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Chapter 28: Problem 47

A strip of copper \(75.0 \mu \mathrm{m}\) thick and \(4.5 \mathrm{~mm}\) wide is placed in a uniform magnetic field \(\vec{B}\) of magnitude \(0.65 \mathrm{~T}\), with \(\vec{B}\) perpendicular to the strip. A current \(i=57 \mathrm{~A}\) is then sent through the strip such that a Hall potential difference \(V\) appears across the width of the strip. Calculate \(V\). (The number of charge carriers per unit volume for copper is \(8.47 \times 10^{28}\) electrons \(/ \mathrm{m}^{3}\).)

Short Answer

Expert verified
The Hall voltage across the copper strip is approximately 364 渭V.

Step by step solution

01

Identify Given Values

We have a copper strip with a thickness of \(75.0 \mu \mathrm{m} = 75.0 \times 10^{-6} \mathrm{~m}\), a width of \(4.5 \mathrm{~mm} = 4.5 \times 10^{-3} \mathrm{~m}\), in a magnetic field of \(B = 0.65 \mathrm{~T}\). The current through the strip is \(i = 57 \mathrm{~A}\), and the number of charge carriers is \(n = 8.47 \times 10^{28} \mathrm{~m}^{-3}\). We need to calculate the Hall voltage \(V\).
02

Recall the Hall Effect Formula

The Hall voltage \(V\) is given by the formula:\[ V = \frac{i B}{n e t} \]Where \(e\) is the elementary charge \(1.6 \times 10^{-19} \mathrm{~C}\), \(n\) is the charge carrier density, \(B\) is the magnetic field, \(i\) is the current, and \(t\) is the thickness of the material.
03

Substitute the Values

Substitute the given values into the Hall voltage formula:\[ V = \frac{57 \times 0.65}{8.47 \times 10^{28} \times 1.6 \times 10^{-19} \times 75.0 \times 10^{-6}} \]
04

Calculate

Perform the calculations:1. Calculate the denominator: \(8.47 \times 10^{28} \times 1.6 \times 10^{-19} \times 75.0 \times 10^{-6} = 1.01716 \times 10^{5}\).2. Calculate the numerator: \(57 \times 0.65 = 37.05\).3. Compute the Hall voltage: \[ V \approx \frac{37.05}{1.01716 \times 10^{5}} \approx 3.64 \times 10^{-4} \mathrm{~V} \].
05

Convert Result to Millivolts

Convert the Hall voltage to millivolts for clarity: \(3.64 \times 10^{-4} \mathrm{~V} = 364 \mathrm{~\mu V}\).

