If resistivity of copper conductor is `1.7xx10^(-8) Omega -m` and electric field is `100Vm^(-1)`, then current density will be

Options

A. `6xx10^(9) Am^(-2)`
B. `1.7xx10^(-6) Am^(-2)`
C. `1.7xx10^(-10) Am^(-2)`
D. `6xx10^(7) Am^(-2)`

Answer

A

Explanation

Current density, `J = sigma E`
where, `sigma` = conductivity `=(1)/("resistivity")=(1)/(rho)=(1)/(1.7xx10^(-8)Omega-m)`
`therefore " " J=(1)/(1.7xx10^(-8))xx100=(100)/(17)xx10^(9)`
`= 6xx10^(9)Am^(-2)`