A current of 5; A is passing through a non-li | vedclass

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(B) The current $I$ is given by the dot product of current density $\vec{J}$ and area vector $\vec{A}$: $I = \vec{J} \cdot \vec{A} = J A \cos(	heta)$

Vedclass · Accepted Answer

$11 	imes 10^{-5}$

Vedclass · Answer

(B) The current $I$ is given by the dot product of current density $\vec{J}$ and area vector $\vec{A}$: $I = \vec{J} \cdot \vec{A} = J A \cos(	heta)$.Given $I = 5\; A$,$A = 0.04\; m^2$,and $	heta = 60^{\circ}$.Substituting the values: $5 = J 	imes 0.04 	imes \cos(60^{\circ})$.Since $\cos(60^{\circ}) = 0.5$,we have $5 = J 	imes 0.04 	imes 0.5 = J 	imes 0.02$.Thus,$J = \frac{5}{0.02} = 250\; A/m^2$.Using Ohm's law in vector form,$\vec{E} = ho \vec{J}$,the magnitude is $E = ho J$.$E = (44 	imes 10^{-8}) 	imes 250 = 11000 	imes 10^{-8} = 11 	imes 10^{-5}\; V/m$.

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