Problem 5 - University Physics with Modern Physics

Question

An electron experiences a magnetic force of magnitude \(4.60 \times 10^{-15} \mathrm{N}\) when moving at an angle of \(60.0^{\circ}\) with respect to a magnetic field of magnitude \(3.50 \times 10^{-3} \mathrm{T.}\) Find the speed of the electron.

Step-by-Step Solution

Answer

The speed of the electron is approximately \( 9.46 \times 10^{6} \, \mathrm{m/s} \).

Step 1: Understanding the Magnetic Force Equation

The magnetic force on a charged particle like an electron moving in a magnetic field can be calculated using the formula \( F = qvB\sin\theta \), where \( F \) is the magnetic force, \( q \) is the charge of the electron, \( v \) is the velocity of the electron, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity vector and the magnetic field vector.

Step 2: Identifying Known Values and Constants

In the exercise, you are given that the magnetic force \( F = 4.60 \times 10^{-15} \, \mathrm{N} \), the magnetic field \( B = 3.50 \times 10^{-3} \, \mathrm{T} \), and the angle \( \theta = 60.0^{\circ} \). The charge of an electron \( q \) is a known constant \( -1.6 \times 10^{-19} \, \mathrm{C} \).

Step 3: Rearranging the Equation to Solve for Velocity

Rearrange the magnetic force equation to solve for the velocity \( v \) of the electron: \[ v = \frac{F}{qB\sin\theta} \]. This manipulation allows you to isolate \( v \) on one side of the equation, making it easier to solve.

Step 4: Substituting Known Values into the Equation

Substitute the given values and constants into the rearranged formula: \[ v = \frac{4.60 \times 10^{-15} \mathrm{N}}{(1.6 \times 10^{-19} \mathrm{C}) (3.50 \times 10^{-3} \mathrm{T})\sin(60.0^{\circ})} \]. Make sure to convert the angle from degrees to radians or use direct sin values from trigonometric tables.

Step 5: Calculating the Velocity

First, determine the value of \( \sin(60.0^{\circ}) \), which is approximately \( 0.866 \). Substitute this into the equation and perform the calculations: \[ v = \frac{4.60 \times 10^{-15}}{1.6 \times 10^{-19} \times 3.50 \times 10^{-3} \times 0.866} \]. Simplifying this will give you the velocity of the electron.

Step 6: Final Answer

After performing the calculations, you will find that the velocity of the electron is approximately \( 9.46 \times 10^{6} \, \mathrm{m/s} \).

Key Concepts

electron velocity

Electron velocity is an essential concept in understanding how charged particles, like electrons, behave in the presence of magnetic fields. When an electron moves, it possesses kinetic energy, which makes it crucial to know its speed or velocity. Velocity is not just the speed of an electron but also describes its direction in space. This is important when considering magnetic forces since the force depends on both the speed and direction of the electron relative to the magnetic field. The magnetic force equation helps to derive velocity from the given magnetic force, charge, and magnetic field strength. Remember that the force equation is given by: \( F = qvB\sin\theta \). Here, \( v \), the electron's velocity, is calculated by rearranging the equation as \( v = \frac{F}{qB\sin\theta} \). By substituting known values, we calculated velocity as approximately \( 9.46 \times 10^{6} \, \mathrm{m/s} \), showcasing the electron's incredible speed.

magnetic field

The magnetic field is a force field created by moving electric charges and magnetic dipoles, and it exerts a force on particles that are moving relative to it, such as electrons. The strength of this field is measured in Teslas (\(\mathrm{T}\)). In our exercise, the magnetic field strength is given as \(3.50 \times 10^{-3} \mathrm{T}\).A magnetic field has both a magnitude and a direction. This vector nature means that the force experienced by a particle, like an electron, depends on its velocity direction relative to the magnetic field. This is why the force is calculated using \(\sin\theta\), where \(\theta\) is the angle between the electron’s velocity and the magnetic field direction. For this exercise, the angle is given as \(60.0^{\circ}\). The trigonometric function \(\sin(60.0^{\circ})\) impacts the magnitude of the force, showing the significance of both angular and field considerations.

charge of electron

The charge of an electron is a fundamental physical constant significant in electromagnetism. It is a negative elementary charge denoted as \( q = -1.6 \times 10^{-19} \, \mathrm{C} \). This uniform charge is one of the building blocks of atoms, determining how electrons interact with electric and magnetic fields.In a magnetic field, the sign and magnitude of an electron's charge are crucially involved in the calculation of the magnetic force using the formula \( F = qvB\sin\theta \). In computations involving magnetic fields, only the magnitude of the charge is typically used, since the direction of force (given by the right-hand rule) accounts for the charge being negative.The constancy of the electron's charge implies that no matter the field it’s in or the speed at which it travels, this value remains unchanged, simplifying many calculations in physics. Understanding the electron's charge helps glean insights into many behavioral properties and interactions within electromagnetic contexts.