I=J⋅A=JAcos(θ)
5=J(1004)×cos(60)
J=5×50=250Am−2
Now, E=ρ×J
=44×10−8×250=11×10−5Vm−1
JEE Main 2021 — Physics Electromagnetism
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)
Held on 20 Jul 2021 · Verified 6 Jul 2026.
11×10−2Vm−1
11×10−7Vm−1
11×10−5Vm−1
11×10−3Vm−1
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