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Learning objectives

By the end of this section, you will be able to:

  • Describe the Hall effect.
  • Calculate the Hall emf across a current-carrying conductor.

The information presented in this section supports the following AP® learning objectives and science practices:

  • 3.C.3.1 The student is able to use right-hand rules to analyze a situation involving a current-carrying conductor and a moving electrically charged object to determine the direction of the magnetic force exerted on the charged object due to the magnetic field created by the current-carrying conductor. (S.P. 1.4)

We have seen effects of a magnetic field on free-moving charges. The magnetic field also affects charges moving in a conductor. One result is the Hall effect, which has important implications and applications.

[link] shows what happens to charges moving through a conductor in a magnetic field. The field is perpendicular to the electron drift velocity and to the width of the conductor. Note that conventional current is to the right in both parts of the figure. In part (a), electrons carry the current and move to the left. In part (b), positive charges carry the current and move to the right. Moving electrons feel a magnetic force toward one side of the conductor, leaving a net positive charge on the other side. This separation of charge creates a voltage ε size 12{ε} {} , known as the Hall emf    , across the conductor. The creation of a voltage across a current-carrying conductor by a magnetic field is known as the Hall effect    , after Edwin Hall, the American physicist who discovered it in 1879.

Figure a shows an electron with velocity v moving toward the left. The magnetic field B is oriented out of the page. The current I is running toward the right. The force vector on the electron points downward. An illustration of the right hand rule shows the right thumb pointing left with the v vector, the fingers pointing toward 7 o’clock with the B vector, the force vector on a positive charge pointing up and the force vector on a negative charge pointing down. Figure b shows a positive charge moving toward the right. The magnetic field lines are coming out of the page. The current I is running toward the right. The force on the positive charge is down. An illustration of the right hand rule shows the thumb pointing in the direction of the charge’s velocity, the fingers pointing in the direction of B, and F pointing down away from the palm.
The Hall effect. (a) Electrons move to the left in this flat conductor (conventional current to the right). The magnetic field is directly out of the page, represented by circled dots; it exerts a force on the moving charges, causing a voltage ε , the Hall emf, across the conductor. (b) Positive charges moving to the right (conventional current also to the right) are moved to the side, producing a Hall emf of the opposite sign, –ε . Thus, if the direction of the field and current are known, the sign of the charge carriers can be determined from the Hall effect.

One very important use of the Hall effect is to determine whether positive or negative charges carries the current. Note that in [link] (b), where positive charges carry the current, the Hall emf has the sign opposite to when negative charges carry the current. Historically, the Hall effect was used to show that electrons carry current in metals and it also shows that positive charges carry current in some semiconductors. The Hall effect is used today as a research tool to probe the movement of charges, their drift velocities and densities, and so on, in materials. In 1980, it was discovered that the Hall effect is quantized, an example of quantum behavior in a macroscopic object.

The Hall effect has other uses that range from the determination of blood flow rate to precision measurement of magnetic field strength. To examine these quantitatively, we need an expression for the Hall emf, ε size 12{ε} {} , across a conductor. Consider the balance of forces on a moving charge in a situation where B size 12{B} {} , v size 12{v} {} , and l size 12{l} {} are mutually perpendicular, such as shown in [link] . Although the magnetic force moves negative charges to one side, they cannot build up without limit. The electric field caused by their separation opposes the magnetic force, F = qvB size 12{F= ital "qvB"} {} , and the electric force, F e = qE size 12{F rSub { size 8{e} } = ital "qE"} {} , eventually grows to equal it. That is,

Questions & Answers

sound waves can be modeled as a change in pressure ,why is the change on in pressure used and not the actual pressure
Dotto Reply
what is the best
Kelly Reply
Water,air,fire
Maung
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Maung
two charges qA and qB are separated by a distance x. if we double the distance between the charges and triple the magnitude of the charge A, what happens to the magnitude of the force that charge A exerts on charge B. what happens to the magnitude of the force that charge B exerts on charge A
tanla Reply
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Owen Reply
what is force
Ibrahim Reply
force is a pull or push action on an object or a body.
joseph
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Frederick
numerical chapter number 3
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joined
Ibrahim
a reflected ray on a mirror makes an angle of 20degree with the incident ray when the mirror is rotated 15degree what angle will the incident ray now make with the reflected ray
Akinyemi Reply
what is simple harmonic motion
Solomon Reply
how vapour pressure of a liquid lost through convection
Yomzi Reply
Roofs are sometimes pushed off vertically during a tropical cyclone, and buildings sometimes explode outward when hit by a tornado. Use Bernoulli’s principle to explain these phenomena.
Aliraza Reply
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Aliraza
what's the basic si unit of acceleration
ELLOIN Reply
Explain why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.
Fabian Reply
Insulators (nonmetals) have a higher BE than metals, and it is more difficult for photons to eject electrons from insulators. Discuss how this relates to the free charges in metals that make them good conductors.
Muhammad Reply
Is the photoelectric effect a direct consequence of the wave character of EM radiation or of the particle character of EM radiation? Explain briefly.
Muhammad
Determine the total force and the absolute pressure on the bottom of a swimming pool 28.0m by 8.5m whose uniform depth is 1 .8m.
Henny Reply
how solve this problem?
Foday
P(pressure)=density ×depth×acceleration due to gravity Force =P×Area(28.0x8.5)
Fomukom
for the answer to complete, the units need specified why
muqaddas Reply
That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
Kyle
Practice Key Terms 2

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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