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Homework Statement
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?
Homework 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.
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.