A 0.60-mm-diameter wire made from an alloy (a combination of different metals) has a conductivity that decreases linearly with distance from the center of the wire: σ(r)=σ0-cr, with σ0=5.0×107Ω-1m-1and c=1.2×1011Ω-1m-2. What is the resistance of a 4.0mlength of this wire?

Short Answer

Expert verified

The resistance of a 4.0mlength of this wire is 0.109mΩ.

Step by step solution

01

Given information

A 0.60-mm-diameter wire has a conductivity that decreases linearly with distance from the center of the wire: σ(r)=σ0-cr, with σ0=5.0×107Ω-1m-1and c=1.2×1011Ω-1m-2.

02

Explanation

The conductance of the small part of the wire is,

dG=σdAL

Here, σis the conductivity of the wire, is the length of the wire, and dAis the area of the wire.

As the wire moves toward the center of wire, its conductivity decreases linearly as follows:

σ(r)=σ0-cr

Here, σ0and care constants and ris the distance of the wire from the center of the wire.

Substituteσ0-crfor σ(r)in the equation dG=σdALand solve for dG.

dG=σ0-crdAL

The area of the wire is,

A=πr2

Differentiate the equation on both sides,

dA=2πrdr

Substitute 2πrdrfor dAin the equation dG=σ0-crdALand solve for dG,.

dG=σ0-cr2πrdrL

Integrate the equation on both sides,

G=0rσ0cr2πrdrL

=2πL0rσ0rcr2dr

role="math" localid="1649160599266" =2πLσ0r22cr330r

=2πLσ0r22cr33

03

Explanation

The relation between the conductance and resistance is,

R=1G

Here, Ris the resistance of the wire.

Substitute 2πLσ0r22-cr33for G,

localid="1649160822320" R=12πLσ0r22-cr33(1)

The relation between diameter and resistance of the wire is,

r=d2

Here, dis the diameter of the wire.

Substitute 0.60mmfor d,

r=0.60mm2

=0.30mm10-3m1mm

=0.30×10-3m

Substitute the values in equation (1)

R=1(4.0m)5.0×107Ω-1m-10.30×10-3m22-1.2×1011Ω-1m-20.30×10-3m33

=0.54Ω

04

Final Answer

Therefore, the resistance of the 40m length wire is 0.54Ω.

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