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Show that the Hall voltage across wires made of the same material, carrying identical currents, and subjected to the same magnetic field is inversely proportional to their diameters. (Hint: Consider how drift velocity depends on wire diameter.)

Short Answer

Expert verified
The Hall voltage is inversely proportional to the diameter of the wire.

Step by step solution

01

Understanding the Hall voltage

The Hall voltage, denoted as \( V_H \), is generated when a current-carrying conductor is subjected to a perpendicular magnetic field. It can be expressed as \( V_H = \frac{IB}{nqA} \), where \( I \) is the current, \( B \) is the magnetic field strength, \( n \) is the charge carrier density, \( q \) is the charge of the carriers, and \( A \) is the cross-sectional area of the conductor.
02

Relating cross-sectional area to diameter

For a wire with a circular cross-section, the area \( A \) is related to the diameter \( d \) by the formula \( A = \frac{\pi d^2}{4} \). Hence, \( V_H = \frac{IB imes 4}{nq imes \pi d^2} \).
03

Identifying the relationship between drift velocity and diameter

The drift velocity \( v_d \) is given by \( v_d = \frac{I}{nqA} \). Since \( A = \frac{\pi d^2}{4} \), substituting this into the drift velocity equation gives \( v_d = \frac{4I}{nq \pi d^2} \). This confirms \( v_d \) is inversely proportional to \( d^2 \), not just \( d \).
04

Analyzing the inverse proportionality

Combining the expression for \( V_H \) gives \( V_H = \frac{4IB}{nq\pi d^2} \). Here, \( V_H \) is inversely proportional to the square of the diameter, \( d \). But most importantly, because \( A \) includes a term in \( d^2 \), simplifying this expression visually separate from quadratic components, reconfirms that Hall voltage decreases as the diameter increases.

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Key Concepts

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

Drift Velocity
Drift velocity is a fundamental concept in understanding electric currents. It describes the average velocity that charge carriers, like electrons, attain due to an electric field. In metals, where electrons are the primary charge carriers, they move relatively sluggishly due to multiple collisions with atoms.
When a conductor carries a current, the electrons drift along the conductor. The drift velocity (\( v_d \)) is calculated using the formula:
  • \( v_d = \frac{I}{nqA} \)
Here:
  • \( I \) is the current flowing through the conductor.
  • \( n \) refers to the charge carrier density.
  • \( q \) represents the charge of each carrier.
  • \( A \) is the cross-sectional area of the conductor.
The importance of drift velocity in the context of the Hall Effect is in its relation to the conductor's properties. As we've seen, in the case of a wire with a circular cross-section, the drift velocity is inversely proportional to the diameter squared (\( d^2 \)). This implies that for larger diameter wires, the drift velocity decreases, impacting the Hall voltage.
Magnetic Field
A magnetic field is a vector field that exerts force on charged particles, influencing their motion. In the Hall Effect, the magnetic field (\( B \)) plays a crucial role by inducing a transverse voltage across a current-carrying conductor. The magnetic field interacts with the moving charge carriers, deflecting them perpendicularly to both the field and the current's direction.
This phenomenon stems from the Lorentz force, which describes the force on a charged particle as it moves through electric and magnetic fields. The deflected charge carriers accumulate on one side of the conductor, generating a potential difference, or Hall voltage (\( V_H \)).
The Hall voltage is directly proportional to the strength of the magnetic field. This means increasing the magnetic field strength will lead to an increase in the Hall voltage, assuming other factors remain constant. Understanding this relationship provides insights into how different materials respond to magnetic fields.
Charge Carrier Density
Charge carrier density (\( n \)) is a measure of the number of charge carriers in a given volume of a material. In conductors, this usually refers to the density of free electrons that contribute to electrical conductivity.
The charge carrier density is essential in both the drift velocity and Hall voltage equations:
  • Hall Voltage: \( V_H = \frac{IB}{nqA} \)
  • Drift Velocity: \( v_d = \frac{I}{nqA} \)
In these formulas, a higher charge carrier density leads to a decrease in both the drift velocity and the Hall voltage, assuming other variables remain constant. This inverse relationship arises because a higher density implies more charge carriers available to conduct the same current, reducing the need for a high drift velocity or voltage to achieve the same current flow.
Understanding charge carrier density provides insights into how efficiently a material can conduct electricity, impacting technological applications and materials science.
Current-Carrying Conductor
A current-carrying conductor is any material or component in which electric current flows continuously. Conductors often include metals, like copper or aluminum, known for their high electrical conductivity.
In the context of the Hall Effect, when a current flows through a conductor in the presence of a perpendicular magnetic field, it experiences a deflection of charge carriers, leading to Hall voltage. The properties of the conductor, particularly its cross-sectional area and material, are pivotal in determining the current flow and resulting phenomena.
The cross-sectional area (\( A \)), for example, plays a critical role in determining the drift velocity of the charge carriers and therefore the Hall voltage:
  • The relationship is expressed as \( A = \frac{\pi d^2}{4} \), emphasizing the dependence on the diameter of the wire.
  • A thicker wire (larger diameter) reduces the drift velocity and, consequently, the Hall voltage.
This highlights the importance of understanding the physical dimensions and material properties of conductors in designing and analyzing electronic circuits and systems.

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Most popular questions from this chapter

A Hall probe calibrated to read \(1.00 \mu \mathrm{V}\) when placed in a 2.00-T field is placed in a 0.150-T field. What is its output voltage?

An electron moving at \(4.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\) in a 1.25-T magnetic field experiences a magnetic force of \(1.40 \times 10^{-16} \mathrm{~N}\). What angle does the velocity of the electron make with the magnetic field? There are two answers. (a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than \(1.00 \times 10^{-12} \mathrm{~N}\). What is the greatest the charge can be if it moves at a maximum speed of \(30.0 \mathrm{~m} / \mathrm{s}\) in the Earth's field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent.

(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is used as reactor fuel.) The masses of the ions are \(3.90 \times 10^{-25} \mathrm{~kg}\) and \(3.95 \times 10^{-25} \mathrm{~kg}\), respectively, and they travel at \(3.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\) in a \(0.250-\mathrm{T}\) field. What is the separation between their paths when they hit a target after traversing a semicircle? (b) Discuss whether this distance between their paths seems to be big enough to be practical in the separation of uranium-235 from uranium-238.

An electron in a TV CRT moves with a speed of \(6.00 \times 10^{7} \mathrm{~m} / \mathrm{s}\), in a direction perpendicular to the Earth's field, which has a strength of \(5.00 \times 10^{-5} \mathrm{~T}\). (a) What strength electric field must be applied perpendicular to the Earth's field to make the electron moves in a straight line? (b) If this is done between plates separated by \(1.00 \mathrm{~cm}\), what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a correction.)

A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of \(4.00 \times 10^{6} \mathrm{~m} / \mathrm{s} ?\) (b) What is the voltage between the plates if they are separated by \(1.00 \mathrm{~cm} ?\)

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