Suppose you wish to fabricate a uniform wire from 1.80 g of copper. The wire is to have a resistance of

and all the copper is to be used.

(b) What must be the diameter of this wire?

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+2 votes

Best answer

Copper is a good conductor because its resistivity is low. We expect to find that this wire is very long and thin.

To solve for the variables, we will use the equation for resistance as a function of resistivity, length, and cross-sectional area.

A = m / (l*p) (*where A=area, m=mass, l=length, p=density)

R = l / (r*A) (*where R=resistance, l=length, r=resistivity, A=area)

Now substitute A into equation for R.

R = l / (r*m / (l*p)) = (r*p*l^{2}) / m

Now solve for L:

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

where m = 1.8×10^{-3} kg

R = 0.52 Ω

r = 1.70 × 10^{−8} Ω · m

p = 8.92 × 10^{3} kg/m^{3}

Plug in values and solve for l:

l = squareroot((1.8×10^{-3} kg * 0.52 Ω) / (1.70 × 10^{−8} Ω · m r * 8.92 × 10^{3} kg/m^{3} ))

l = 2.48m

Part B, the diameter can be found by noting that:

A = pi*r^2 (*r=radius, and r=d/2 d=diameter)

A=pi(d/2)^2 = m/(l*p)

solving for d:

d = squareroot (4*m/(pi*l*p))

plugging in the values we get that d:

d = squareroot (4*1.8×10^{-3} kg / (pi*2.48m * 8.92 × 10^{3} kg/m^{3}))

d = 3.219*10^-4 = 3.21 µm

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