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A copper wire of length L and cross-sectional area A carries a current I. If the specific resistance (resistivity) of copper is rho, what is the magnitude of the electric field inside the wire?

rho*I/A
rho*A/I
I/(rho*A)
rho/L

Correct answer: rho*I/A

Solution

By Ohm's law in point form, E = rho*J = rho*(I/A), independent of the wire's length.

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