364. The conductivity of a material with current density 1 unit and electric field 200 μV is
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d Explanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10-6 = 5000.
d
See lessExplanation: The current density is given by, J = σE. To find conductivity, σ = J/E =
1/200 X 10-6 = 5000.
D Explanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10-6 = 5000.
D
See lessExplanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10-6 = 5000.
Answer: d Explanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10-6 = 5000.
Answer: d
See lessExplanation: The current density is given by, J = σE. To find conductivity, σ = J/E =
1/200 X 10-6 = 5000.
d) Explanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10-6 = 5000.
d)
See lessExplanation: The current density is given by, J = σE. To find conductivity, σ = J/E = 1/200 X 10-6 = 5000.