A copper wire of cross-sectional area 2-0mm-2- resistivity - 1-7times 10-8-Omega m- carries a current of 1 A- The electric field in the copper wire is-8-5times 10-4-V-m8-5times 10-3-V-m8-5times 10-5-V-m8-5times 10-2-V-m

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Toppr Admin · Accepted Answer

From given data A-2mm2-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0- -xA0- -xA0- -xA0- -x3C1-1-7-xD7-10-x2212-8-x3A9-m-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0-xA0- i-1AV-iR - from ohms law-V-i-xD7-x3C1-lAE-l-i-xD7-x3C1-lA

A copper wire of cross-sectional area $2.0 m m^{2}$ , resistivity = $1.7 \times 10^{- 8} Ω m$ , carries a current of 1 A. The electric field in the copper wire is:
$8.5 \times 10^{- 4} V / m$
$8.5 \times 10^{- 3} V / m$
$8.5 \times 10^{- 5} V / m$
$8.5 \times 10^{- 2} V / m$

from given data $A = 2 m m^{2}$
$ρ = 1.7 \times 10^{- 8} Ω m .$
$i = 1 A$
$V = i R$ ( from ohms law.)
$V = i \times \frac{ρ l}{A}$
$E . l = i \times \frac{ρ l}{A}$

A copper wire of cross-sectional area 2.0mm2, resistivity = 1.7×10−8Ωm, carries a current of 1 A. The electric field in the copper wire is:8.5×10−4V/m8.5×10−3V/m8.5×10−5V/m8.5×10−2V/m

from given data A=2mm2 ρ=1.7×10−8Ωm. i=1AV=iR ( from ohms law.)V=i×ρlAE.l=i×ρlA

A copper wire of cross-sectional area $2.0 m m^{2}$ , resistivity = $1.7 \times 10^{- 8} Ω m$ , carries a current of 1 A. The electric field in the copper wire is:
$8.5 \times 10^{- 4} V / m$
$8.5 \times 10^{- 3} V / m$
$8.5 \times 10^{- 5} V / m$
$8.5 \times 10^{- 2} V / m$

from given data $A = 2 m m^{2}$
$ρ = 1.7 \times 10^{- 8} Ω m .$
$i = 1 A$
$V = i R$ ( from ohms law.)
$V = i \times \frac{ρ l}{A}$
$E . l = i \times \frac{ρ l}{A}$