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Fabricate uniform wire out of 2.1g of copper w/ 0.3 ohm resist. - wire diameter?

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose you wish to fabricate a uniform wire out of 2.1 grams of copper. If the wire is to have a resistance of 0.3 ohms and if all of the copper is to be used, what must the diameter of this wire (in mm) be?


    2. Relevant equations
    mass (in kg), cross-sectional area extruded by a length (in m) multiplied by mass density: [tex]m = \rho_{m} \cdot LA[/tex]
    Resistance as product of resisitivity and length divided by cross-sectional area: [tex]R = \rho_{r} \cdot \frac{L}{A}[/tex]


    Re-written density relation in terms of L to be plugged into resistance equation:[tex]\frac{m}{\rho_{m} A} = L[/tex]

    [tex]\rho_{m}: Mass \ density[/tex]
    [tex]\rho_{r}: Resistivity[/tex]

    Where m is mass in kg, L is length of wire in meters, A is cross-sectional area of wire in m2, R is resistance in ohms, and rho is resistivity in ohm meters.


    3. The attempt at a solution
    Use resistance relation with L replaced by re-written form to solve for radius of cross-sectional area. A is replaced by cross-sectional area of a cylinder being pi * r2.

    New Equation:

    [tex]R = \rho_{r} \cdot \frac{\frac{m}{\rho_{m} \pi r^{2}}}{\pi r^{2}}[/tex]

    Plug and chug values from problem (copper density taken to be 8.92 kilograms per meter cubed, resistivity of copper taken to be 1.7E-8 ohm meters, 0.3 ohms is target resistance, and 0.0021 kg from converted 2.1 grams of available copper mass):

    [tex]0.3 = \frac{\frac{1.7E-8 * 0.0021}{8.92\pi r^{2}}}{\pi r^{2}}[/tex]

    Solving on TI-89 gives 1.0783E-3 which when converted from m to mm gives just 1.0783 as the cross-sectional radius. Then multiplying this by 2 to get final answer of 2.1565 mm is incorrect but close. I am not sure where I messed up. I am sure it must be simple at this point but I am just not seeing it at the moment. Can anyone give me a hint?

    I've looked at similar posts and I seem to be on the right track but cannot see my error.

    Thank you for your time.
     
  2. jcsd
  3. Apr 1, 2009 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi student_fun,

    I have not looked too closely at the rest of your post, but this is not right. Water has a density of 1000 kg/m^3, and copper would have a higher density than that. Does fixing this give you the right answer?
     
  4. Apr 1, 2009 #3
    ah geeze, thanks for pointing that out. The proper density I should have used was 8940 kg / m3. That solved it. Thanks for pointing out my simple error :) Was up till 2 AM which explains the silly error.
     
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