = 6\,A$.**

**Note:**Current density vector is defined as the vector of magnitude is the electric current per cross section area at a certain point mentioned in the space.the Current flow is divided by the area of cross section. Area of the cross section is parallel to either top or bottom of the solid, the product of length and height or the product of width and height are the area of the area cross sectional. The S.I unit of area of cross section is meter square.

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