Problem 9 - University Physics with Modern Physics

Question

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 \(\mathrm{km} / \mathrm{s}\) im the \(+x\) -direction experiences a force of \(2.25 \times 10^{-16} \mathrm{N}\) in the \(+y\) -direction, and an electron moving at 4.75 \(\mathrm{km} / \mathrm{s}\) in the \(-\mathrm{z}\) -direction experiences a force of \(8.50 \times 10^{-16} \mathrm{N}\) . (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\) -direction at 3.2 \(\mathrm{km} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

The magnetic field is 0.09375 T in the +z direction; force on the electron is 4.8 x 10^-16 N in the -x direction.

Step 1: Identify the Relevant Equation

To solve the problem, we use the formula for the magnetic force on a charged particle: \[ F = q(v \times B) \]where \( F \) is the force, \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field.

Step 2: Calculate the Magnetic Field Using the Proton

For the proton:- Charge of a proton \( q_p = 1.6 \times 10^{-19} \mathrm{C} \).- Velocity \( v_p = 1.50 \times 10^3 \mathrm{m/s} \) in the \(+x\) direction.- Force \( F_p = 2.25 \times 10^{-16} \mathrm{N} \) in the \(+y\) direction.Using the right-hand rule, the magnetic field \( B \) is directed along the \(+z\) axis.Thus, \[ F_p = q_p v_p B \]\[ B = \frac{F_p}{q_p v_p} = \frac{2.25 \times 10^{-16}}{1.6 \times 10^{-19} \times 1.5 \times 10^3} = 9.375 \times 10^{-2} \mathrm{T} \]

Step 3: Verify Using the Electron's Motion

For the electron:- Charge of an electron \( q_e = -1.6 \times 10^{-19} \mathrm{C} \).- Velocity \( v_e = 4.75 \times 10^3 \mathrm{m/s} \) in the \(-z\) direction.- Force \( F_e = 8.50 \times 10^{-16} \mathrm{N} \).Using the force direction (same direction implies opposite charge) and right-hand rule, the calculated field should hold true:\[ F_e = |q_e| v_e B \]Replacing the values, \( B \) should agree with the previously calculated value.

Step 4: Calculate Force on Electron in -y Direction

Now that we have \( B = 9.375 \times 10^{-2} \mathrm{T} \), let's calculate the force on an electron moving in the \(-y\) direction:- Velocity \( v'_e = 3.2 \times 10^3 \mathrm{m/s} \) in the \(-y\) direction.The magnetic force \( F'_e = q_e v'_e B \) will act in the \(-x\) direction by the right-hand rule.\[ F'_e = 1.6 \times 10^{-19} \times 3.2 \times 10^3 \times 9.375 \times 10^{-2} \]\[ F'_e = 4.8 \times 10^{-16} \mathrm{N} \]

Step 5: Summarize the Results

- The magnitude of the magnetic field is \( 9.375 \times 10^{-2} \mathrm{T} \) and is directed in the \(+z\)-direction.- The force on a second electron moving at 3.2 \( \mathrm{km/s} \) in the \(-y\)-direction is \( 4.8 \times 10^{-16} \mathrm{N} \) in the \(-x\)-direction.

Key Concepts

Magnetic Field

A magnetic field is an invisible but very real force field around magnetic materials or moving electric charges. It exerts forces on other nearby moving charges and magnetic dipoles. You can visualize it like the wind: invisible, but you can feel its effects.
The direction and magnitude of a magnetic field are important characteristics. They determine how strong the force will be on charged particles moving within the field. For example, in the given exercise, the magnetic field causes a proton and an electron to experience forces. It's important to understand how these forces manifest, which is largely governed by the direction and magnitude of the magnetic field.
The magnetic field (denoted by the symbol \( B \)) is measured in teslas (T) and has both direction and magnitude, making it a vector quantity.

Right-Hand Rule

The right-hand rule is a simple way to find the direction of the magnetic force acting on a moving charged particle.
Using the right-hand rule, you point your fingers in the direction of the particle's velocity, align your palm such that it faces towards the magnetic field direction, and your thumb then points in the direction of the acting force. For a positive charge like a proton, this rule applies directly.
For a negative charge, like an electron, the force direction is exactly opposite to the one predicted by the right-hand rule for positive charges. In the provided exercise, this rule helps us determine the direction of force on the proton in the \(+y\) direction and the magnetic field in the \(+z\) direction.

Charged Particles

Charged particles, like electrons and protons, interact with magnetic fields in interesting ways. Depending on the charge and direction of motion, they will experience different magnitudes and directions of force.
Protons have a positive charge, while electrons are negatively charged. The charge of a proton and electron is exactly equal in magnitude but opposite in sign, approximately \(1.6 \times 10^{-19} \mathrm{C}\). This equal charge magnitude causes them to experience similar but opposite forces when moving in the same magnetic field.
Within a magnetic field, the force on a charged particle can be calculated using the formula \( F = q(v \times B) \), where \( F \) is the force experienced, \( q \) is the charge, \( v \) is the velocity vector, and \( B \) is the magnetic field vector.

Vector Calculations

Vector calculations are essential for solving problems involving magnetic forces, as both the velocity of the particle and the magnetic field are vector quantities.
Vectors have both magnitude and direction, and calculating the net result often involves vector cross products. The cross product in magnetic force calculations is used to determine the direction of the resulting force. The resultant of the vector cross product of the velocity \( v \) and the magnetic field \( B \) vector is also a vector force \( F \).
The formula \( F = q(v \times B) \) signifies that the force is perpendicular to both the velocity and the magnetic field. This concept is crucial when dealing with forces on charged particles, as demonstrated in the solution to the problem. Understanding how to compute vector cross products is key to predicting the correct direction and magnitude of these forces.