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Key Concepts

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

Hall voltage
The Hall voltage is an intriguing concept that arises in the study of the Hall effect. It represents the voltage that is generated across a conductor when a magnetic field is applied perpendicular to the flow of electric charge. This phenomenon occurs because the magnetic field exerts a force on the charge carriers, pushing them to one side of the material. This build-up of charge creates an electric field that opposes further charge motion, ultimately leading to a measurable potential difference, known as Hall voltage.
In simple terms, when electrons move through a conductor in the presence of a magnetic field, they are deflected to one side, causing a voltage to appear across the conductor. This voltage can be calculated using the formula:
  • \( V = \frac{iB}{net} \)
This formula considers the current \(i\), magnetic field \(B\), thickness of the material \(t\), the number density of charge carriers \(n\), and the charge of each carrier \(e\).
Hence, Hall voltage is an essential measurement in understanding how materials respond to magnetic fields and allows determination of important properties like charge carrier density.
magnetic field
A magnetic field plays a crucial role in the Hall effect. It is essentially a region around a magnet or a current-carrying conductor where magnetic forces can be detected. In our scenario, the magnetic field is applied perpendicularly to the copper strip, and it influences the flow of charge carriers moving through the strip.
When current flows through the material, the magnetic field exerts a Lorentz force on the moving charges. This force causes the charges to deflect to one side, setting up a transverse potential difference, which is the Hall voltage.
  • Magnetic field strength is typically measured in Tesla (T). In the exercise, the field strength was given as \(0.65\ \text{T}\).
The magnetic field's orientation and magnitude directly impact the Hall voltage observed, as stronger fields will exert a greater force, thereby increasing the potential difference across the material. This relationship is fundamental to the functionality of devices such as Hall sensors, which rely on the detection of magnetic fields to operate properly.
charge carriers
Charge carriers are the particles that carry electric charge through a conductor. In metals like copper, the charge carriers are primarily electrons. The density of these charge carriers, denoted as \(n\), is a significant factor in determining the Hall voltage.
Electrons moving through a conductor in the presence of a magnetic field experience a force perpendicular to their direction of motion, caused by the magnetic field. This deflection of electrons leads to the build-up of a charge on one side of the conductor.
  • The number of charge carriers per unit volume is a crucial parameter that can be found from the Hall effect, and in this example, it is given as \(8.47 \times 10^{28} \ \text{carriers/m}^3\).
This density, combined with the magnetic field and current, helps in calculating the Hall voltage, which provides insights into the electrical behavior of the material. This is particularly useful for studying and characterizing materials in terms of their conductivity properties.
current density
Current density is an important concept in understanding electrical flow in conductors. It represents the amount of electric current flowing per unit area of a cross-section of the material. Expressed in amperes per square meter (A/m虏), it is given by the formula:
  • \( j = \frac{i}{A} \)
Where \(i\) is the electric current, and \(A\) is the cross-sectional area through which the current flows. In the Hall effect, this property is crucial as it interacts with the magnetic field to produce the Hall voltage.
In the given exercise, the copper strip's current density helps determine how the magnetic field affects the charge carriers. When these carriers drift due to the applied current and magnetic field, the result is the measurable Hall voltage across the strip.
This understanding of current density not only aids in calculating Hall voltages but also has applications in electrical engineering and material science, where knowing the distribution of current can impact the design and function of electronic components.

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Most popular questions from this chapter

A wire \(66.0 \mathrm{~cm}\) long carries a \(0.750 \mathrm{~A}\) current in the positive direction of an \(x\) axis through a magnetic field \(\vec{B}=(3.00 \mathrm{mT}) \hat{j}+\) \((14.0 \mathrm{mT}) \hat{k}\). In unit-vector notation, what is the magnetic force on the wire?

A conducting rectangular solid of dimensions \(d_{x}=5.00 \mathrm{~m}, d_{y}=3.00 \mathrm{~m}\), and \(d_{z}=2.00 \mathrm{~m}\) moves with a constant velocity \(\vec{v}=(20.0 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) through a uniform magnetic field \(\vec{B}=(40.0 \mathrm{mT}) \hat{\mathrm{j}}\) (Fig. 28-22). What are the resulting (a) electric field within the solid, in unit- vector notation, and (b) potential difference across the solid? (c) Which face becomes negatively charged?

A proton circulates in a cyclotron, beginning approximately at rest at the center. Whenever it passes through the gap between dees, the electric potential difference between the dees is \(350 \mathrm{~V}\). (a) By how much does its kinetic energy increase with each passage through the gap? (b) What is its kinetic energy as it completes 100 passes through the gap? Let \(r_{100}\) be the radius of the proton's circular path as it completes those 100 passes and enters a dee, and let \(r_{101}\) be its next radius, as it enters a dee the next time. (c) By what percentage does the radius increase when it changes from \(r_{100}\) to \(r_{101}\) ? That is, what is percentage increase \(=\frac{r_{101}-r_{100}}{r_{100}} 100 \% ?\)

A proton travels through uniform magnetic and electric fields. The magnetic field is \(\vec{B}=-3.25 \hat{\mathrm{i}} \mathrm{mT}\). At one instant the velocity of the proton is \(\vec{v}=2000 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}\). At that instant and in unit-vector notation, what is the net force acting on the proton if the electric field is (a) \(4.00 \mathrm{k} \mathrm{V} / \mathrm{m}\), (b) \(-4.00 \mathrm{k} \mathrm{V} / \mathrm{m}\), and (c) \(4.00 \hat{\mathrm{i}} \mathrm{V} / \mathrm{m}\) ?

An electron of kinetic energy \(600 \mathrm{eV}\) circles in a plane perpendicular to a uniform magnetic field. The orbit radius is \(12.5 \mathrm{~cm}\). Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion. (e) Through what potential difference would the electron have to be accelerated from rest to reach this kinetic energy?
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