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 : (Resistivity of magnesium \\rho=44 \\times10^{-8} \\Omega{m} ) [20 Jul 2021 Shift 1]

Correct Answer is : 11×10−5V∕m

JEE Main 20 July 2021 Shift 1 Solved Paper

Section: Physics

Question : 13 of 90

Marks: +1, -0

A current of 5A is passing through a non-linear magnesium wire of cross-section 0.04m2. 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 :
(Resistivity of magnesium ρ=44×10−8Ωm )

[20 Jul 2021 Shift 1]

11×10−2V∕m
11×10−7V∕m
11×10−5V∕m
11×10−3V∕m

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