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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

A particle with charge \(2.15 \mu \mathrm{C}\) and mass \(3.20 \times 10^{-11} \mathrm{~kg}\) is initially traveling in the \(+y\) -direction with a speed \(v_{0}=1.45 \times 10^{5} \mathrm{~m} / \mathrm{s} .\) It then enters a region containing a uniform magnetic field that is directed into, and perpendicular to, the page in Fig. \(\mathrm{P} 27.81 .\) The magnitude of the field is \(0.420 \mathrm{~T}\). The region extends a distance of \(25.0 \mathrm{~cm}\) along the initial direction of travel; \(75.0 \mathrm{~cm}\) from the point of entry into the magnetic field region is a wall. The length of the field-free region is thus \(50.0 \mathrm{~cm} .\) When the charged particle enters the magnetic field, it follows a curved path whose radius of curvature is \(R .\) It then leaves the magnetic field after a time \(t_{1},\) having been deflected a distance \(\Delta x_{1} .\) The particle then travels in the field-free region and strikes the wall after undergoing a total deflection \(\Delta x\). (a) Determine the radius \(R\) of the curved part of the path. (b) Determine \(t_{1}\), the time the particle spends in the magnetic field. (c) Determine \(\Delta x_{1}\), the horizontal deflection at the point of exit from the field. (d) Determine \(\Delta x\), the total horizontal deflection.

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section 39.3 ), in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\). (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I ?\) (c) What is the magnetic moment of the atom due to the motion of the electron?

A plastic circular loop has radius \(R,\) and a positive charge \(q\) is distributed uniformly around the circumference of the loop. The loop is then rotated around its central axis, perpendicular to the plane of the loop, with angular speed \(\omega .\) If the loop is in a region where there is a uniform magnetic field \(\vec{B}\) directed parallel to the plane of the loop, calculate the magnitude of the magnetic torque on the loop.

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is \(155 \mathrm{~V} / \mathrm{m}\) and the magnetic field is \(0.0315 \mathrm{~T}\). The ions next enter a uniform magnetic field of magnitude \(0.0175 \mathrm{~T}\) that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is \(17.5 \mathrm{~cm},\) what is their mass?

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 \(t=0.0200 \mathrm{~s} ?\)
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