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On applying an electric field of 15xx 10^(-6) vm^(-1)across a conductor , current density through it is 3.0 Am^(-2). The resistivity of the conductor is ....... |
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Answer» `45 XX 10^(-6) Omega m` `J = sigma ErArr (E)/(RHO)` `THEREFORE rho= (E)/(J) = (15 xx 10^(-6))/(3) = 5 xx 10^(-6) Omega`m |
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