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Chapter 26: Problem 11

11\. Longer Wire A wire with a resistance of \(6.0 \Omega\) is drawn out through a die so that its new length is three times its original length. Find the resistance of the longer wire, assuming that the resistivity and density of the material are unchanged.

Short Answer

Expert verified
The resistance of the longer wire is \( 54 \Omega \).

Step by step solution

01

Understand the Original Resistance Formula

The resistance of a wire is given by the formula: \[ R = \rho \frac{L}{A} \] where \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area.
02

Relate the New Length to Its Original

The length of the new wire is three times the original length. Hence, if the original length is \( L \), the new length is \( 3L \).
03

Understand Volume Conservation of the Wire

The volume of the wire remains constant when stretched. Volume \( V = A \times L \). Hence, \[ A_1 L = A_2 (3L) \] where \( A_1 \) is the original cross-sectional area and \( A_2 \) is the new cross-sectional area.
04

Derive the New Cross-Sectional Area

Since the volume is constant, we can solve for the new cross-sectional area: \[ A_2 = \frac{A_1}{3} \]
05

Substitute into the Resistance Formula

We substitute the new length and new cross-sectional area into the resistance formula: \[ R_2 = \rho \frac{3L}{\frac{A_1}{3}} = \rho \frac{3L \times 3}{A_1} = 9 \rho \frac{L}{A_1} \]
06

Relate to Original Resistance

Recognize that the original resistance \( R = \rho \frac{L}{A_1} \). Therefore: \[ R_2 = 9R \]
07

Substitute the Given Values

Given the original resistance \( R = 6 \Omega \), the new resistance is: \[ R_2 = 9 \times 6 \Omega = 54 \Omega \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

resistivity
Resistivity is a material's inherent property that indicates how much it resists the flow of electric current. It is represented by the symbol \( \rho \).
Resistivity is measured in ohm-meters (\( \Omega \cdot m \)). Different materials have different resistivities.
A lower resistivity means the material conducts electricity better.
In the formula for resistance, \( R \), resistivity is a crucial factor:
\[ R = \rho \frac{L}{A} \]
Here, \( L \) is the wire's length, and \( A \) is its cross-sectional area. The resistivity will not change if the material remains the same through the process.
cross-sectional area
The cross-sectional area of a wire, denoted as \( A \), is the area of a slice of the wire cut perpendicular to its length.
It affects the wire's resistance, as seen in the formula:
\[ R = \rho \frac{L}{A} \]
If the cross-sectional area decreases, the resistance increases.
In the exercise, the wire's length is tripled, but its volume remains the same.
This means the cross-sectional area must decrease since the volume (\( V \)) is calculated by \( V = A \times L \).
When the volume is conserved and length increases, the new cross-sectional area \( A_2 \) becomes one-third of the original \( A_1 \):
\[ A_2 = \frac{A_1}{3} \]
length of wire
The length of the wire is another crucial factor in determining resistance.
In the resistance formula:
\[ R = \rho \frac{L}{A} \]
\( L \) represents the wire鈥檚 length.
If the length increases, the resistance increases proportionally.
In the given problem, the wire鈥檚 length is tripled.
Thus, the new length \( L_2 \) is:
\[ L_2 = 3L \]
The longer the wire, the more resistance it offers to the flow of current due to more collisions between the electrons and atoms.
volume conservation
Volume conservation means that even though the wire is stretched, its total volume remains the same.
This principle is vital in solving the problem and understanding how cross-sectional area and length are related.
The wire volume \( V \) is calculated by:
\( V = A \times L \)
Since the volume doesn鈥檛 change, we have:
\[ A_1 \times L = A_2 \times 3L \]
\( A_1 \) is the original area, and \( A_2 \) is the new area after stretching.
Solving for \( A_2 \), we get:
\[ A_2 = \frac{A_1}{3} \]
This relationship shows how the cross-sectional area decreases by one-third when the length is tripled.
Understanding this conservation is key to solving the resistance in the new wire comprehensively.

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Most popular questions from this chapter

33\. A Fuse A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to \(440 \mathrm{~A} / \mathrm{cm}^{2} .\) What diameter of cylindrical wire should be used to make a fuse that will limit the current to \(0.50 \mathrm{~A}\) ?

31\. A Beam A beam contains \(2.0 \times 10^{8}\) doubly charged positive ions per cubic centimeter, all of which are moving north with a speed of \(1.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). (a) What are the magnitude and direction of the current density \(\vec{J}\) ? (b) Can you calculate the total current \(i\) in this ion beam? If not what additional information is needed?

41\. Gas Discharge Tube A current is established in a gas discharge tube when a sufficiently high potential difference is applied across the two electrodes in the tube. The gas ionizes; electrons move toward the positive terminal and singly charged positive ions toward the negative terminal. (a) What is the magnitude of the current in a hydrogen discharge tube in which \(3.1 \times 10^{18}\) electrons and \(1.1 \times\) \(10^{18}\) protons move past a cross- sectional area of the tube each second? (b) What is the direction of the current density \(\vec{J}\) ?

27\. Linear Accelerator A linear accelerator produces a pulsed beam of electrons. The pulse current is \(0.50 \mathrm{~A}\), and each pulse has a duration of \(0.10 \mu \mathrm{s}\) (a) How many electrons are accelerated per pulse? (b) What is the average current for an accelerator operating at 500 pulses/s? (c) If the electrons are accelerated to an energy of \(50 \mathrm{MeV}\), what are the average and peak powers of the accelerator?

42\. A Block A block in the shape of a rectangular solid has a crosssectional area of \(3.50 \mathrm{~cm}^{2}\) across its width, a front-to-rear length of \(15.8 \mathrm{~cm}\), and a resistance of \(935 \Omega\). The material of which the block is made has \(5.33 \times 10^{22}\) conduction electrons/m \(^{3}\). A potential difference of \(35.8 \mathrm{~V}\) is maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its value? (c) What is the average or drift speed of the conduction electrons? (d) What is the magnitude of the electric field in the block?
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