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Chapter 27: Problem 21

An electron in the beam of a cathode-ray tube is accelerated by a potential difference of \(2.00 \mathrm{kV}\). Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius \(0.180 \mathrm{~m} .\) What is the magnitude of the field?

Short Answer

Expert verified
The magnitude of the magnetic field is \(3.69 \times 10^{-2} T\).

Step by step solution

01

Convert the accelerating voltage to electron kinetic energy

First, let's calculate the kinetic energy of the electron which is given by its potential energy gained. An electron accelerated through a voltage \(V\) gains the energy expressed as: \(KE = qV\), where \(q\) is the charge of the electron equal to \(1.6 x 10^{-19} C\) and \(V = 2.00 kV = 2.00 x 10^3 V\). So, \(KE = 1.6 x 10^{-19} C \times 2.00 x 10^3 V = 3.2 x 10^{-16} J\).
02

Find the electron's speed by relating KE to \(0.5mv^2\)

The kinetic energy of the electron is also expressed by the formula: \(KE = 0.5mv^2\). Equate this to the energy we found in step 1, we get: \(3.2 x 10^{-16} J = 0.5 \times 9.11 x 10^{-31} kg \times v^2\). Solving for \(v\), we obtain: \(v = \sqrt{(3.2 x 10^{-16} / 0.5) / 9.11 x 10^{-31}} = 2.64 x 10^7 m/s\).
03

Equate magnetic force to centripetal force to find the magnetic field

Since circular motion is maintained by the magnetic force, this force must equal the centripetal force: \(qvB = mv^2/r\). We want to isolate \(B\), the magnetic field, so we find \(B = (mv^2/r)/qv = ((9.11 x 10^{-31} kg) \times (2.64 x 10^7 m/s)^2 / (0.180 m)) / (1.6 x 10^{-19} C \times 2.64 x 10^7 m/s ) = 3.69 x 10^-2 T\).
04

Provide the answer in standard SI units.

The magnitude of the magnetic field is expressed in Tesla (T). Therefore, the final answer is: \(B = 3.69 \times 10^{-2} T\).

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Key Concepts

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

Kinetic Energy of Electrons
When we talk about the kinetic energy of electrons in the context of a cathode-ray tube, we are essentially discussing how fast these particles are moving after being accelerated by a potential difference. An electron, being a negatively charged particle, gains kinetic energy when it's accelerated through an electric field.

The potential difference in a cathode-ray tube acts like a 'push' for the electrons, accelerating them to high speeds. This potential difference (voltage) is directly related to the amount of kinetic energy an electron will gain. Mathematically, this can be represented as KE = qV, where q is the charge of the electron (-1.6 x 10-19 Coulombs) and V is the potential difference.

The kinetic energy imparted to the electron due to this acceleration is crucial because it determines how the electron will behave when it encounters other forces, such as a magnetic field. In simple terms: the faster the electron is moving, the more kinetic energy it has. This kinetic energy can then be converted into other forms of motion, like circular paths, which leads us to our next concept.
Magnetic Force on Charged Particles
Imagine a charged particle moving into the invisible force field created by a magnet. This is somewhat how electrons in a cathode-ray tube experience magnetic force. Charged particles like electrons, when moving through a magnetic field, experience a force that acts perpendicular to the direction of their motion and the magnetic field. This force is known as the magnetic Lorenz force.

The magnitude of this force can be expressed by the equation F = qvB sin(胃), where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and 胃 is the angle between the velocity vector and the magnetic field. In the case of the cathode-ray tube, where the magnetic field is perpendicular to the motion of electrons, 胃 is 90 degrees, and the sine function equals 1, simplifying our equation to F = qvB.

This concept of magnetic force is crucial when it comes to understanding how to control the path of the electron within the tube. Adjusting the magnetic field allows us to change the trajectory of electrons, which is fundamentally how old television screens and oscilloscopes worked.
Circular Motion in a Magnetic Field
When an electron moves through a magnetic field, it can undergo circular motion if the force exerted by the field is always perpendicular to the velocity of the electron. This ties back to our previous concept: the magnetic force acting as a centripetal force, which is always directed towards the center of the electron's circular path.

The magnetic force necessitates that the electron follows a curved path rather than a straight line, much like a car turning in a circular track due to the frictional force.

In the context of the cathode-ray tube exercise, we equate the magnetic force (qvB) to the centripetal force required for circular motion (mv2/r). As the forces are equal in magnitude but opposite in direction, we derive B = mv2/(rqv). This formula allows us to calculate the precise magnetic field needed to maintain an electron's circular path with a given radius. If the charged particle weren鈥檛 under the influence of a magnetic field, it would continue to move straight, as per Newton's first law of motion. Thus, the magnetic field plays a pivotal role in shaping the path of electrons within a cathode-ray tube.

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Most popular questions from this chapter

One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed \(8.50 \mathrm{~km} / \mathrm{s}\), and you want to bend them within a uniform magnetic field in a semicircle of diameter \(25.0 \mathrm{~cm}\) for the \({ }^{12} \mathrm{C}\). The measured masses of these isotopes are \(1.99 \times 10^{-26} \mathrm{~kg}\left({ }^{12} \mathrm{C}\right)\) and \(2.16 \times 10^{-26} \mathrm{~kg}\left({ }^{13} \mathrm{C}\right) .\) (a) What strength of magnetic field is required? (b) What is the diameter of the \({ }^{13} \mathrm{C}\) semicircle? (c) What is the separation of the \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\vec{B}\) fills the region between the rails (Fig. \(\mathbf{P 2 7 . 5 9}\) ). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m,\) find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{~km} / \mathrm{s}) .\) Let \(B=0.80 \mathrm{~T}, I=2.0 \times 10^{3} \mathrm{~A}, m=25 \mathrm{~kg}\) and \(L=50 \mathrm{~cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

A small particle with positive charge \(q=+3.75 \times 10^{-4} \mathrm{C}\) and mass \(m=5.00 \times 10^{-5} \mathrm{~kg}\) is moving in a region of uniform electric and magnetic fields. The magnetic field is \(B=4.00 \mathrm{~T}\) in the \(+z\) -direction. The electric field is also in the \(+z\) -direction and has magnitude \(E=60.0 \mathrm{~N} / \mathrm{C}\). At time \(t=0\) the particle is on the \(y\) -axis at \(y=+1.00 \mathrm{~m}\) and has velocity \(v=30.0 \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction. Neglect gravity. (a) What are the \(x\) -, \(y\) and \(z\) -coordinates of the particle at \(t=0.0200 \mathrm{~s} ?\) (b) What is the speed of the particle at \(t=0.0200 \mathrm{~s} ?\)

A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses (see Section 27.5 ). In one design for such an instrument, ions with mass \(m\) and charge \(q\) are accelerated through a potential difference \(V\). They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicircular path of radius \(R .\) A detector measures where the ions complete the semicircle and from this it is easy to calculate \(R\). (a) Derive the equation for calculating the mass of the ion from measurements of \(B, V, R,\) and \(q\). (b) What potential difference \(V\) is needed so that singly ionized \({ }^{12} \mathrm{C}\) atoms will have \(R=50.0 \mathrm{~cm}\) in a 0.150 T magnetic field? (c) Suppose the beam consists of a mixture of \({ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C}\) ions. If \(v\) and \(B\) have the same values as in part \((\mathrm{b}),\) calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished? (Make the assumption described in Problem 27.53 for the masses of the ions.)

A straight, \(2.5 \mathrm{~m}\) wire carries a typical household current of \(1.5 \mathrm{~A}\) (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?
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