Problem 17 - University Physics with Modern Physics

Question

A \(150-\) g ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125-\mathrm{m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 \(\mathrm{T}\) and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

Step-by-Step Solution

Answer

The magnetic force is approximately \(7.94 \times 10^{-10}\, \text{N}\) directed south.

Step 1: Convert Mass to Kilograms

The mass of the ball is given as 150 grams. First, we need to convert this mass to kilograms. Since 1 kilogram is equal to 1000 grams, the mass in kilograms is \[m = \frac{150\, \text{g}}{1000} = 0.150 \, \text{kg}.\]

Step 2: Calculate the Electric Charge

The ball has \(4.00 \times 10^8\) excess electrons. Each electron has a charge of \(-1.6 \times 10^{-19}\, \text{C}\). Therefore, the total charge \(q\) of the ball is calculated by multiplying the number of electrons by the charge of one electron:\[q = 4.00 \times 10^8 \times (-1.6 \times 10^{-19}\, \text{C}) = -6.4 \times 10^{-11}\, \text{C}.\]

Step 3: Calculate the Velocity of the Ball

The ball is dropped from a height of 125 meters. Since air resistance is negligible, we can use the formula for the final velocity of an object in free fall: \[v = \sqrt{2gh}\]where \(g = 9.81\, \text{m/s}^2\) is the acceleration due to gravity and \(h = 125\, \text{m}\). Plugging in these values gives:\[v = \sqrt{2 \times 9.81\, \text{m/s}^2 \times 125\, \text{m}} = \sqrt{2452.5} \approx 49.52 \, \text{m/s}.\]

Step 4: Determine the Magnetic Force

The formula for the magnetic force \(F\) acting on a charge moving in a magnetic field is given by the Lorentz force equation: \[F = |q|vB \sin \theta\]where \(q = -6.4 \times 10^{-11}\, \text{C}\), \(v = 49.52\, \text{m/s}\), \(B = 0.250\, \text{T}\), and \(\theta = 90^\circ\) since the velocity is vertical and the magnetic field is horizontal. Thus, \(\sin 90^\circ = 1\). Substituting these values gives:\[F = |-6.4 \times 10^{-11}| \times 49.52 \times 0.250 \times 1 = 7.94 \times 10^{-10} \, \text{N}.\]

Step 5: Determine the Direction of the Force

According to the right-hand rule, the force direction is perpendicular to both the velocity of the ball and the magnetic field. The ball falls vertically downward, and the magnetic field is directed from east to west. Using the right-hand rule, the force will be directed either north or south. Since the charge is negative, it reverses the direction of the force predicted by the right-hand rule, resulting in a force directed to the south.

Key Concepts

Lorentz force equation

The Lorentz force equation is a fundamental principle in electromagnetism that describes how electrically charged particles interact with magnetic and electric fields. The equation captures the total force exerted on a charged particle and is expressed as:\[ F = q(E + v \times B) \]In this formula:

• \( F \) is the Lorentz force.
• \( q \) represents the charge of the particle.
• \( E \) stands for the electric field.
• \( v \) is the velocity of the particle.
• \( B \) denotes the magnetic field.

The first part of the equation, related to the electric field, is not applicable in this problem since only a magnetic field is involved. The second part, \( q(v \times B) \), focuses on the magnetic component. Here, the velocity \( v \) and the magnetic field \( B \) interact in a cross product, making the force direction dependent on their perpendicularity. The magnitude of force is influenced by the sine of the angle \( \theta \) between these vectors. In practical scenarios, as in this exercise, the angle is often \( 90^\circ \), maximizing the magnetic force.

free fall velocity

Free fall velocity is a crucial concept when analyzing objects that are solely influenced by gravity, excluding any other forces like air resistance. When an object is dropped from a certain height, it accelerates due to gravity. The velocity just before it impacts is found using the equation for free fall:\[ v = \sqrt{2gh} \]Here,

• \( v \) is the final velocity of the falling object.
• \( g \) represents the acceleration due to gravity, approximately 9.81 m/s².
• \( h \) is the height from which the object is dropped.

This formula assumes ideal conditions with negligible air resistance. In the given problem, a ball dropped from 125 meters would achieve a velocity of approximately 49.52 m/s just before encountering the magnetic field. Understanding free fall velocity aids in determining how fast the object is moving as it transitions from gravitational influence to interacting with a magnetic field.

charge of electron

The concept of electron charge is fundamental to understanding interactions involving magnetic and electric forces. An electron is a subatomic particle with a negative charge, denoted as \( -1.6 \times 10^{-19} \, \text{C} \). This charge is essential in calculating the total electric charge possessed by a body with excess or deficit electrons.In this exercise, the ball has \( 4.00 \times 10^8 \) excess electrons. To determine the total charge \( q \) on the ball:\[ q = \text{number of electrons} \times \text{charge of one electron} \]Substituting values, we find:\[ q = 4.00 \times 10^8 \times (-1.6 \times 10^{-19} \, \text{C}) = -6.4 \times 10^{-11} \, \text{C} \]Understanding the charge of an electron allows us to predict the ball's interaction with a magnetic field, as charges are crucial in calculating forces according to the Lorentz force equation. The negative charge also indicates the direction of the force will be opposite to that predicted by conventional right-hand rule application.

right-hand rule

The right-hand rule is a helpful mnemonic used in physics to determine the direction of force on a charged particle moving in a magnetic field. When applying this rule, conduct the following steps:

• Align your right hand so that your thumb points in the direction of the positive particle's velocity \( v \).
• Let your fingers point in the direction of the magnetic field \( B \).
• The force \( F \) direction is shown by your palm pushing outward.

However, this rule is directly used only for positive charges. Because the problem involves a negative charge (electrons), the direction indicated by your palm must be reversed. In the exercise, after application of the right-hand rule, the force from a positive charge would be north. For the negative charge, which is opposite, it becomes south. Understanding the right-hand rule is essential in visualizing and predicting the dynamic interactions between charged particles and magnetic fields.