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A current of $5 \mathrm{~A}$ is passing through a non-linear magnesium wire of cross-section $0.04 \mathrm{~m}^{2}$. 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:
(Resistivity of magnesium $\rho=44 \times 10^{-8} \Omega \mathrm{m}$ )
Correct Option: , 3
$\mathrm{I}=\overrightarrow{\mathrm{J}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{JA} \cos (\theta)$
$5=J\left(\frac{4}{100}\right) \times \cos (60)$
$\mathrm{J}=5 \times 50=250 \mathrm{~A} / \mathrm{m}^{2}$
Now, $\overrightarrow{\mathrm{E}}=\rho \cdot \overrightarrow{\mathrm{J}}$
$=44 \times 10^{-8} \times 250=11 \times 10^{-5} \mathrm{~V} / \mathrm{m}$
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