Fabricating a Uniform Wire: Length & Diameter Calculation

[pkujulo2]

Homework Statement

Suppose you wish to fabricate a uniform wire from a mass m of a metal with density $$\rho$$m 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 = $$\pi$$r^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) / ($$\rho$$m*$$\rho$$) )

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$$\rho$$l) / $$\pi$$*R )What did I do wrong for b)?

 

[Apphysicist]

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.

 

[SammyS]

...

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.

 

[pkujulo2]

Thanks everyone.