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Chapter 27: Problem 20

(a) An 16 nucleus (charge \(+8 e )\) moving horizontally from west to east with a speed of 500 \(\mathrm{km} / \mathrm{s}\) experiences a magnetic force of 0.00320 \(\mathrm{nN}\) vertically downward. Find the magnitude and direction of the weakest magnetic field required to produce this force. Explain how this same force could be caused by a larger magnetic field. (b) An electron moves in a uniform, horizontal, 2.10 - T magnetic field that is toward the west. What must the magnitude and direction of the minimum velocity of the electron be so that the magnetic force on it will be 4.60 \(\mathrm{pN}\) , vertically upward? Explain how the velocity could be greater than this minimum value and the force still have this same magnitude and direction.

Short Answer

Expert verified
(a) 0.05 mT, north; by aligning the magnetic field differently. (b) 13.7 Mm/s, south; by adjusting velocity component direction.

Step by step solution

01

Understanding the Magnetic Force Equation

The magnetic force on a moving charge is given by the equation \( F = qvB \sin \theta \), where \( F \) is the magnetic force, \( q \) is the charge, \( v \) is the velocity of the particle, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and the magnetic field. Given: \( F = 0.00320 \text{ nN} = 3.20 \times 10^{-12} \text{ N} \), charge of 16 nucleus with charge \(+8e\) (\( e = 1.6 \times 10^{-19} \text{ C} \)), and \( v = 500 \times 10^3 \text{ m/s} \). The force is downward, suggesting \( \theta = 90^\circ \) or \( 270^\circ \) with \( \sin \theta = 1 \).
02

Calculating the Magnetic Field

Using the equation \( F = qvB \), we can solve for \( B \) as follows:\[ B = \frac{F}{qv} = \frac{3.20 \times 10^{-12} \text{ N}}{8 \times 1.6 \times 10^{-19} \text{ C} \times 500 \times 10^3 \text{ m/s}} \]Calculating gives \( B = 5 \times 10^{-5} \text{ T} \). The direction will be north based on the right-hand rule.
03

Scenario with Larger Magnetic Field

A larger magnetic field could also produce the same force if the velocity component perpendicular to the magnetic field is smaller than 500 km/s. This could happen if the magnetic field aligns differently with the velocity but perpendicular to the moving charge's path, maintaining the force magnitude and direction.
04

Calculate Minimum Velocity for Electron

For the electron, the magnetic force is given by the same formula. Let \( q = e = 1.6 \times 10^{-19} \text{ C} \), \( F = 4.60 \text{ pN} = 4.60 \times 10^{-12} \text{ N} \), and \( B = 2.10 \text{ T} \). Solving for the velocity:\[ v = \frac{F}{qB} = \frac{4.60 \times 10^{-12} \text{ N}}{1.6 \times 10^{-19} \text{ C} \times 2.10 \text{ T}} \]Calculate \( v \approx 1.37 \times 10^7 \text{ m/s} \). The direction is toward the south due to the right-hand rule.
05

Effect of Greater Velocity

If the electron moves faster than this minimum velocity with some component in the direction of the magnetic field, the perpendicular component could remain to produce the same force. This will maintain the magnitude and direction of the force as upward.

