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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  • 1 decade ago
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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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  • 1 decade ago

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

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  • Anonymous
    1 decade ago

    if the CSA of the cable is 2mm and the amps flowing are 5 then you need to know what the resistence is first in ohms.

    Then amps x resistence = voltage. And if your talking about emf that's calculated in micro webbers.

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