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In an Iron mod of area $2\mathrm{~m}^2$ having resistivity $1 \times 10^{-7} \Omega$ . Current having magnitude 10A is following at an angle of $30^{\circ}$ from axis of rod. Calculate electric field in the rod.

Option: 1
$5.3 \times 10^{-6} \mathrm{~N} / \mathrm{C}$

Option: 2
$4.3 \times 10^{-7} \mathrm{~N} / \mathrm{C}$

Option: 3
$4.3 \times 10^{-5} \mathrm{~N} / \mathrm{C}$

Option: 4
$5.3 \times 10^{-7} \mathrm{~N} / \mathrm{C}$

Answers (1)

best_answer

V=IR

$V=I \frac{\rho \cdot l}{A_{\text {effective }}} \quad [R=\rho \frac{ l}{A}]$

A is Area in direction of current

$\frac{V}{l}=I \frac{\rho}{A_{effective}}$

$E=\frac{10A \times1 \times 10^{-7} \Omega \,m}{\frac{2}{\cos 30^{\circ}}} \\$ $E=\frac{10A \times 1 \times 10^{-7} \Omega}{\frac{2}{\cos 30^{\circ}}} \\$ $E=\frac{10A \times 1 \times 10^{-7} \Omega \, m}{\frac{2}{\cos 30^{\circ}}} \\$

$E=\frac{10^{-6} \cos 30^{\circ}}{2}=\frac{10^{-6}}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4} \times 10^{-6} \\$

$E=0.43 \times 10^{-6} \mathrm{~N} / \mathrm{C} \\$

$E=4.3 \times 10^{-7} \mathrm{~N} / \mathrm{C}$

Posted by

manish painkra

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