A current of 5   A is passing through a non-linear magnesium wire of cross-section 0 . 04   m 2 .…

A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m2. 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=44×10-8 Ωm)
  1. 11×10-2 V m-1
  2. 11×10-7 V m-1
  3. 11×10-5 V m-1
  4. 11×10-3 V m-1

Solution

I=J→·A→=JAcos(θ)
5=J4100×cos(60)
J=5×50=250 A m-2
Now, E→=ρ×J→
=44×10-8×250=11×10-5 V m-1

Asked in: JEE Main 2021 (20 Jul Shift 1)

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