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Chapter 29: Problem 28

Strip of Copper A strip of copper \(150 \mu \mathrm{m}\) 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 \(i=23 \mathrm{~A}\) is then sent through the strip such that a Hall potential difference \(\Delta V\) appears across the width of the strip. Calculate \(\Delta V\).(The number of charge carries per unit volume for copper is \(8.47 \times 10^{2 \mathrm{~s}}\) electrons/m \(^{3}\).)

Short Answer

Expert verified
The Hall potential difference is approximately 7.364 脳 10^(-5) V.

Step by step solution

01

- Identify known values

List the given values:Width of copper strip, w = 150 碌m (converted to meters: 150 脳 10^(-6) m)Magnetic field, B = 0.65 TCurrent, i = 23 ANumber of charge carriers per unit volume, n = 8.47 脳 10^28 electrons/m鲁Elementary charge, e = 1.6 脳 10^(-19) C
02

- Use Hall effect formula

The Hall voltage is given by the formula \[ \Delta V = \frac{Bi}{n e w} \] Substitute the known values into the formula.
03

- Substitute known values

Substitute B = 0.65 T, i = 23 A, n = 8.47 脳 10^28 electrons/m鲁, e = 1.6 脳 10^(-19) C, w = 150 脳 10^(-6) m into the formula: \[ \Delta V = \frac{0.65 T \times 23 A}{8.47 \times 10^{28} \text{ electrons/m}^3 \times 1.6 \times 10^{-19} C \times 150 \times 10^{-6} m} \]
04

- Calculate the Hall voltage

Complete the calculation: \[ \Delta V = \frac{0.65 \times 23}{8.47 \times 10^{28} \times 1.6 \times 10^{-19} \times 150 \times 10^{-6}} \] \[ \Delta V = \frac{14.95}{2.03 \times 10^5} \] \[ \Delta V \approx 7.364 \times 10^{-5} V \]

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Key Concepts

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

Magnetic Field
A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. In our example, the magnetic field strength is given as 0.65 Tesla (T). When a conductor, such as a strip of copper, is placed in this field and an electric current passes through it, the interaction between the magnetic field and the moving charge carriers (electrons) creates a phenomenon known as the Hall Effect. This essentially causes a voltage difference across the conductor, perpendicular to both the current and magnetic field direction.
Charge Carriers
Charge carriers are particles that carry electric charge, which can be either positive or negative. In metals like copper, the charge carriers are typically electrons. The number of charge carriers is a crucial factor in the Hall effect. For the copper strip in this exercise, we are given the number of charge carriers per unit volume, which is 8.47 脳 10^28 electrons/m鲁. This value, combined with the charge of each electron (1.6 脳 10^(-19) C), helps us calculate the Hall voltage.
Hall Voltage Calculation
The Hall voltage (\textDelta V) is the voltage difference generated across a conductor due to the Hall Effect. The formula for calculating the Hall voltage is: \[ \Delta V = \frac{Bi}{ne w} \] where: \( B \) is the magnetic field strength, \( i \) is the current flowing through the conductor, \( n \) is the number of charge carriers per unit volume, \( e \) is the elementary charge, and \( w \) is the width of the conductor. By plugging in the values: \( B = 0.65 \) T, \( i = 23 \) A, \( n = 8.47 \times 10^{28} \) electrons/m鲁, \( e = 1.6 \times 10^{-19} \) C, and \( w = 150 \times 10^{-6} \) m, we get: \[ \Delta V = \frac{0.65 \times 23}{8.47 \times 10^{28} \times 1.6 \times 10^{-19} \times 150 \times 10^{-6}} \] When you solve this, you get around \( 7.364 \times 10^{-5} \) V. This is the Hall voltage for the given copper strip under the specified conditions.
Current in Conductor
Current refers to the flow of electric charge in a conductor. In this case, a current of 23 Amperes (A) is being passed through the copper strip. When current flows through a conductor placed in a perpendicular magnetic field, it causes charge carriers to be deflected to one side of the conductor due to the Lorentz force. This movement of electrons results in the Hall voltage. The size of the current directly affects the magnitude of the Hall voltage, as seen in the Hall voltage formula.
Elementary Charge
The elementary charge is the charge carried by a single proton, which is equal to \( 1.6 \times 10^{-19} \) Coulombs (C). It is a fundamental physical constant and is essential in many physics equations, including the Hall voltage calculation. In the context of the Hall Effect, the elementary charge helps determine how the charge carriers (electrons in copper, in this case) will interact with the magnetic field and current to produce the Hall voltage. Every single electron carries this elementary charge and, when combined with the number of charge carriers per unit volume, plays a key role in calculating the Hall voltage.

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Most popular questions from this chapter

Heavy Ions Physicist S. A. Goudsmit devised a method for measuring the masses of heavy ions by timing their periods of revolution in a known magnetic field. A singly charged ion of iodine makes \(7.00\) rev in a field of \(45.0 \mathrm{mT}\) in \(1.29 \mathrm{~ms}\). Calculate its mass, in atomic mass units. (Actually, the method allows mass measurements to be carried out to much greater accuracy than these approximate data suggest.)

Current Loop A single-turn current loop, carrying a current of \(4.00 \mathrm{~A}\), is in the shape of a right triangle with sides \(50.0,120\), and \(130 \mathrm{~cm}\). The loop is in a uniform magnetic field of magnitude \(75.0 \mathrm{mT}\) whose direction is parallel to the current in the \(130 \mathrm{~cm}\) side of the loop. (a) Find the magnitude of the magnetic force on each of the three sides of the loop. (b) Show that the total magnetic force on the loop is zero.

Potential Difference An electron is accelerated through a potential difference of \(1.0 \mathrm{kV}\) and directed into a region between two parallel plates separated by \(20 \mathrm{~mm}\) with a potential difference of \(100 \mathrm{~V}\) between them. The electron is moving perpendicular to the electric field of the plates when it enters the region between the plates. What magnitude of uniform magnetic field, applied perpendicular to both the electron path and the electric field, will allow the electron to travel in a straight line?

Zero Total Force A closed wire loop with current \(i\) is in a uni. form magnetic field \(\vec{B}\), with the plane of the loop at angle \(\theta\) to the direction of \(\vec{B}\). Show that the total magnetic force on the loop is zero. Does your proof also hold for a nonuniform magnetic field?

Proton, Deuteron, Alpha A proton, a deuteron \((q=+e, m=\) \(2.0 \mathrm{u})\), and an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) with the same kinetic energies enter a region of uniform magnetic field \(\vec{B}\), moving perpendicular to \(\vec{B}\). Compare the radii of their circular paths.
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