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Chapter 21: Problem 27

A positively charged particle of mass \(7.2 \times 10^{-8} \mathrm{kg}\) is traveling due east with a speed of \(85 \mathrm{m} / \mathrm{s}\) and enters a \(0.31-\mathrm{T}\) uniform magnetic field. The particle moves through one- quarter of a circle in a time of \(2.2 \times 10^{-3} \mathrm{s},\) at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

Short Answer

Expert verified
(a) The magnetic force is \(4.36 \times 10^{-4} \text{ N}\). (b) The charge is \(1.66 \times 10^{-5} \text{ C}\).

Step by step solution

01

Understand the Problem

We have a charged particle moving in a magnetic field and completing one-quarter of a circular path. We need to find the magnetic force and the charge of the particle. The given values are: mass \(m = 7.2 \times 10^{-8} \, \text{kg}\), velocity \(v = 85 \, \text{m/s}\), magnetic field \(B = 0.31 \, \text{T}\), and time for one-quarter of a circle \(t = 2.2 \times 10^{-3} \, \text{s}\).
02

Calculate Radius of Circular Path

Since the particle travels one-quarter of a circle in \(2.2 \times 10^{-3} \, \text{s}\), the time for a full circle is \(4 \cdot 2.2 \times 10^{-3} = 8.8 \times 10^{-3} \, \text{s}\). The period \(T\) is \(8.8 \times 10^{-3} \, \text{s}\). The circular speed is given, allowing us to calculate the radius \(r\) using \(v = \frac{2\pi r}{T}\): \[ r = \frac{vT}{2\pi} = \frac{85 \times 8.8 \times 10^{-3}}{2\pi} \approx 0.119 \, \text{m} \].
03

Use Centripetal Force Formula to Find Magnetic Force

The magnetic force provides the centripetal force, \(F = \frac{mv^2}{r}\), needed to keep the particle moving in a circular path. Substitute the known values: \[ F = \frac{(7.2 \times 10^{-8})(85)^2}{0.119} \approx 4.36 \times 10^{-4} \, \text{N} \]. Thus, the magnitude of the magnetic force is \(4.36 \times 10^{-4} \, \text{N}\).
04

Calculate Charge Using Magnetic Force Equation

The magnetic force is also described by \(F = qvB\), where \(q\) is the charge. Rearrange the equation to solve for \(q\): \[ q = \frac{F}{vB} = \frac{4.36 \times 10^{-4}}{85 \times 0.31} \approx 1.66 \times 10^{-5} \, \text{C} \]. The magnitude of the charge is \(1.66 \times 10^{-5} \, \text{C}\).

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Key Concepts

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

Charged Particle Motion
When a charged particle moves through a magnetic field, it experiences a force known as the Lorentz force, which can cause its path to curve. This type of motion is a classic example of how magnetic fields influence charged particles. The force depends on several factors:
  • The charge of the particle: More charge means a larger force.
  • The speed of the particle: Faster movement results in a greater force.
  • The strength of the magnetic field: A stronger field increases the force.
In the given exercise, a positively charged particle travels through one-quarter of a circle, demonstrating circular motion within a uniform magnetic field. The constant change in direction is a key aspect of magnetic interactions with charged particles.
Uniform Magnetic Field
A uniform magnetic field is characterized by having the same magnetic force at every point within it. This means that a charged particle entering this field will experience a consistent magnetic influence regardless of its position in the field. In this exercise, the field has a strength of 0.31 T (Tesla). When calculating the effects on the particle, we can rely on this consistent field strength, simplifying the calculations of magnetic force and subsequent motion. Uniform magnetic fields are often used in experiments and applications because they provide predictable behavior, essential for understanding fundamental physics concepts.
Centripetal Force
The centripetal force is crucial in circular motion, acting as the necessary force that keeps an object moving along a circular path. In the context of the exercise, the magnetic force acts as a centripetal force. This is calculated using the formula:\[ F = \frac{mv^2}{r} \]where:
  • \( m \) is the mass of the particle
  • \( v \) is the velocity
  • \( r \) is the radius of the circular path
By plugging in the known values, you determine how much force is acting on the particle to keep it in its curved path. Grasping centripetal force dynamics helps explain why the particle follows a circular trajectory within the magnetic field.
Charge Calculation
Calculating the charge of a particle in a magnetic field involves understanding the relationship between the magnetic force and the particle's charge. In the exercise, we've used the formula:\[ F = qvB \], where:
  • \( F \) is the magnetic force
  • \( q \) is the charge
  • \( v \) is the velocity
  • \( B \) is the magnetic field strength
Rearranging this equation allows us to solve for the charge,\[ q = \frac{F}{vB} \]. This direct relationship truly showcases the nature of magnetic fields and charged particles. By applying this method, you derive the particle's charge as 1.66 x 10^-5 C. Understanding this calculation helps in determining various properties of the motion in real-life particles' interactions with magnetic fields.

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Most popular questions from this chapter

A horizontal wire of length \(0.53 \mathrm{m}\), carrying a current of \(7.5 \mathrm{A}\), is placed in a uniform external magnetic field. When the wire is horizontal, it experiences no magnetic force. When the wire is tilted upward at an angle of \(19^{\circ},\) it experiences a magnetic force of \(4.4 \times 10^{-3} \mathrm{N}\). Determine the magnitude of the external magnetic field.

The drawing shows a parallel plate capacitor that is moving with a speed of \(32 \mathrm{m} / \mathrm{s}\) through a \(3.6-\mathrm{T}\) magnetic field. The velocity \(\overrightarrow{\mathbf{v}}\) is perpendicular to the magnetic field. The electric field within the capacitor has a value of \(170 \mathrm{N} / \mathrm{C},\) and each plate has an area of \(7.5 \times 10^{-4} \mathrm{m}^{2} .\) What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

The ion source in a mass spectrometer produces both singly and doubly ionized species, \(X^{+}\) and \(X^{2+} .\) The difference in mass between these species is too small to be detected. Both species are accelerated through the same electric potential difference, and both experience the same magnetic field, which causes them to move on circular paths. The radius of the path for the species \(\mathrm{X}^{+}\) is \(r_{1}\), while the radius for species \(\mathrm{X}^{2+}\) is \(r_{2}\). Find the ratio \(r_{1} / r_{2}\) of the radii.

Particle 1 and particle 2 have masses of \(m_{1}=2.3 \times 10^{-8} \mathrm{kg}\) and \(m_{2}=5.9 \times 10^{-8} \mathrm{kg},\) but they carry the same charge \(q .\) The two particles accelerate from rest through the same electric potential difference \(V\) and enter the same magnetic field, which has a magnitude \(B\). The particles travel perpendicular to the magnetic field on circular paths. The radius of the circular path for particle 1 is \(r_{1}=12 \mathrm{cm} .\) What is the radius (in \(\mathrm{cm}\) ) of the circular path for particle \(2 ?\)

A charge of \(-8.3 \mu \mathrm{C}\) is traveling at a speed of \(7.4 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is \(52^{\circ} .\) A force of magnitude \(5.4 \times 10^{-3} \mathrm{N}\) acts on the charge. What is the magnitude of the magnetic field?
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