Question

Current density j at an area a=(2i+3j)mm2 is j=(3j+4k).Current through the area is

Answer 1

we know, relation between current (i) , current density ($\vec{J}$) and cross sectional area ($\vec{A}$) is given by,
$i=\vec{J}.\vec{A}$

here it is given that , $\vec{A}$ = (2i + 3j+ 0k) mm² = (2i + 3j) × 10^-6 m²

and $\vec{J}$ = (0i + 3j + 4k) A/m²

now, i = (2i + 3j + 0k) × 10^-6 .(0i + 3j + 4k) A

= (2 × 0 + 3 × 3 + 0 × 4) × 10^-6

= 9 × 10^-6 A = 9$\mu A$

hence, current through the area is 9$\mu A$

Answer 2

Answer:

hope it's helpful for you

Explanation:

we know, relation between current (i) , current density (\vec{J}

J

) and cross sectional area (\vec{A}

A

) is given by,

i=\vec{J}.\vec{A}i=

here it is given that , \vec{A}

A

= (2i + 3j+ 0k) mm² = (2i + 3j) × 10^-6 m²

and \vec{J}

J

= (0i + 3j + 4k) A/m²

now, i = (2i + 3j + 0k) × 10^-6 .(0i + 3j + 4k) A

= (2 × 0 + 3 × 3 + 0 × 4) × 10^-6

= 9 × 10^-6 A = 9\mu AμA

hence, current through the area is 9\mu AμA