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Key Concepts

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

Magnetic Field
A magnetic field is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. Understanding the core concept of magnetic fields is essential, as they are integral to the exercise at hand. In the given scenario, we calculate the magnetic field using the equation for magnetic force: \( F = qvB \sin \theta \), with \( \theta \) being the angle between the velocity of the charged particle and the magnetic field direction. Here, we assumed \( \theta = 90^\circ \), which simplified our equation to \( F = qvB \) because \( \sin 90^\circ = 1 \).
To determine the magnetic field's magnitude, we rearranged the formula to \( B = \frac{F}{qv} \). The solution revealed that the magnetic field required to cause the given magnetic force was \( 5 \times 10^{-5} \text{ T} \). The concept of magnetic fields was further explored by examining if a larger field could produce the same force by altering charge velocity to have minimal perpendicular components.
  • Magnetic fields exert force only on moving charges.
  • The direction is determined by the right-hand rule.
  • The force is maximum when velocity is perpendicular to the field.
Charge-Velocity Relationship
When it comes to magnetic force, the relationship between the charge's velocity and the magnetic field is crucial. The force a charged particle experiences in a magnetic field not only depends on the strength of the field and the charge but also heavily on the velocity of the charge. This force can be expressed as \( F = qvB \sin \theta \).
This equation shows that the force is directly proportional to both the charge's speed and the magnitude of the magnetic field. In our problem, we see that manipulating the velocity of a charged particle can alter the direction and magnitude of the force it experiences.
For the charged nucleus moving from west to east, its velocity was a key factor in determining the magnetic force downward. In contrast, the electron required a specific velocity to ensure the force direction was vertically upward. Such precise control over velocity highlights how sensitive the relationship is between a charge's velocity and magnetic force. The calculation for minimum velocity reassures that even with a larger field, similar force effects are obtained by altering velocity directions.
Right-hand Rule
The right-hand rule is a simple technique used to predict the direction of the magnetic force on a positive moving charge within a magnetic field. It's an essential tool to visualize the relationship between magnetic fields and moving charges. For this exercise, it helped us deduce the direction of the magnetic field and the magnetic force experienced by the particles. For a positively charged particle, like the nucleus in the exercise, you align your right thumb with the velocity (from west to east) and curl your fingers in the direction of the magnetic force (vertically downward). Your palm then points towards the direction of the magnetic field (north).
Similarly, for the electron moving through a horizontal magnetic field toward the west:
  • The thumb represents the direction of velocity.
  • The fingers indicate the direction of the magnetic field.
  • The force emerges perpendicular from the palm.
By using the right-hand rule, even complex scenarios become manageable, ultimately enabling us to predict the force's magnitude and direction with integrity and ease. This rule is invaluable for solving physics problems regarding forces on charges in magnetic fields.

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Most popular questions from this chapter

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.

A cycloron is to accelerate protons to an energy of 5.4 MeV. The superconducting electromagnet of the cyclotron produces a \(3.5-\) T magnetic field perpendicular to the proton orbits. (a) When the protons have achieved a kinetic energy of 2.7 \(\mathrm{MeV}\) , what is the radius of their circular orbit and what is their angular speed? (b) Repeat part (a) when the protons have achieved their final kinetic energy of 5.4 \(\mathrm{MeV}\) .

An insulated wire with mass \(m=5.40 \times 10^{-5} \mathrm{kg}\) is bent into the shape of an inverted U such that the horizontal part has a length \(l=15.0 \mathrm{cm} .\) The bent ends of the wire are partially immersed in two pools of mercury, with 2.5 \(\mathrm{cm}\) of each end below the mercury's surface. The entire structure is in a region containing a uniform \(0.00650-\mathrm{T}\) magnetic field directed into the page (Fig. 27.71\()\) . An electrical connection from the mercury pools is made through the ends of the wires. The mercury pools are connected to a \(1.50-\mathrm{V}\) battery and a switch \(\mathrm{S}\) . When switch \(\mathrm{S}\) is closed, the wire jumps 35.0 \(\mathrm{cm}\) into the air, measured from its initial position. (a) Determine the speed \(v\) of the wire as it leaves the mercury. (b) Assuming that the current \(I\) through the wire was constant from the time the switch was closed until the wire left the mercury, determine \(I\) (c) Ignoring the resistance of the mercury and the circuit wires, determine the resistance of the moving wire.

An electromagnet produces a magnetic field of 0.550 T in a cylindrical region of radius 2.50 \(\mathrm{cm}\) between its poles. A straight wire carrying a current of 10.8 \(\mathrm{A}\) 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 is exerted on the wire?

Determining the Mass of an Isotope. The electric field between the plates of the velocity selector in a Bainbridge mass spectrometer (see Fig. 27.22) is 1.12 \(\times 10^{5} \mathrm{V} / \mathrm{m}\) , and the magnetic field in both regions is 0.540 T. A stream of singly charged selenium ions moves in a circular path with a radius of 31.0 \(\mathrm{cm}\) in the magnetic field. Determine the mass of one selenium ion and the mass number of this selenium isotope. (The mass number is equal to the mass of the isotope in atomic mass units, rounded to the nearest integer. One atomic mass unit \(=1 \mathbf{u}=1.66 \times 10^{-27} \mathrm{kg} .\) .
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