What must be the length and the diameter of this wire?

Suppose you wish to fabricate a uniform wire from a mass m of a metal with density ρm and resistivity ρ. If the wire is to have a resistance of R and all the metal is to be used, what must be the length and the diameter of this wire? (Use any variable or symbol stated above as necessary.)

(a) l =

(b) d =

asked Feb 3, 2012 in Physics

Start out by writing out density as:

(pm) = m / V   (pm = density, m = mass, V = volume)

Now we can reaarange this and get that:

m = (pm)*V,   then Volume can further be replaced by A*l (where A=area, l=length)

m = (pm)*A*l

solving for A we get:

A = m /(pm)*l

Now write out formula for resistivity,

R = p*l / A (where R=resistance, p=resistivity of material, l=length, A=area)

Substituting the area into this equation we get:

R = (p*l) / (m /(pm)*l)  =  (p*(pm)*l^2 )/ m

Now we can solve for the length l:

l = squareroot(m*R / p*(pm))

***note that the p should actually be the greek symbol and not p.   And also the for the density (pm), the m should be in the the subscript.***

Part B, the diameter can also be determined through manipulating formulas.

begin with volume V,

V = m/(pm)

volume can be replaced by the formula for the area of the cylinder:

pi * r^2 * l  =  m/(pm)

now replace radius r with diameter d.

r = 1/2 d, so

pi * (1/2 d )^2 * l  =  m/(pm)

you can also replace length l, with the answer from part A, and get:

pi * (1/2 d )^2 * [squareroot(m*R / p*(pm))] =  m/(pm)

now rearrange the equation so you are solving for d:

d = 2 * squareroot( m/(pm)*pi)  * [squareroot(m*R / p*(pm))] )

To avoid confusion, this is what the answer should look like(*remember p is actually the greek symbol):

answered Feb 9, 2012 by ~Expert~ (3,020 points)
selected Feb 10, 2012 by Joey33