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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-m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 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.

Short Answer

Expert verified
The magnetic force is approximately \( 7.92 \times 10^{-10} \text{ N} \) upward.

Step by step solution

01

Determine Charge of the Ball

First, we need to find the total charge on the ball due to excess electrons. The charge of one electron is approximately \[ q_e = -1.6 \times 10^{-19} \text{ C}. \]The total excess charge is \[ Q = n \times q_e = 4.00 \times 10^8 \times (-1.6 \times 10^{-19}) \text{ C} = -6.4 \times 10^{-11} \text{ C}. \]
02

Identify Velocity Direction and Magnetic Field Direction

As the ball is in free fall through the shaft, it moves vertically downward. This motion is at the instant it just enters the magnetic field, which is directed from east to west.
03

Use the Magnetic Force Equation

To find the magnetic force, use the formula\[ F = |Q|vB \sin(\theta), \]where \( F \) is the magnetic force, \( Q \) is the charge, \( v \) is the velocity of the ball just as it enters the field, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field. Here, \( \theta = 90^\circ \) because the velocity is vertical and the magnetic field is horizontal. Thus, \( \sin(\theta) = 1 \).
04

Calculate the Velocity of the Ball Just Before Entering the Field

The ball falls a distance of 125 m, so we apply the equation of motion \[ v = \sqrt{2gh}, \]where \( g = 9.8 \text{ m/s}^2 \) (acceleration due to gravity) and \( h = 125 \text{ m}. \) Thus, \[ v = \sqrt{2 \times 9.8 \times 125} \approx 49.5 \text{ m/s}. \]
05

Calculate Magnitude of Magnetic Force

Substitute the known values into the magnetic force equation:\[ F = |6.4 \times 10^{-11} \text{ C}| \times 49.5 \text{ m/s} \times 0.250 \text{ T} \times 1 = 7.92 \times 10^{-10} \text{ N}. \]
06

Determine the Direction of Magnetic Force

The direction of the magnetic force is determined using the right-hand rule. With your fingers pointing in the direction of velocity (downward) and curling towards the magnetic field (east to west), the thumb points in the direction of force, which will be out of the surface of the Earth (upward), opposite to the motion of gravity.

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Key Concepts

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

Charge of Electrons
Electrons carry a fundamental negative charge. This charge is known as the elementary charge, denoted by the symbol \( e \). The value of this charge is approximately \( -1.6 \times 10^{-19} \) coulombs.
To determine the total charge on an object due to excess electrons, you multiply the number of excess electrons by the charge of a single electron.
  • If you know the number of excess electrons, \( n \), in an object, the total charge, \( Q \), is calculated as:
\[ Q = n \times (-1.6 \times 10^{-19} \text{ C}) \]
In the case of a ball with \( 4.00 \times 10^8 \) excess electrons, the calculation gives a total charge of \( -6.4 \times 10^{-11} \) coulombs. This indicates that the ball is negatively charged.
Understanding how the charge of electrons contributes to the overall charge of an object is crucial when determining how it interacts with electric and magnetic fields.
Right-hand Rule
The right-hand rule is a simple way to determine the direction of the magnetic force on a charged particle or object moving within a magnetic field.
When you apply this rule, you use your right hand to guide you:
  • Point your fingers in the direction of the velocity of the charged object.
  • Next, curl your fingers toward the direction of the magnetic field lines.
  • Your thumb will point in the direction of the magnetic force on a positive charge.
  • For a negative charge, like our ball, the force direction is opposite to where your thumb points.
In the example problem, the velocity of the ball is downward, and the magnetic field is directed from east to west.
Therefore, by the right-hand rule, even though your thumb would point upward for a positive charge, the force on the negatively charged ball will be downward against the free fall.
Uniform Magnetic Field
A uniform magnetic field is one where the magnetic field lines are parallel and equidistant from each other, indicating a constant magnetic field strength at every point in the region.
Some important characteristics of a uniform magnetic field:
  • The magnitude of the magnetic field, denoted by \( B \), is the same everywhere in that field.
  • Because the field is uniform, the force calculated on a charged particle is steady when it moves perpendicular to these field lines.
  • The intensity of the magnetic field is measured in tesla (T).
In the exercise above, the uniform magnetic field has a strength of \( 0.250 \) T, running horizontally from east to west.
It's this uniformity that ensures that when the ball enters the magnetic field, it experiences a consistent force, which can be calculated using the formula \( F = |Q|vB \sin(\theta) \), where \( \theta = 90^\circ \) in this scenario, as the directions are perpendicular.

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Most popular questions from this chapter

A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses (see Section 27.5). In one design for such an instrument, ions with mass \(m\) and charge \(q\) are accelerated through a potential difference \(V\). They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicircular path of radius \(R\). A detector measures where the ions complete the semicircle and from this it is easy to calculate \(R\). (a) Derive the equation for calculating the mass of the ion from measurements of \(B\), \(V\), \(R\), and \(q\). (b) What potential difference \(V\) is needed so that singly ionized \(^{12}\)C atoms will have \(R =\) 50.0 cm in a 0.150-T magnetic field? (c) Suppose the beam consists of a mixture of \(^{12}\)C and \(^{14}\)C ions. If \(v\) and \(B\) have the same values as in part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished? (Make the assumption described in Problem 27.59 for the masses of the ions.)

An electron moves at 1.40 \(\times\) 10\(^6\) m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.40 \(\times\) 10\(^{-2}\) T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

\(\textbf{Determining Diet.}\) One method for determining the amount of corn in early Native American diets is the \(stable\) \(isotope\) \(ratio\) \(analysis\) (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the \(^{12}\)C and \(^{13}\)C isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed 8.50 km /s, and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0 cm for the \(^{12}\)C. The measured masses of these isotopes are 1.99 \(\times\) 10\(^{-26}\) kg (\(^{12}\)C) and 2.16 \(\times\) 10\(^{-26}\) kg (\(^{13}\)C). (a) What strength of magnetic field is required? (b) What is the diameter of the \(^{13}\)C semicircle? (c) What is the separation of the \(^{12}\)C and \(^{13}\)C ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

A circular area with a radius of 6.50 cm lies in the \(xy\)-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field \(B =\) 0.230 T (a) in the \(+z\)-direction; (b) at an angle of 53.1\(^\circ\) from the \(+z\)-direction; (c) in the \(+y\)-direction?

An electromagnet produces a magnetic field of 0.550 T in a cylindrical region of radius 2.50 cm between its poles. A straight wire carrying a current of 10.8 A passes through the center of this region and is perpendicular to both the axis of the cylindrical region and the magnetic field. What magnitude of force does this field exert on the wire?
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