Fabricating a Uniform Wire: Length & Diameter Calculation

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Homework Help Overview

The problem involves calculating the length and diameter of a uniform wire fabricated from a given mass of metal, considering its density and resistivity, while aiming for a specific resistance. The subject area includes concepts from physics related to electrical resistance and material properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various equations related to resistance, length, and cross-sectional area, exploring substitutions and relationships between the variables. There are attempts to derive expressions for length and diameter based on the provided parameters, with some questioning the correctness of specific calculations.

Discussion Status

The discussion is ongoing, with participants providing different approaches to derive the necessary equations. Some guidance has been offered regarding the relationships between the variables, but there is no explicit consensus on the correct method or final expressions for the diameter.

Contextual Notes

Participants note discrepancies in calculations and question assumptions related to the setup of the problem, particularly regarding the use of variables and the interpretation of the equations involved.

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Homework Statement


Suppose you wish to fabricate a uniform wire from a mass m of a metal with density \rhom and resistivity \rho.

If the wire is to have a resistance of R and all the metal is to be used, what must be the
a)length and
b) the diameter of this wire?
(Use any variable or symbol stated above as necessary.)

Homework Equations



\rho - resistivity,l -length, r -radius, m - mass, d - density.

a)
R = \rho l/A
d = m/Al
l = Rm/ d\rho

b) A = \pir^2 =\rho

The Attempt at a Solution


a)I have done this part already and got the question right
For this I ended up solving for l^2 and getting:
l = sqrt( (Rm) / (\rhom*\rho) )

b) for the cross sectional area, I used \pi*r^2 and substituted that for \pi*(d/2)^2

And I responded with the following answer that was marked incorrect:
d = sqrt( (4\rhol) / \pi*R )What did I do wrong for b)?
 
Last edited:
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By the same token as (a), in which you solved for A=m/(L*density), you can say:

L=m/(A*density)

Then R=(rho*m)/(density*A2)

Solve for A, and then also note A=Pi*diameter2/4

I got a double radical as an answer, but the units under the second radical was m4, which indeed, raised to the (1/4) is meters for diameter.
 
pkujulo2 said:
...

R = ρ ℓ/A
d = m/Aℓ

Solve one of the the above equations for A and substitute the value you got for ℓ.

Then solve A = π(a/2)2 for d substitute the previous value for A into that.

As Apphysicist said, you should have a radical of a quantity with a radical.
 
Thanks everyone.
 

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