A current of 5A is passing through a non-linear magnesium wire of cross-section 0.04 m² . At every point, the direction of current density is at an angle of 60° with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is (Take, resistivity of magnesium, $\mathrm{\rho }=44\times {10}^{-8}\mathrm{\Omega }-\mathrm{m}$)

Given, current, I = 5 A

Area of cross-section of wire, A = 0.04 m²

We know that, $\mathrm{J}=\frac{\mathrm{I}}{\mathrm{A}}$

$\Rightarrow \mathrm{I}=\mathrm{JA}$

or I = J . A or $\mathrm{I}=\mathrm{JAcos}\mathrm{\theta }$

where, J = current density.

$\begin{array}{c}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}5=\mathrm{J}\left(\frac{4}{100}\right)\times \cos \left({60}^{\circ }\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[\because \mathrm{Given},\mathrm{\theta }={60}^{\circ }\right]\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{J}=500\times \frac{1}{2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left[\because \cos {60}^{\circ }=\frac{1}{2}\right]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{J}=250{\mathrm{Am}}^{-2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\end{array}$

The relation between electric field, current density and resistivity can be given as

$\begin{array}{c}\mathrm{E}=\mathrm{\rho }\cdot \mathrm{J}\\ =44\times {10}^{-8}\times 250\left[\because \text{ Resistivity, }\mathrm{\rho }=44\times {10}^{-8}\mathrm{\Omega }-\mathrm{m}\right]\\ =11\times {10}^{-5}\mathrm{V}/\mathrm{m}\end{array}$