Problem 28 - University Physics with Modern Physics

Question

(a) What is the speed of a beam of electrons when the simultancous 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 \(\overrightarrow{\boldsymbol{v}}, \overrightarrow{\boldsymbol{E}},\) and \(\overrightarrow{\boldsymbol{B}} .\) (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 of electrons is \(3.38 \times 10^6 \, \text{m/s}\). (b) Use right-hand rule to orient vectors perpendicular. (c) Orbit radius is approximately 0.041 m, and the period is approximately \(7.6 \times 10^{-8} \, \text{s}\).

Step 1: Understand the Concept

An electron beam experiences no deflection when the electric and magnetic forces cancel each other out. This happens when \[ \text{Force due to electric field (} F_E \text{)} = - \text{Force due to magnetic field (} F_B \text{)} \]. To calculate velocity, the equations \( F_E = eE \) and \( F_B = evB \) are used, where \( e \) is the electron charge, \( v \) is the velocity, \( E \) is the electric field, and \( B \) is the magnetic field.

Step 2: Calculate Speed with no Deflection

Since forces cancel out, \( F_E = F_B \). Substitute the expressions for these forces: \[ eE = evB \].Solve for \( v \), the speed:\[ v = \frac{E}{B} \].Plug in the known values,\[ v = \frac{1.56 \times 10^{4} \, \text{V/m}}{4.62 \times 10^{-3} \, \text{T}} \approx 3.38 \times 10^{6} \, \text{m/s} \].

Step 3: Diagram - Vector Orientation

In the configuration where there's no deflection, the vectors are oriented such that:- \( \overrightarrow{v} \) is the velocity vector of the electrons,- \( \overrightarrow{E} \) is the electric field vector,- \( \overrightarrow{B} \) is the magnetic field vector.If the electrons move along the x-axis, \( \overrightarrow{E} \) would point in the y-axis and \( \overrightarrow{B} \) into the z-axis or vice versa, keeping them perpendicular to both \( \overrightarrow{v} \) and each other.

Step 4: Calculate Radius of Electron Orbit

When 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. Thus,\[ evB = \frac{mv^2}{r} \], where \( m \) is mass of an electron and \( r \) is the radius of the orbit.Solve for \( r \):\[ r = \frac{mv}{eB} \].Substituting \( v = 3.38 \times 10^{6} \, \text{m/s} \), \( m = 9.11 \times 10^{-31} \, \text{kg} \), \( e = 1.6 \times 10^{-19} \, \text{C} \), \( B = 4.62 \times 10^{-3} \, \text{T} \),\[ r = \frac{9.11 \times 10^{-31} \times 3.38 \times 10^{6}}{1.6 \times 10^{-19} \times 4.62 \times 10^{-3}} \approx 0.041 \text{ m} \].

Step 5: Calculate Period of Electron Orbit

The period \( T \) of the orbit is the time for one complete revolution: \[ T = \frac{2\pi r}{v} \].Using \( r = 0.041 \, \text{m} \) and \( v = 3.38 \times 10^6 \, \text{m/s} \), we find:\[ T = \frac{2\pi \times 0.041}{3.38 \times 10^{6}} \approx 7.6 \times 10^{-8} \, \text{s} \].

Key Concepts

Velocity of Electrons

The velocity of electrons in electric and magnetic fields is vital to understanding the behavior of charged particles under simultaneous field influences. When an electron moves through an electric field (\( E \)) and a magnetic field (\( B \)), it experiences forces due to each field. The crucial condition where there's no deflection occurs when these forces exactly cancel each other out. This is expressed mathematically where the force from the electric field, \( F_E = eE \), equals the force from the magnetic field, \( F_B = evB \), noting \( e \) is the electron charge and \( v \) is the electron velocity. In the case without deflection, the relationship simplifies to \[ v = \frac{E}{B} \]. Plugging in the given values: \( E = 1.56 \times 10^{4} \, \text{V/m} \) and \( B = 4.62 \times 10^{-3} \, \text{T} \), the velocity of electrons in this specific setup is found to be approximately \( 3.38 \times 10^{6} \, \text{m/s} \). Key points include:

• Electron velocity is determined by balancing electric and magnetic field forces.
• Speed remains constant as path deflection is neutralized.
• Velocity calculation becomes a simple operation involving division of field magnitudes.

Magnetic Force

The magnetic force on a moving electron in a magnetic field acts perpendicularly to both the velocity of the electron and the magnetic field lines. This is a fundamental principle known as the Lorentz force law. The force can be expressed as: \[ F_B = evB \sin\theta \], where \( \theta \) is the angle between the velocity vector \( \overrightarrow{v} \) and the magnetic field vector \( \overrightarrow{B} \). In situations where the electron's path is perpendicular to the magnetic field, \( \theta = 90^\circ \) and \( \sin\theta = 1 \), simplifying our formula.When no deflection is observed, this magnetic force is perfectly balanced by the electric force. If the electric field is removed but the magnetic field is maintained, the electrons instead follow a circular path due to the magnetic force. The expression for the radius \( r \) of this path is derived from balancing centripetal force with the magnetic force:\[ evB = \frac{mv^2}{r} \]which rearranges to:\[ r = \frac{mv}{eB} \]. This shows how:

• Magnetic forces redirect electron motion under field influence, causing circular or spiral paths.
• Force magnitude can determine path curvature or deviation.

Electric Field

Electric fields are regions around a charged particle where forces are exerted on other charges. In the context of electron motion, the electric field exerts a force that can alter the trajectory of the electron. The force from an electric field is given by: \[ F_E = eE \], where \( e \) is the magnitude of the electron charge and \( E \) is the electric field strength.Understanding the interaction between an electric field and electrons provides insights into how they can be manipulated or controlled. For instance, in devices where precise control over electron paths is required, such as cathode ray tubes or electron microscopes, the electric field plays a pivotal role.Key insights regarding electric fields include:

• They apply a constant force on charges, influencing acceleration and direction.
• Magnitude and direction of electric fields determine how and where electrons move within a field.