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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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