Current Electricity

JEE Main level, single correct, medium

22 of 380

Question 22 of 380

A current of 5A5A is passing through a non-linear magnesium wire of cross-section 0.04m20.04{m}^{2}. At every point the direction of current density is at an angle of 60∘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 ρ=44×10−8Ωm\rho =44\times {10}^{-8}\Omega m)

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Answer: (C) 11×10−5Vm−111\times {10}^{-5}V{m}^{-1}

I=J⃗⋅A⃗=JAcos⁡(θ)I=\vec{J}\cdot \vec{A}=JA\cos (\theta )
5=J(4100)×cos⁡(60)5=J\left(\dfrac{4}{100}\right)\times \cos (60)
J=5×50=250Am−2J=5\times 50=250A{m}^{-2}
Now, E⃗=ρ×J⃗\vec{E}=\rho \times \vec{J}
=44×10−8×250=11×10−5Vm−1=44\times {10}^{-8}\times 250=11\times {10}^{-5}V{m}^{-1}

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