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Chapter 27: Q. 21 (page 762)

21. A 0.0075V/m electric field creates a 3.9mA current in a 1.0-mm-diameter wire. What material is the wire made of?

Short Answer

Expert verified

The material is the wire made of Nichrome.

Step by step solution

01

Given information

The electric field in the wire is 0.0075V/m.The current through the wire is 3.9mA. The diameter of the wire is 1.0-mm-diameter wire.

So, the radius isr=0.5mm=5×10-4m.

02

Explanation

Determine the material from its conductivity σ by,
J=σE(1)
whereJ=Iπr2is the current density . That means the current per square meter passing through a cross section of the wire -- andE is the electric field strength in the material.
Plugging the formula for the current density into equation (1) and solving for σ,
σ=IEπr2(2)

03

Explanation

The Calculation: Now we can simply plug these values into equation (2):σ=IEπr2
=3.9×10−3A(0.0075V/m)π5×10−4m2

=6.6×105Ω−1×m−1

The only material with a conductivity near6.6×105Ω-1×m-1 is Nichrome.
04

Final answer

The material the wire is made from Nichrome.

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Most popular questions from this chapter

The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivities of muscle and fat are 13 Ω m and 25 Ω m, respectively. One person’s upper leg is 82% muscle, 18% fat. What current is measured if a 1.5 V potential difference is applied between the person’s hip and knee?

A20-cm-long hollow nichrome tube of inner diameter 2.8mm, outer diameter3.0mmis connected, at its ends, to a 3.0Vbattery. What is the current in the tube?

The two wires in FIGURE P27.62 are made of the same material. What are the current and the electron drift speed in the 2.0-mm-diameter segment of the wire?

23. A 0.50-mm-diameter silver wire carries a 20mAcurrent. What are (a) the electric field and (b) the electron drift speed in the wire?

56. A hollow metal cylinder has inner radius a, outer radius b, length L, and conductivityσ. The current Iis radially outward from the inner surface to the outer surface.
a. Find an expression for the electric field strength inside the metal as a function of the radius r from the cylinder's axis.
b. Evaluate the electric field strength at the inner and outer surfaces of an iron cylinder if a=1.0cm,b=2.5cm,L=10cm,and I=25A.
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