Ch 27: Magnetic Field and Magnetic Forces

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 27: Magnetic Field and Magnetic Forces

Problem 22

Chapter 27, Problem 22

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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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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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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A long wire carrying 4.50 A of current makes two 90° bends, as shown in Fig. E27.35. The bent part of the wire passes through a uniform 0.240 T magnetic field directed as shown in the figure and confined to a limited region of space. Find the magnitude and direction of the force that the magnetic field exerts on the wire.

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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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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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