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 enrers the field.

Step-by-Step Solution

Answer

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

Step 1: Calculate Charge of the Electrons

To find the charge of the electrons, we use the charge of a single electron, which is approximately \[ e = 1.602 \times 10^{-19} \, \text{Coulombs}. \]The total excess charge on the ball is then\[ Q = 4.00 \times 10^{8} \times 1.602 \times 10^{-19} \approx 6.408 \times 10^{-11} \, \text{Coulombs}. \]

Step 2: Apply the Magnetic Force Formula

The magnetic force on a moving charge in a magnetic field is given by the formula \[ F = qvB \sin(\theta), \]where \( q \) is the charge, \( v \) is the velocity of the ball as it enters the magnetic field, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and the magnetic field. Here, \( \theta = 90^\circ \) since the magnetic field is horizontal and the motion is vertical.

Step 3: Calculate the Velocity of the Ball

The ball is dropped from a height of 125 m, so its velocity when it reaches the bottom of the shaft can be found using the equation of motion:\[ v^2 = u^2 + 2gh, \]where \( u = 0 \) (initial velocity), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 125 \, \text{m}. \)Plugging in the values:\[ v^2 = 0 + 2 \times 9.81 \times 125 \v^2 = 2452.5 \v = \sqrt{2452.5} \approx 49.51 \, \text{m/s}. \]

Step 4: Calculate the Magnetic Force

Now substitute the values into the magnetic force formula:\[ F = QvB \sin(\theta) = 6.408 \times 10^{-11} \, \text{C} \times 49.51 \, \text{m/s} \times 0.250 \, \text{T} \times \sin(90^\circ). \]Calculating the force:\[ F = 6.408 \times 10^{-11} \times 49.51 \times 0.250 \times 1 \approx 7.94 \times 10^{-10} \, \text{N}. \]

Step 5: Determine the Direction of the Force

Using the Right-hand Rule for the cross product in magnetic fields, with the velocity towards the magnetic field and charge negative, the force direction is towards the north, perpendicular to both the vertical velocity and the horizontal magnetic field.

Key Concepts

Excess Electrons

In this exercise, the concept of excess electrons is central to understanding the charge of the ball. An electron carries a negative elementary charge of approximately \( 1.602 \times 10^{-19} \ \text{Coulombs} \). When we say a ball is carrying 'excess electrons', it means there are more electrons than protons in the material, resulting in a net negative charge.

• To find the total charge on the ball, we multiply the number of excess electrons by the charge of one electron.
• For \(4.00 \times 10^{8}\) excess electrons, the total charge \( Q \) can be calculated as:

\[ Q = 4.00 \times 10^{8} \times 1.602 \times 10^{-19} \approx 6.408 \times 10^{-11} \text{Coulombs} \] Understanding how excess electrons contribute to the net charge is crucial when calculating forces in any electromagnetic context, as the charge directly influences how the object interacts with electric and magnetic fields.

Magnetic Fields

A magnetic field is a region where a magnetic force can act on moving charges, such as electrons in the ball. This field is characterized by its direction and strength, measured in Teslas (T). In this problem, the magnetic field has a magnitude of 0.250 T, horizontally from east to west.

• Magnetic fields interact with charged particles, exerting a force perpendicular to both the velocity of the particle and the direction of the field.
• In this situation, the magnetic field affects the ball as it gains velocity moving vertically down the shaft and then enters a horizontal magnetic field.

The field strength, combined with the charge and velocity of the object, determines the magnitude and direction of the magnetic force, a principle central to electromagnetism.

Electromagnetism Concepts

Electromagnetism is the branch of physics that studies the interactions between electric charges and currents with magnetic and electric fields. One of the key relationships in electromagnetism is how magnetic forces are exerted on moving charges.

The magnetic force \( F \) on a charged particle moving through a magnetic field is given by the equation \( F = qvB \sin(\theta) \), where:

• \( q \) is the charge.
• \( v \) is the velocity of the particle.
• \( B \) is the magnetic field strength.
• \( \theta \) is the angle between the velocity and the magnetic field direction.
• Since the motion is vertically downwards and the field is horizontal, \( \theta = 90^\circ \) ensuring that \( \sin(\theta) = 1 \).

The magnetic force experienced by the ball is calculated based on this law. The direction of the force can be determined using the right-hand rule, which helps in identifying the direction based on the velocity vector, magnetic field direction, and the negative charge of the electrons. Understanding these principles enables accurate prediction and control of how objects behave in various electromagnetic environments.