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° 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 =44 × 10-8, m)

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JEE Mains · Physics · STD 12 - Current electricity

JEE Mains 2021
Hard
MCQ

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 \(\rho=44 \times 10^{-8}\, \Omega m\))

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Answer & Solution

Correct Answer

(B) \(11 \times 10^{-5}\)

Step-by-step Solution

Detailed explanation

\(I=\vec{J} \cdot \vec{A}=J A \cos (\theta)\)

\(5=J\left(\frac{4}{100}\right) \times \cos (60)\)

\(J=5 \times 50=250\, {A} / {m}^{2}\)

Now, \(\vec{E}=\rho \cdot \vec{J}\)

\(=44 \times 10^{-8} \times 250=11 \times 10^{-5}\, {V} / {m}\)

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