# What electric field strenght is needed to create a 5.0 A current in a 2.0 mm diameter iron wire?

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Cutnell, John & Johnson, Kenneth. Physics Fourth Edition. NY: John Wiley and Sons Inc., 1998: 591. lists the resitivity of iron as 97 nanoOhm meters.

Wiki lists it as 100 nOhm m

We'll use 100 nOhm meters.

Since you need an answer in Volts/meter, that fits well with not knowing the exact length of the iron wire, because all we know is the resistance per unit length (Ohms/meter). So, ohms law becomes:

Volts/meter = Amps * Ohms/meter.

All we need to find out is the resistance per unit length of iron wire.

R = resistivity * length/area

R/length = resistivity / area

R/meter = 100*10^-9 Ohms meter / pi * 0.001 meters^2

R/meter = 32 microOhms / meter

Now you can plug that number back into the "Ohms law per meter" equation and get your electric field.

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• The length of the wire also factors in. You've got too many variables for anyone to answers this question.

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