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Two ends of a metallic wire having cross sectional are $1 {mm}^{2}$ are connected to a battery. Current 10 A is flowing through the wire. If resistivity of the material of the wire is $10^{- 4} ohm - m$ , the intensity of electric field inside the wire is

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0.1 V/m

$10^{3}$ V/m

$10^{2} V / m$

1 V/m

answer is B.

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Detailed Solution

J = Current density = $\frac{i}{A} = \frac{10}{10^{- 6}} A / m^{2} = 10^{7} A / m^{2}$

$Now \vec{E} = ρ \vec{J} \Rightarrow E = ρJ = 10^{- 4} x 10^{7} V / m = 10^{3} V / m .$

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