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 \(\mathrm{km} / \mathrm{s}\) in 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 \(-z\) -direction experiences a force of \(8.50 \times\) \(10^{-16} \mathrm{N}\) in the \(+y\) -direction. (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 3.20 \(\mathrm{km} / \mathrm{s}\) ?

Step-by-Step Solution

Answer

Magnetic field is 0.9375 T in the +z-direction. Magnetic force on the electron is 4.80 x 10^{-16} N in the -x-direction.

Step 1: Understanding the Lorentz Force

The magnetic force experienced by a charged particle moving in a magnetic field is given by the Lorentz force equation: \( F = q(v \times B) \), where \( q \) is the charge of the particle, \( v \) is the velocity vector, and \( B \) is the magnetic field vector.

Step 2: Force on the Proton

A proton experiences a force of \( 2.25 \times 10^{-16} \, \text{N} \) in the \( +y \)-direction when moving at \( 1.50 \, \text{km/s} \) in the \( +x \)-direction. The charge of a proton is \( q_p = 1.6 \times 10^{-19} \, \text{C} \). Using the Lorentz force formula, the equation becomes \( 2.25 \times 10^{-16} = 1.6 \times 10^{-19} \cdot 1500 \cdot B \).

Step 3: Solving for Magnetic Field from Proton's Data

We solve for \( B \): \[ B = \frac{2.25 \times 10^{-16}}{1.6 \times 10^{-19} \times 1500} \approx 0.9375 \, \text{T} \].

Step 4: Component Direction Confirmation using Electron

An electron moving at \(4.75 \, \text{km/s}\) in the \(-z\)-direction experiences a force \(8.50 \times 10^{-16} \, \text{N}\) in the \(+y\)-direction. The charge of an electron is \( q_e = -1.6 \times 10^{-19} \, \text{C} \). Using \( F = q(v \times B) \), and considering the negative charge: \[8.50 \times 10^{-16} = (-1.6 \times 10^{-19})(-4750)B \]. Solving confirms \( B \) and dictates the z-component of \(v \times B\) is \(+y\), confirming \( B = 0.9375 \, \text{T} \, (+z)\).

Step 5: Solve for Force on Electron in -y Direction

To find the force on an electron moving in the \(-y\)-direction, use \( F = q(v \times B) \), \( v = 3200 \times 10^3 \, \text{m/s} \), and \( q = -1.6 \times 10^{-19} \). The resultant force direction will be in the \(-x\)-direction. Magnitude is \[ F = |(-1.6 \times 10^{-19})(-3200 \times 0.9375)| \]. Calculating gives: \[ F = 4.80 \times 10^{-16} \, \text{N}\].

Key Concepts

Magnetic Field Direction

Understanding the direction of a magnetic field is crucial when analyzing the motion of charged particles. According to the right-hand rule, if you point your thumb in the direction of a positively charged particle's velocity and your fingers in the direction of the magnetic force experienced by the particle, your palm faces the direction of the magnetic field.

In the given problem, a proton moves in the \(+x\)-direction and experiences a force in the \(+y\)-direction. This implies that the magnetic field is directed along the \(+z\)-axis. For other particles, like electrons, similar reasoning applies, keeping in mind their negative charge which results in a magnetic force in the opposite direction.

Charged Particles Motion

The motion of charged particles in a magnetic field is profoundly influenced by the Lorentz force. This force is perpendicular to both the velocity of the particle and the magnetic field, causing charged particles to move in a circular or helical path depending on their initial velocity and the uniformity of the magnetic field.

In our context, when a proton or an electron enters a magnetic field, it doesn’t move in a straight line but curves based on the field's direction. The curvature of the path depends on the particle’s charge, speed, and the magnetic field strength. For instance, the movement of our proton and electron in the exercise confirms these principles.

Magnetic Force Calculation

Magnetic force calculation is done using the Lorentz force equation: \( F = q(v \times B) \).Here, the force depends on:

• \( q \), the charge of the particle.
• \( v \), the velocity of the particle.
• \( B \), the magnetic field.

In the exercise, calculations were performed for both a proton and an electron.
For instance, the force on a proton was calculated by rearranging the formula to find the magnetic field strength, knowing the force and velocity.
Thus, \( B = \frac{F}{q \cdot v} = \frac{2.25 \times 10^{-16}}{1.6 \times 10^{-19} \cdot 1500} \), resulting in a field strength of approximately 0.9375 T. This consistency with the electron’s data confirmed the field's magnitude and direction.

Proton and Electron Charge

The charge of subatomic particles plays a significant role in calculating magnetic forces. Protons, with a positive charge of \( +1.6 \times 10^{-19} \, \text{C} \), and electrons, with a negative charge of \( -1.6 \times 10^{-19} \, \text{C} \), experience forces in opposite directions under identical conditions.

In the given exercise, these charges were pivotal in determining the forces acting on them when they moved through the magnetic field. The positive charge of a proton leads to a force in the expected direction, while an electron, due to its negative charge, experiences a force in the opposite direction relative to the conventional right-hand rule predictions.