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Find the resistivity of a conductor in which a current density of 2.5 `A m^(-2)` is found to exist, when an electric field of `15 V m^(-1)` is applied on it. |
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Answer» Here, `J=2.5 Am^(-2) ,E=15 Vm^(-1)` Resistivity, `rho = (RA)/(l) = (V)/(I) (A)/(l) =(V//l)/(I//A) = (E)/(J) = (15)/(2.5)` `= 6 Omega m` |
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