Problem 29 - Fundamentals of Physics

Question

An electron follows a helical path in a uniform magnetic field of magnitude \(0.300 \mathrm{~T}\). The pitch of the path is \(6.00 \mu \mathrm{m},\) and the magnitude of the magnetic force on the electron is \(2.00 \times 10^{-15} \mathrm{~N}\). What is the electron's speed?

Step-by-Step Solution

Answer

The electron's speed is approximately \( 4.17 \times 10^7 \text{ m/s} \).

Step 1: Understand the problem components

We have an electron moving in a helical path within a magnetic field, which means the electron is influenced by the Lorentz force. We also know the magnetic field magnitude and the pitch of the path. The problem asks for the electron's speed.

Step 2: Identify the relation for magnetic force

The magnetic force acting on a charged particle moving in a magnetic field is given by the equation \( F = qvB \sin \theta \), where \( F \) is the force, \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field and \( \theta \) is the angle between \( v \) and \( B \). For a helical path, \( \sin \theta = 1 \). Thus, \( F = qvB \).

Step 3: Rearrange the formula to solve for velocity

Rearrange this equation in terms of the electron's velocity \( v \): \( v = \frac{F}{qB} \).

Step 4: Plug in known values

The charge of an electron \( q \) is approximately \( 1.60 \times 10^{-19} \text{ C} \), the given force \( F \) is \( 2.00 \times 10^{-15} \text{ N} \), and the magnetic field \( B \) is \( 0.300 \text{ T} \). Substitute these values into the equation: \( v = \frac{2.00 \times 10^{-15} \text{ N}}{1.60 \times 10^{-19} \text{ C} \times 0.300 \text{ T}} \).

Step 5: Calculate the velocity

Perform the calculation: \( v = \frac{2.00 \times 10^{-15}}{1.60 \times 10^{-19} \times 0.300} \approx 4.17 \times 10^7 \text{ m/s} \). The electron's speed is approximately \( 4.17 \times 10^7 \text{ m/s} \).

Key Concepts

Lorentz Force

The Lorentz force is an essential concept when discussing the motion of charged particles like electrons in electromagnetic fields. This force describes how charged particles move in response to electric and magnetic environments. It is a pivotal aspect of understanding electron dynamics in magnetic fields. The Lorentz force formula is pivotal for calculating the movement of electrons in a magnetic field:

• It is defined as \( F = q(E + v \times B) \), where:
• \( F \) is the force exerted on the particle,
• \( q \) is the charge of the particle,
• \( E \) is the electric field,
• \( v \) is the velocity of the particle,
• \( B \) is the magnetic field.

For our specific problem, the electric field is not influencing the electron, so the formula simplifies to \( F = qvB \sin \theta \), where \( \theta \) is the angle between the velocity and the magnetic field vectors. In a helical path, \( \theta = 90^\circ \) simplifies the equation since \( \sin 90^\circ = 1 \). So, it becomes \( F = qvB \). This highlights how the Lorentz force plays a role in determining the path of electrons when exposed to a magnetic field.

Helical Path

When an electron moves through a magnetic field, it can follow a helical path, which is a combination of circular and linear motion. This occurs because of the perpendicular (to the magnetic field) component of the velocity induces a circular motion, and the parallel component continues as linear motion through the field. The overall effect of these motions results in a spiral or corkscrew path. The helix can be characterized by:

• The radius of the circle (r), which is due to the perpendicular velocity component. It is derived from \( r = \frac{mv_{\perp}}{qB} \), where \( m \) is the mass and \( v_{\perp} \) is the velocity perpendicular to the magnetic field.
• The pitch of the helix, which is determined by the parallel movement to the magnetic field lines. It is calculated as \( \text{pitch} = v_{||}T \), with \( v_{||} \) as the parallel velocity and \( T \) being the period of the circular motion.

Understanding the helical path is crucial for accurately predicting the trajectory of charged particles in magnetic fields. This is especially useful in applications like particle accelerators and magnetic confinement in fusion reactors, where controlled motion is essential.

Magnetic Force Calculation

To calculate the velocity of a charged particle like an electron in a magnetic field, you need to understand the relationship between magnetic force and velocity. The magnetic force acts perpendicular to the velocity of the charge and the magnetic field. This force can be calculated using the formula:

• \( F = qvB \)

Where:

• \( F \) is the magnetic force, given in newtons (N),
• \( q \) is the charge, which for an electron is \( 1.60 \times 10^{-19} \text{ C} \),
• \( v \) is the velocity of the charge in meters per second (m/s),
• \( B \) is the magnetic field strength in teslas (T).

Rearranging the formula allows you to solve for velocity as follows: \( v = \frac{F}{qB} \). In the exercise, substituting known values for the force, charge, and magnetic field allows direct calculation of the speed of the electron, resulting in a velocity of approximately \( 4.17 \times 10^7 \text{ m/s} \). This approach helps in quantifying how fast an electron travels under its specific conditions and highlights the practical application of mathematical concepts in solving physics problems.