   Chapter 17, Problem 58AP

Chapter
Section
Textbook Problem

A 50.0-g sample of a conducting material is all that is available. The resistivity of the material is measured to be 11 × 10−8 Ω · m, and the density is 7.86 g/cm3. The material is to be shaped into a solid cylindrical wire that has a total resistance of 1.5 Ω. (a) What length of wire is required? (b) What must be the diameter of the wire?

(a)

To determine
The length of conducting material in shape of solid cylindrical wire.

Explanation

Given Info: The mass of wire is 50.0g . The resistivity of material is 11×108Ωm and the density is 7.86gcm-3 . The resistance of wire is 1.5Ω Wire has the shape of solid cylinder.

Explanation:

Formula to calculate the volume is,

V=MD

• V is the volume of wire,
• M is the mass of the wire,
• D is the density of the wire,

Formula to calculate the length of wire is,

L=RAρ

• L is the length of wire,
• R is the resistance of the wire,
• A is the circular cross section of the wire,
• ρ is the resistivity of wire,

Equation to find A from V is given by,

A=VL

Use M/D for V in the above equation to rewrite A.

A=MDL

Substitute M/DL for A in the L=RA/ρ to rewrite L .

L=Rρ(MDL)L=RρD

Substitute 50

(b)

To determine
The diameter of the conducting material in shape of solid cylindrical wire.

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