A current of $5\; {A}$ is passing through a non-linear magnesium wire of cross-section $0.04\; {m}^{2}$. At every point the direction of current density is at an angle of $60^{\circ}$ with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is ....${V} / {m}$ (Resistivity of magnesium is $\rho=44 \times 10^{-8}\, \Omega m$)
JEE MAIN 2021, Diffcult
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$I=\vec{J} \cdot \vec{A}=J A \cos (\theta)$
$5=J\left(\frac{4}{100}\right) \times \cos (60)$
$J=5 \times 50=250\, {A} / {m}^{2}$
Now, $\vec{E}=\rho \cdot \vec{J}$
$=44 \times 10^{-8} \times 250=11 \times 10^{-5}\, {V} / {m}$
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