Problem 22 - College 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) The speed is approximately \(3.38 \times 10^{6} \, \mathrm{m/s}\). (b) The vectors form a right-handed coordinate system. (c) The radius is \(4.18 \times 10^{-2} \, \mathrm{m}\), and the period is \(7.76 \times 10^{-9} \, \mathrm{s}\).

Step 1: Understanding the Condition for No Deflection

For the electron beam to experience no deflection, the electric force must equal the magnetic force on the electrons. The electric force is given by \( F_E = eE \) and the magnetic force is \( F_B = evB \), where \( e \) is the charge of an electron, \( v \) is the speed, \( E \) is the electric field, and \( B \) is the magnetic field. Equate \( eE = evB \) to find that the velocity \( v = \frac{E}{B} \).

Step 2: Calculate the Speed of Electrons

Plug the known values into the formula \( v = \frac{E}{B} \) to find the speed. Use \( E = 1.56 \times 10^{4} \, \mathrm{V/m} \) and \( B = 4.62 \times 10^{-3} \, \mathrm{T} \). Calculate \( v = \frac{1.56 \times 10^{4}}{4.62 \times 10^{-3}} \approx 3.38 \times 10^{6} \, \mathrm{m/s} \).

Step 3: Sketch the Orientation of Vectors

In a right-handed coordinate system, draw the velocity vector \( \vec{v} \) along the positive \( x \)-axis. Set the electric field vector \( \vec{E} \) along the positive \( y \)-axis, and the magnetic field vector \( \vec{B} \) along the positive \( z \)-axis. This configuration ensures that the forces are perpendicular to each other, causing no deflection when they are balanced.

Step 4: Determine the Radius of Electron Orbit

With the electric field removed, the electrons move in a circular path due to the magnetic force. Use the formula \( r = \frac{mv}{eB} \) for the radius, where \( m = 9.11 \times 10^{-31} \, \mathrm{kg} \) is the electron mass, \( v = 3.38 \times 10^{6} \, \mathrm{m/s} \), and \( B = 4.62 \times 10^{-3} \, \mathrm{T} \). Calculate \( r \approx \frac{(9.11 \times 10^{-31})(3.38 \times 10^{6})}{(1.6 \times 10^{-19})(4.62 \times 10^{-3})} \approx 4.18 \times 10^{-2} \mathrm{m} \).

Step 5: Calculate the Period of the Orbit

The period \( T \) is the time it takes for the electron to complete one full circle. This is given by \( T = \frac{2\pi r}{v} \). Use the radius \( r \approx 4.18 \times 10^{-2} \mathrm{m} \) and \( v = 3.38 \times 10^{6} \, \mathrm{m/s} \). Calculate \( T \approx \frac{2\pi (4.18 \times 10^{-2})}{3.38 \times 10^{6}} \approx 7.76 \times 10^{-9} \, \mathrm{s} \).

Key Concepts

Electric Field

An electric field is a region around a charged particle where forces are exerted on other charged particles. Imagine it as the "space of influence" created by the charge.

• The electric field ( \(E\)) is defined as a force per unit charge, represented mathematically as \(E = \frac{F}{q}\). Here, \(F\) is the force experienced by a small positive charge \(q\).
• In the provided exercise, the electric field has a strength of \(1.56 \times 10^{4} \, \mathrm{V/m}\).

This field applies a force on the electrons in the beam, pushing them in a direction perpendicular to the field lines. When combined with magnetic fields, it creates interesting phenomena, such as no deflection observed in the exercise provided if balanced correctly.

Magnetic Field

A magnetic field is a vector field surrounding a magnetic material or a moving electric charge, within which the force of magnetism acts. It's often represented by the symbol \(B\), and in the studied exercise, it has a magnitude of \(4.62 \times 10^{-3} \, \mathrm{T}\) (Tesla).

• Magnetic fields exert a force on moving charges, influencing their trajectory. This force is perpendicular to both the velocity of the charge and the field itself, known as the Lorentz Force law.
• The magnetic force experienced by an electron moving through the magnetic field in this scenario is given by \(F_B = evB\), where \(e\) is the charge of an electron and \(v\) its velocity.

In the exercise, the balance between this magnetic force and the electric force ensures no deflection of the electrons, as they move perpendicularly to both fields.

Circular Motion

When electrons move under the influence of a magnetic field without any electric field, they tend to travel in circular paths. This is due to the magnetic force acting as a centripetal force, always pointing towards the center of the circular path.

• The circular motion occurs because the magnetic force is perpendicular to the velocity of the electrons, constantly changing the direction of motion while maintaining force perpendicular to velocity.
• The radius \(r\) of this circular orbit can be calculated using the equation \(r = \frac{mv}{eB}\), where \(m\) is the mass of the electron, \(v\) is the velocity, \(e\) is the charge of the electron, and \(B\) is the magnetic field strength.

In the solution, the radius determined for these conditions is approximately \(4.18 \times 10^{-2} \, \mathrm{m}\). This calculation helps in comprehending how electrons navigate through magnetic fields.

Electron Velocity

The velocity of an electron beam is a critical aspect in determining how these particles interact with electric and magnetic fields.

• Given no deflection condition, the velocity \(v\) can be obtained from the formula \(v = \frac{E}{B}\). This relationship equates the electric force to the magnetic force, as seen in balanced conditions.
• For the exercise provided, with an electric field of \(1.56 \times 10^{4} \, \mathrm{V/m}\) and a magnetic field of \(4.62 \times 10^{-3} \, \mathrm{T}\), the calculated velocity \(v\) is approximately \(3.38 \times 10^{6} \, \mathrm{m/s}\).

Understanding electron velocity in this context allows for predictions about particle paths, necessary for applications in fields like electron microscopy or particle accelerators.

Electron Orbit Period

The orbit period of an electron moving in a magnetic field is the time taken to complete one full circle.

• This concept helps in understanding oscillatory motion in magnetic fields. It's calculated with the formula \(T = \frac{2\pi r}{v}\), where \(r\) is the radius of the electron's orbit and \(v\) is its velocity.
• In the specific problem analyzed, using the previously determined radius \(r \approx 4.18 \times 10^{-2} \, \mathrm{m}\) and velocity \(v \approx 3.38 \times 10^{6} \, \mathrm{m/s}\), the calculated period \(T\) is approximately \(7.76 \times 10^{-9} \, \mathrm{s}\).

Understanding this period helps in applications involving cyclic motion of electrons, such as in cyclotrons or in defining semiconductor behaviors.