current density j at an area A=(2i+3j)mm2 is j=(3j+4k).current through The area is
1)9mewA. 2)zero 3)18mewA 4)12mew A

Dear Student,

Please find below the solution to the asked query:

The current density can be defined as,

$J=\frac{dI\left(A\right)}{dA} \Rightarrow I=\overrightarrow{J}.\overrightarrow{A}$
Given that, 
Area vector as $\overrightarrow{A}=\left(2\hat{i}+3\hat{j}\right) m{m}^2=\left(2\hat{i}+3\hat{j}\right) \times {10}^{-6} m^2$ and current density vector is $\overrightarrow{J }=\left(3\widehat{j }+4\widehat{k }\right) \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$. Therefore,

$$\begin{gathered} I=\overrightarrow{J }.\overrightarrow{A }=\left(0\widehat{i }+3\widehat{j }+4\hat{k}\right).\left(2\widehat{i }+3\widehat{j }+0\widehat{k }\right)\times {10}^{-6}=\left(0\times 2+3\times 3+4\times 0\right)\times {10}^{-6} \\ \Rightarrow I=9 \mu A \end{gathered}$$

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