Conductivity σ=5×107Sm−1
Radius r=0.5mm=5×10−4m
E=10×10−3Vm−1
J=σE=10×10−3×5×107
⇒J=5×105
⇒Ai=5×105
i=5×105×πr2
=5×105×π×(5×10−4)2
=125π×10−3A
Therefore, i=125πmA
Hence, x=5
JEE Main 2021 — Physics Electromagnetism
A cylindrical wire of radius 0.5mm and conductivity 5×107Sm−1 is subjected to an electric field of 10mVm−1. The expected value of current in the wire will be x3πmA. The value of x is _________.
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