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 24

Chapter 27, Problem 24

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?

Verified step by step guidance

Identify the relevant physics principle: The motion of charged particles in a magnetic field is governed by the Lorentz force, which causes the particles to move in a circular path when the velocity is perpendicular to the magnetic field.

Use the formula for the radius of the circular path: The radius \( r \) of the path is given by \( r = \frac{mv}{qB} \), where \( m \) is the mass of the proton, \( v \) is the velocity, \( q \) is the charge of the proton, and \( B \) is the magnetic field strength.

Relate the distance traveled to the radius: Since the beam exits the field perpendicular to its original direction, it travels a quarter of a circle. Therefore, the distance \( d \) traveled is \( d = \frac{1}{4} \, 2\pi r \).

Solve for the radius \( r \): Rearrange the equation \( d = \frac{1}{4} \, 2\pi r \) to find \( r = \frac{2d}{\pi} \). Substitute the given distance \( d = 1.18 \text{ cm} \) into this equation.

Calculate the magnetic field \( B \): Substitute the expression for \( r \) into the radius formula \( r = \frac{mv}{qB} \) and solve for \( B \). Use the known values for the mass and charge of a proton, and the given velocity.

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Key Concepts

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

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through a magnetic field. It is given by the equation F = q(v × B), where q is the charge, v is the velocity, and B is the magnetic field. This force is perpendicular to both the velocity of the particle and the magnetic field, causing the particle to move in a circular path.

Circular Motion in Magnetic Fields

When a charged particle moves perpendicular to a magnetic field, it experiences a centripetal force that causes it to move in a circular path. The radius of this path is determined by the balance between the magnetic force and the centripetal force, given by r = mv/(qB), where m is the mass, v is the velocity, q is the charge, and B is the magnetic field.

Magnetic Field and Path Deflection

The deflection of a charged particle in a magnetic field depends on the strength of the field and the velocity of the particle. The particle exits the field at a direction perpendicular to its entry due to the circular motion induced by the magnetic force. The distance traveled in the field relates to the arc length of the circular path, which can be used to calculate the magnetic field's magnitude.

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Textbook Question

A straight, 2.5 m wire carries a typical household current of 1.5 A (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

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Textbook Question

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?

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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

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 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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