Ch 27: Magnetic Field and Magnetic Forces

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 27: Magnetic Field and Magnetic Forces

Problem 22

Chapter 27, Problem 22

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm. You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV. An alpha particle has charge q = +2e and mass m = 6.64 x 10^-27 kg. If the magnetic field isn't changed, what will be the orbital radius of the alpha particles?

Verified step by step guidance

Understand the problem: We need to find the orbital radius of alpha particles in a cyclotron given their energy and the fact that the magnetic field remains unchanged. We know the orbital radius of protons with the same energy.

Recall the formula for the radius of a charged particle in a magnetic field: \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field.

Since the energy \( E \) is given, use the kinetic energy formula \( E = \frac{1}{2}mv^2 \) to solve for the velocity \( v \) of the alpha particles: \( v = \sqrt{\frac{2E}{m}} \).

Substitute the values for the alpha particle's mass \( m = 6.64 \times 10^{-27} \) kg, charge \( q = +2e \), and energy \( E = 300 \) keV into the formulas. Note that \( 1 \) keV = \( 1.602 \times 10^{-16} \) J.

Calculate the orbital radius \( r \) using the formula \( r = \frac{mv}{qB} \) with the velocity \( v \) obtained from the kinetic energy formula and the known magnetic field \( B \) from the proton's orbital radius.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Cyclotron Motion

Cyclotron motion refers to the circular path charged particles take when moving perpendicular to a uniform magnetic field. The magnetic force acts as the centripetal force, causing the particle to move in a circle. The radius of this orbit depends on the particle's mass, charge, velocity, and the strength of the magnetic field.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through a magnetic field. It is given by F = q(v x B), where q is the charge, v is the velocity, and B is the magnetic field. This force is responsible for the circular motion of particles in a cyclotron, and its magnitude determines the radius of the orbit.

Kinetic Energy and Velocity

Kinetic energy (KE) of a particle is related to its velocity by the equation KE = 0.5mv^2, where m is the mass and v is the velocity. For particles in a cyclotron, knowing the kinetic energy allows us to calculate the velocity, which is crucial for determining the orbital radius when combined with the Lorentz force equation.

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Related Practice

Textbook Question

A 150 V battery is connected across two parallel metal plates of area 28.5 cm² and separation 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64 x 10^-27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field, as shown in Fig. E27.29. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?

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Textbook Question

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accelerate H^- (the hydride ion, which has one proton and two electrons) to an energy of 5 MeV to 20 MeV. This ion has a mass very close to that of a proton because the electron mass is negligible — about 1/2000 of the proton's mass. A typical magnetic field in such cyclotrons is 1.9 T. (a) What is the speed of a 5.0 MeV H^-? (b) If the H^- has energy 5.0 MeV and B = 1.9 T, what is the radius of this ion's circular orbit?

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Textbook Question

A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. E27.24) . The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?

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Textbook Question

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?

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Textbook Question

A 150 g ball containing 4.00 x 10⁸ 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.

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Textbook Question

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 x 10^-27 kg and a charge of +e. The deuteron travels in a circular path with a radius of 6.96 mm in a magnetic field with magnitude 2.50 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

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