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

[student_fun]

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: m = \rho_{m} \cdot LA
Resistance as product of resisitivity and length divided by cross-sectional area: R = \rho_{r} \cdot \frac{L}{A}


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

\rho_{m}: Mass \ density
\rho_{r}: Resistivity

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:

R = \rho_{r} \cdot \frac{\frac{m}{\rho_{m} \pi r^{2}}}{\pi r^{2}}

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):

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

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.

 

[alphysicist]

Hi student_fun,

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:

R = \rho_{r} \cdot \frac{\frac{m}{\rho_{m} \pi r^{2}}}{\pi r^{2}}

Plug and chug values from problem (copper density taken to be 8.92 kilograms per meter cubed,

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?

 

[student_fun]

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.