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Chapter 21: Problem 20

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 cm) of the circular path for particle 2\(?\)

Short Answer

Expert verified
The radius of particle 2's path is approximately 30.78 cm.

Step by step solution

01

Equation for Particle Path in a Magnetic Field

When a charged particle travels perpendicular to a magnetic field, it experiences a centripetal force due to the magnetic field. The equation for the magnetic force that acts as the centripetal force is given by:\[ qvB = \frac{mv^2}{r} \]where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, \( m \) is the particle mass, and \( r \) is the radius of the path.
02

Expression for Velocity from Potential Difference

Both particles accelerate from rest due to an electric potential difference \( V \). This means the potential energy is converted into kinetic energy:\[ qV = \frac{1}{2}mv^2 \]From this, the velocity \( v \) can be expressed as:\[ v = \sqrt{\frac{2qV}{m}} \]
03

Substitute Velocity in Magnetic Force Equation

Substitute the expression for \( v \) into the magnetic force equation:\[ q \cdot \sqrt{\frac{2qV}{m}} \cdot B = \frac{m \left( \sqrt{\frac{2qV}{m}} \right)^2}{r} \]Simplifying gives:\[ q \cdot \sqrt{\frac{2qV}{m}} \cdot B = \frac{m(2qV/m)}{r} \]
04

Solving for Radius

Rearrange the equation to solve for \( r \):\[ r = \frac{m \cdot \sqrt{\frac{2qV}{m}}}{qB} \]Since the kinetic energy is determined by \( qV \), the radius simplifies to:\[ r = \frac{m \cdot v}{qB} \]
05

Ratio of Radii for the Two Particles

Using the derived formula for radius, the ratio of radii \( r_1/r_2 \) for particle 1 and particle 2 is:\[ \frac{r_1}{r_2} = \frac{\frac{m_1 \cdot v}{qB}}{\frac{m_2 \cdot v}{qB}} = \frac{m_1}{m_2} \]Thus, the radius for particle 2 can be expressed in terms of known quantities:\[ r_2 = r_1 \cdot \frac{m_2}{m_1} \]
06

Calculate the Radius of Particle 2

Substitute the given values into the ratio equation:\[ r_2 = 12 \text{ cm} \times \frac{5.9 \times 10^{-8} \text{ kg}}{2.3 \times 10^{-8} \text{ kg}} \]\[ r_2 = 12 \text{ cm} \times \frac{5.9}{2.3} \approx 30.78 \text{ cm} \]

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Key Concepts

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

Centripetal Force
When a charged particle moves perpendicular to a magnetic field, it experiences a force that keeps it moving in a circular path. This force is known as the centripetal force. It's essential in maintaining the particle's constant circular motion within the magnetic field.
The centripetal force arises from the magnetic force, which is described by the equation \( qvB = \frac{mv^2}{r} \). Here, \( q \) represents the charge, \( v \) the velocity, \( B \) the magnetic field strength, \( m \) the particle's mass, and \( r \) the radius of its path.
  • This relationship tells us that the magnetic force acting on the particle also acts as the centripetal force.
  • Such forces are responsible for the curved paths taken by charged particles in magnetic fields.
By understanding the concept of centripetal force, one can predict the movement behavior of charged particles in various magnetic and electric environments.
Electric Potential Difference
Electric potential difference, commonly referred to as voltage, impacts how charged particles behave. When a charged particle accelerates from rest through an electric potential difference, it gains kinetic energy. The relationship between potential difference \( V \) and kinetic energy is provided by \( qV = \frac{1}{2}mv^2 \).
From this equation, we can see how the electric potential difference directly influences the particle's velocity by converting stored electrical energy into kinetic energy.
  • The speed of the particle can be calculated as \( v = \sqrt{\frac{2qV}{m}} \), showcasing the influence of mass and charge.
  • This acceleration is what allows the particles to move and ultimately interact with the magnetic field to form circular paths.
Understanding how electric potential difference affects particle velocity is vital for analyzing movements and interactions in fields.
Particle Mass Ratios
Mass ratios of particles help in determining the dynamics of their paths within magnetic fields. In cases where identical charges pass through the same electric potential difference and enter a magnetic field, the path radius of each particle is determined by its mass.
The key equation is \( r = \frac{m \cdot v}{qB} \), which implies:\
  • Particles with greater mass generally have larger path radii, given the same velocity and charge.
  • The ratio of path radii between two particles can be compared using their mass ratio \( \frac{m_1}{m_2} \).
This concept allows us to predict and compare the paths of particles in magnetic fields based on their mass. For example, if you know the mass and path radius of one particle, you can calculate the radius for another by using their mass ratio. This understanding is crucial for practical applications, such as separating isotopes or analyzing particle trajectories in physics experiments.

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Most popular questions from this chapter

(a) A proton, traveling with a velocity of \(4.5 \times 10^{6} \mathrm{m} / \mathrm{s}\) due east, experiences a magnetic force that has a maximum magnitude of \(8.0 \times 10^{-14} \mathrm{N}\) and a direction of due south. What are the magnitude and direction of the magnetic field causing the force? (b) Repeat part (a) assuming the proton is replaced by an electron.

Suppose that a uniform magnetic field is everywhere perpendicular to this page. The field points directly upward toward you. A circular path is drawn on the page. Use Amp猫re's law to show that there can be no net current passing through the circular surface.

The ion source in a mass spectrometer produces both singly and doubly ionized species, \(\mathrm{X}^{+}\) and \(\mathrm{X}^{2+}\) . 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 ficld, which causes them to move on circular paths. The radius of the path for the species \(X^{+}\) is \(r_{1},\) while the radius for species \(X^{2+}\) is \(r_{2}\) . Find the ratio \(r_{1} / r_{2}\) of the radii.

In the model of the hydrogen atom created by Niels Bohr, the electron moves around the proton at a speed of \(2.2 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a circle of radius \(5.3 \times 10^{-11} \mathrm{m}\) . Considering the orbiting electron to be a small current loop, determine the magnetic moment associated with this motion. Hint: The electron travels around the circle in a time equal to the period of the motion.)

A magnetic field has a magnitude of \(1.2 \times 10^{-3} \mathrm{T}\) , and an electric field has a magnitude of \(4.6 \times 10^{3} \mathrm{N} / \mathrm{C}\) . Both fields point in the same direction. A positive 1.8\(\mu \mathrm{C}\) charge moves at a speed of \(3.1 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.
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