Question:
A current of 5 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 \(\rho = 44 \times 10^{-8}\ \Omega\mathrm{m}\))
A current of 5 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 \(\rho = 44 \times 10^{-8}\ \Omega\mathrm{m}\))
(Resistivity of magnesium \(\rho = 44 \times 10^{-8}\ \Omega\mathrm{m}\))
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Current \(I = \vec J\cdot\vec A = JA\cos 60^\circ\), and \(E = \rho J\).
Updated On: Nov 4, 2026
- \(11 \times 10^{-5}\) V/m
- \(11 \times 10^{-3}\) V/m
- \(11 \times 10^{-7}\) V/m
- \(11 \times 10^{-2}\) V/m
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The Correct Option is A
Solution and Explanation
The current through the cross-section is \(I = JA\cos\theta\) with \(\theta = 60^\circ\).
\(J = \frac{I}{A\cos 60^\circ} = \frac{5}{0.04 \times 0.5} = 250\ \mathrm{A/m^2}\).
Electric field \(E = \rho J = 44 \times 10^{-8} \times 250 = 1.1 \times 10^{-4}\ \mathrm{V/m}\).
\(E = 11 \times 10^{-5}\ \mathrm{V/m}\).
Answer: (1) \(11 \times 10^{-5}\) V/m
\(J = \frac{I}{A\cos 60^\circ} = \frac{5}{0.04 \times 0.5} = 250\ \mathrm{A/m^2}\).
Electric field \(E = \rho J = 44 \times 10^{-8} \times 250 = 1.1 \times 10^{-4}\ \mathrm{V/m}\).
\(E = 11 \times 10^{-5}\ \mathrm{V/m}\).
Answer: (1) \(11 \times 10^{-5}\) V/m
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