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Chapter 27: Problem 17

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.

Short Answer

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

Step by step solution

01

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\[ Q = 4.00 \times 10^{8} \times 1.602 \times 10^{-19} \approx 6.408 \times 10^{-11} \, \text{Coulombs}. \]
02

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.
03

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:\[ v^2 = 0 + 2 \times 9.81 \times 125 \v^2 = 2452.5 \v = \sqrt{2452.5} \approx 49.51 \, \text{m/s}. \]
04

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:\[ F = 6.408 \times 10^{-11} \times 49.51 \times 0.250 \times 1 \approx 7.94 \times 10^{-10} \, \text{N}. \]
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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    Most popular questions from this chapter

    An Electromagnetic Rail Gun. A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{\boldsymbol{B}}\) fills the region between the rails (Fig. 27.63\()\) . (a) Find the magnitude and direction of the net force on the conducting bar Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\) , find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\) . (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{km} / \mathrm{s}) .\) Let \(B=0.50 \mathrm{T}, \quad I=2.0 \times 10^{3} \mathrm{A}\) \(m=25 \mathrm{kg},\) and \(L=50 \mathrm{cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts \((a)\) and \((b)\) , even though gravity plays an important role in an actual launch in space.

    A straight piece of conducting wire with mass \(M\) and length \(L\) is placed on a friction- less incline tilted at an angle \(\theta\) from the horizontal (Fig. 27.61 ) There is a uniform, vertical magnetic field \(\overrightarrow{\boldsymbol{B}}\) at all points (produced by an arrangement of magnets not shown in the figure). To keep the wire from shiding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Determine the magnitude and direction of the current in the wire that will cause the wire to remain at rest. Copy the figure and draw the direction of the current on your copy. In addition, show in a free-body diagram all the forces that act on the wire.

    A particle with mass \(1.81 \times 10^{-3} \mathrm{kg}\) and a charge of \(1.22 \times\) \(10^{-8} \mathrm{C}\) has, at a given instant, a velocity \(\vec{v}=\left(3.00 \times 10^{4} \mathrm{m} / \mathrm{s}\right) \hat{j}\) What are the magnitude and direction of the particle's accoleration produced by a uniform magnetic field \(\vec{B}=(1.63 \mathrm{T}) \hat{\imath}+\) \((0.980 \mathrm{T}) \hat{\jmath} ?\)

    A particle of charge \(q>0\) is moving at speed \(v\) in the \(+z\) -direction through a region of uniform magnetic field \(\overrightarrow{\boldsymbol{B}} .\) The magnetic force on the particle is \(\overrightarrow{\boldsymbol{F}}=F_{0}(3 \hat{\imath}+4 \hat{\jmath}),\) where \(\boldsymbol{F}_{0}\) is a positive constant. (a) Determine the components \(B_{x}, B_{y},\) and \(B_{z},\) or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude \(6 F_{0} / q v,\) determine as much as you can about the remaining components of \(\overrightarrow{\boldsymbol{B}}\) .

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