Problem 86 - Fundamentals of Physics

Question

An electron has velocity \(\vec{v}=(32 \hat{\mathrm{i}}+40 \mathrm{j}) \mathrm{km} / \mathrm{s}\) as it enters a uniform magnetic field \(\vec{B}=60 \hat{\mathrm{i}} \mu \mathrm{T}\). What are (a) the radius of the helical path taken by the electron and (b) the pitch of that path? (c) To an observer looking into the magnetic field region from the entrance point of the electron, does the electron spiral clockwise or counterclockwise as it moves?

Step-by-Step Solution

Answer

(a) Radius is calculated from Step 4; (b) Pitch from Step 5; (c) Spirals clockwise.

Step 1: Identify Components of Velocity

The given velocity vector \(\vec{v} = (32\hat{\mathrm{i}} + 40\hat{\mathrm{j}})\, \text{km/s}\) has components \(v_x = 32\, \text{km/s}\) and \(v_y = 40\, \text{km/s}\). The magnetic field is \(\vec{B} = 60\hat{\mathrm{i}}\, \mu \text{T} = 60 \times 10^{-6} \, \text{T}\).

Step 2: Calculate the Perpendicular Velocity

The velocity component perpendicular to the magnetic field is \(v_y = 40\, \text{km/s}\) (since \(\vec{B}\) is only in the \(\hat{\mathrm{i}}\) direction).

Step 3: Calculate Cyclotron Frequency

The cyclotron frequency is given by \(\omega_c = \frac{qB}{m}\), where \(q = -1.6 \times 10^{-19} \text{C}\) is the electron charge and \(m = 9.11 \times 10^{-31} \text{kg}\) is the electron mass. Substituting the values, we get \(\omega_c = \frac{1.6 \times 10^{-19} \times 60 \times 10^{-6}}{9.11 \times 10^{-31}}\, \text{rad/s}\).

Step 4: Calculate the Radius of Helical Path

Using \(r = \frac{m v_y}{|q|B}\), substitute values to find \(r\). This calculation yields \(r = \frac{(9.11 \times 10^{-31}) \times (40 \times 10^3)}{1.6 \times 10^{-19} \times 60 \times 10^{-6}}\, \text{m}\).

Step 5: Calculate the Pitch of the Helix

The pitch \(p\) is given by \(p = v_x \times \text{period}\). The period \(T\) of revolution is \(\frac{2\pi}{\omega_c}\). Substitute to find the pitch \(p = 32 \times 10^3 \times \frac{2\pi \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 60 \times 10^{-6}}\).

Step 6: Determine Direction of Spiral

Use the right-hand rule. Since \(\vec{v}\) and \(\vec{B}\) span a plane in which \(\vec{F}_m = -q(\vec{v} \times \vec{B})\), the force direction gives a clockwise spiral if viewed from the electron's entrance.

Key Concepts

Velocity Components

When an electron enters a magnetic field with a specific velocity, we can break down that velocity into components to better understand its motion. The velocity vector \[\vec{v} = (32 \hat{\mathrm{i}} + 40 \hat{\mathrm{j}}) \text{km/s}\]represents the electron's speed and direction as it moves through space.

- **Component in x-direction (\(v_x\))**: This is the part of the electron's velocity parallel to the magnetic field direction (in this case, along the \(\hat{\mathrm{i}}\) axis), which is 32 km/s.- **Component in y-direction (\(v_y\))**: This portion is perpendicular to the magnetic field and has a value of 40 km/s.

Understanding these components is crucial because they determine different aspects of the electron's path. The parallel component (\(v_x\)) influences how far the electron travels along the magnetic field, while the perpendicular component (\(v_y\)) affects the circular motion plane, leading to a helical trajectory.

Cyclotron Frequency

The cyclotron frequency describes how quickly a charged particle like an electron orbits in a magnetic field. It depends on the charge, mass of the particle, and the magnetic field's strength.
The formula for calculating cyclotron frequency (\(\omega_c\)) is given by:\[\omega_c = \frac{qB}{m}\]Where:- \(q\) is the electron's charge, \(-1.6 \times 10^{-19} \text{C}\),- \(B\) is the magnetic field strength, \(60 \times 10^{-6} \text{T}\),- \(m\) is the electron's mass, \(9.11 \times 10^{-31} \text{kg}\).

This frequency tells us how fast the electron completes one full circle in its orbital path within the magnetic field. The higher the cyclotron frequency, the more times per second the electron will orbit.

The cyclotron frequency is critical because it's involved in calculations that determine other properties of the motion, such as the radius of the helical path and the pitch of the electron's spiral.

Magnetic Field

In this scenario, the magnetic field (\(\vec{B}\)) is a uniform vector pointing along the \(\hat{\mathrm{i}}\) direction with a magnitude of \(60 \mu \text{T}\). The uniformity implies that the strength and direction of the field are constant throughout the region the electron moves through.

A magnetic field exerts a force on moving charged particles like electrons, influencing their motion significantly. This force is perpendicular to both the magnetic field direction and the particle's velocity. This is governed by the right-hand rule and the Lorentz force equation:\[\vec{F}_m = q(\vec{v} \times \vec{B})\]Where:- \(\vec{F}_m\) is the magnetic force,- \(\vec{v}\) is the velocity of the particle,- \(\vec{B}\) is the magnetic field.

Because of these perpendicular forces, the electron's trajectory becomes helical or spiral. The force constantly changes the electron direction, creating circular motion in a plane perpendicular to the magnetic field, combined with constant linear motion parallel to the magnetic field.