/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 When \(100.0 \mathrm{V}\) is app... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When \(100.0 \mathrm{V}\) is applied across a 5 -gauge (diameter 4.621 \(\mathrm{mm}\) ) wire that is \(10 \mathrm{m}\) long, the magnitude of the current density is \(2.0 \times 10^{8} \mathrm{A} / \mathrm{m}^{2}\). What is the resistivity of the wire?

Short Answer

Expert verified
The resistivity of the wire is \(5.00 \times 10^{-6}\, \Omega\,\text{m}\).

Step by step solution

01

1. Write down the given values

Voltage (V): \(100.0\,\text{V}\) Length of wire (L): \(10\,\text{m}\) Diameter of the wire (d): \(4.621\,\text{mm}\) Current density (J): \(2.0 \times 10^8\,\text{A/m}^2\)
02

2. Calculate the cross-sectional area of the wire

Since the wire is cylindrical, its cross-sectional area (A) can be calculated using the formula for the area of a circle: \(A = \pi (\frac{d}{2})^2 \) But first, we need to convert the diameter from millimeters to meters: \(d = 4.621\,\text{mm} = 4.621 \times 10^{-3}\,\text{m}\) Then, plug in the value for d and calculate the area: \(A = \pi (\frac{4.621 \times 10^{-3}}{2})^2 = 1.675 \times 10^{-5}\,\text{m}^2\)
03

3. Calculate the current (I) based on the current density

By definition, current density (J) is the relationship between the current (I) and the cross-sectional area (A), so we can use the following equation to find the current: \(J = \frac{I}{A}\) Solving for I, we get: \(I = J \times A = (2.0 \times 10^8\,\text{A/m}^2) \times (1.675 \times 10^{-5}\,\text{m}^2) = 3350\,\text{A}\)
04

4. Use Ohm's Law to find resistance

Now that we have the current, we can use Ohm's Law to find the resistance (R) of the wire: \(V = I \times R\) Solving for R: \(R = \frac{V}{I} = \frac{100.0\,\text{V}}{3350\,\text{A}} = 0.02985\,\Omega\)
05

5. Calculate the resistivity

Finally, we can use the formula for resistivity (\(\rho\)) to solve the problem: \(\rho = \frac{R \times A}{L}\) Plugging in the values we know: \(\rho = \frac{0.02985\,\Omega \times 1.675 \times 10^{-5}\,\text{m}^2}{10\,\text{m}} = 5.00 \times 10^{-6}\, \Omega\,\text{m}\) Thus, the resistivity of the wire is \(5.00 \times 10^{-6}\, \Omega\,\text{m}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Ohm's Law
Understanding Ohm's Law is crucial to solving problems related to electrical circuits. It's a fundamental principle that relates three important quantities: voltage (V), current (I), and resistance (R). Ohm's Law is succinctly described by the equation:

\( V = I \times R \)

This tells us that the voltage across a conductor is directly proportional to the current flowing through it, provided the resistance remains constant. As such, if you know any two of these values, you can calculate the third. In the context of our textbook problem, once we determined the current flowing through the wire, we were able to use Ohm’s Law to find the wire's resistance by rearranging the equation to \( R = \frac{V}{I} \). This step is vital to ultimately finding the resistivity of the material, which informs us about the intrinsic property of the material to resist current flow.
Current Density
Current density is a measure of how much current is flowing through a unit area of material and is denoted by the symbol \( J \). It is calculated by the formula:

\( J = \frac{I}{A} \)

where \( I \) is the current and \( A \) is the cross-sectional area through which the current flows. Hence, it has units of amperes per square meter (\(\text{A/m}^2\)). In simpler terms, current density allows us to understand the distribution of current flow across different areas of a conductor. The problem provided required us to reverse-engineer the current using the given current density and the area. Higher current density implies a higher amount of current in a smaller area, which could lead to higher heat generation and potential risks if the material is not suited to handle such conditions.
Electrical Resistance
Electrical resistance is the property of a material that impedes the flow of electric current. It is often simply referred to as resistance and is denoted by the symbol \( R \). The higher the resistance, the lower the current for a given voltage, according to Ohm's Law. In practical terms, resistance can be thought of as the friction that electrons face when moving through a conductor.

The resistance of a wire is influenced by several factors: the material's intrinsic resistivity (\( \rho \) ), the length of the wire (\( L \) ), and the cross-sectional area (\( A \) ):

\( R = \frac{\rho \times L}{A} \)

In our problem, after calculating the wire's resistance using Ohm's Law, we arrived at its resistivity. Resistivity is a fundamental property that does not change with the dimensions of the material, unlike resistance, which varies with the size and shape of the object. By determining the resistivity, we gain insight into the material's capability to conduct electricity, which has practical applications in selecting materials for different electrical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A conductor carries a current that is decreasing exponentially with time. The current is modeled as \(I=I_{0} e^{-t / \tau},\) where \(I_{0}=3.00 \mathrm{A}\) is the current at time \(t=0.00 \mathrm{s}\) and \(\tau=0.50 \mathrm{s}\) is the time constant. How much charge flows through the conductor between \(t=0.00 \mathrm{s}\) and \(t=3 \tau ?\)

It was stated that the motion of an electron appears nearly random when an electrical field is applied to the conductor. What makes the motion nearly random and differentiates it from the random motion of molecules in a gas?

A resistor rated at \(250 \mathrm{k} \Omega\) is connected across two D cell batteries (each \(1.50 \mathrm{V}\) ) in series, with a total voltage of 3.00 V. The manufacturer advertises that their resistors are within \(5 \%\) of the rated value. What are the possible minimum current and maximum current through the resistor?

An aluminum wire 1.628 mm in diameter (14-gauge) carries a current of 3.00 amps. (a) What is the absolute value of the charge density in the wire? (b) What is the drift velocity of the electrons? (c) What would be the drift velocity if the same gauge copper were used instead of aluminum? The density of copper is \(8.96 \mathrm{g} / \mathrm{cm}^{3}\) and the density of aluminum is \(2.70 \mathrm{g} / \mathrm{cm}^{3} .\) The molar mass of aluminum is \(26.98 \mathrm{g} / \mathrm{mol}\) and the molar mass of copper is \(63.5 \mathrm{g} / \mathrm{mol} .\) Assume each atom of metal contributes one free electron.

Of what material is a resistor made if its resistance is \(40.0 \%\) greater at \(100.0^{\circ} \mathrm{C}\) than at \(20.0^{\circ} \mathrm{C} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.