Problem 28 - University Physics with Modern Physics

Question

(a) What is the speed of a beam of electrons when the simultaneous influence of an electric field of \(1.56 \times 10^{4} \mathrm{V} / \mathrm{m}\) and a magnetic field of \(4.62 \times 10^{-3} \mathrm{T},\) with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors \(\vec{\boldsymbol{v}}, \vec{\boldsymbol{E}},\) and \(\vec{\boldsymbol{B}}\) . (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

Step-by-Step Solution

Answer

(a) Speed: \(3.38 \times 10^{6} \ \text{m/s}\). (b) Orientation: \(\vec{v}\) right, \(\vec{E}\) up, \(\vec{B}\) into page. (c) Radius: \(4.18 \times 10^{-2} \ \text{m}\), Period: \(7.76 \times 10^{-11} \ \text{s}\).

Step 1: Understand the problem

The problem describes an electron beam subjected to perpendicular electric and magnetic fields, where no deflection occurs. This condition suggests that the electric and magnetic forces balance each other out, allowing us to calculate the speed of the electrons. We will also determine the orientation of these fields and find the characteristics of the electron's motion when the electric field is removed.

Step 2: Equate Electric and Magnetic Forces

When no deflection occurs, the electric force is balanced by the magnetic force. The electric force is given by \(F_E = eE\) and the magnetic force by \(F_B = evB\), where \(e\) is the charge of the electron, \(E\) is the electric field strength, \(v\) is the speed of the electron, and \(B\) is the magnetic field strength. Set \(eE = evB\) and solve for \(v\):\[ v = \frac{E}{B} \]

Step 3: Calculate Speed of Electrons

Substitute the given values into the formula: \[ v = \frac{1.56 \times 10^{4} \ \text{V/m}}{4.62 \times 10^{-3} \ \text{T}} \approx 3.38 \times 10^{6} \ \text{m/s} \] This is the speed of the electrons under the given fields.

Step 4: Draw the Vector Diagram

In the diagram, the velocity vector \(\vec{v}\) should be perpendicular to both the electric field vector \(\vec{E}\) and the magnetic field vector \(\vec{B}\). If \(\vec{E}\) is in the upward direction, \(\vec{B}\) could be directed into the page, using the right-hand rule, \(\vec{v}\) would then be directed to the right.

Step 5: Analyze Motion Without Electric Field

Once the electric field is removed, electrons move in a circular path due to the magnetic field. The centripetal force is provided by the magnetic force: \(F_B = \frac{mv^2}{r}\). Equating the magnetic force \(evB\) to the centripetal force gives \(r = \frac{mv}{eB}\).

Step 6: Calculate Radius of Orbit

Assume the electron has mass \(m = 9.11 \times 10^{-31} \ \text{kg}\) and charge \(e = 1.6 \times 10^{-19} \ \text{C}\). Substitute these values into the formula for radius:\[ r = \frac{(9.11 \times 10^{-31} \ \text{kg})(3.38 \times 10^{6} \ \text{m/s})}{(1.6 \times 10^{-19} \ \text{C})(4.62 \times 10^{-3} \ \text{T})} \approx 4.18 \times 10^{-2} \ \text{m} \]

Step 7: Calculate Period of Orbit

The period \(T\) is the time for one complete orbit. Given \(T = \frac{2\pi r}{v}\), substitute \(r\) and \(v\):\[ T = \frac{2\pi (4.18 \times 10^{-2} \ \text{m})}{3.38 \times 10^{6} \ \text{m/s}} \approx 7.76 \times 10^{-11} \ \text{s} \]

Key Concepts

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 the context of the Lorentz force and electron motion, a magnetic field can exert a force on electrons moving through it.
This force is crucial for understanding electron behavior in various applications, such as particle accelerators and electronic devices. The magnetic field exerts a force according to the equation:

• Magnetic Force ( \( F_B \)) = \( qvB \sin\theta \)

where \( q \) is the charge of the electron, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the field and the velocity vector.
For uniform fields and charged particles, the magnetic force acts perpendicular to the motion, often resulting in circular or helical paths. For electrons, because they have a negative charge, the direction of this force is determined using the right-hand rule. Moving electrons in a magnetic field will cause them to experience a perpendicular force that alters their straight path into a circular one.

Electric Field

An electric field is the region around charged particles or objects in which an electric charge experiences a force. It's measured in volts per meter (V/m) and is vital in understanding how charged particles move. An electric field \( \vec{E} \) acts on an electron to produce a force \( F_E \):

• Electric Force ( \( F_E \)) = \( eE \)

where \( e \) is the charge of an electron.
Electric fields influence particle motion significantly. When balanced with a magnetic field, they can keep a particle moving in a straight line without deflection. In scenarios where both electric and magnetic fields act perpendicular to each other, as in the problem related to Lorentz Force, the condition \( eE = evB \) arises. Here, the fields balance each other, resulting in a net zero force, maintaining the particle's uniform motion.

Electron Motion

Electron motion in the presence of electric and magnetic fields reveals the fascinating interplay of these forces. When electrons encounter a perpendicular electric and magnetic field, like in our exercise, they experience forces from both fields.
The observed result of these combined effects, as demonstrated in the problem, is orbital motion upon removal of the electric field. In the absence of deflection conditions, speed \( v \) of electrons is determined by \( v = \frac{E}{B} \), indicating no net force exists.
Once the electric field is removed, only the magnetic field influences the electrons, causing them to follow a circular path. This path's radius \( r \) is determined by the formula:

• \( r = \frac{mv}{eB} \)

This change in motion depends significantly on the particle's speed, mass, and the magnetic field strength. As the particles loop in circles, the period \( T \), or time for a complete orbit, is given by:

• \( T = \frac{2\pi r}{v} \)

This intricate dance of electron motion showcases the Lorentz force application in manipulating particle trajectories using electromagnetic fields.