Given, radius of cylindrical wire, r=0.5mm=0.5×10−3m Conductivity, σ=5×107S∕m Electric field, E=10mV∕m=10×10−3V∕m We know that current density, ∴J=σE =5×107×10×10−3=5×105A∕m2 Also, J=I∕A⇒I=JA ⇒I=5×105×π×(0.5×10−3)2 =5×105×π×25×10−8=125π×10−3 ⇒x3πmA=125πmA⇒x3=53 ⇒x=5