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 25

Chapter 27, Problem 25

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?

Verified step by step guidance

First, understand that for the alpha particles to emerge undeflected, the magnetic force must balance the electric force acting on them. The electric force \( F_e \) is given by \( F_e = qE \), where \( q \) is the charge of the alpha particle and \( E \) is the electric field between the plates.

Calculate the electric field \( E \) between the plates using the formula \( E = \frac{V}{d} \), where \( V \) is the potential difference across the plates (150 V) and \( d \) is the separation between the plates (8.20 mm).

The magnetic force \( F_m \) on a moving charge in a magnetic field is given by \( F_m = qvB \), where \( v \) is the velocity of the alpha particle and \( B \) is the magnetic field. For the particle to be undeflected, set \( F_e = F_m \), leading to \( qE = qvB \).

Solve for the velocity \( v \) of the alpha particles using the kinetic energy gained from the potential difference they were accelerated through: \( \frac{1}{2}mv^2 = qV_{acc} \), where \( V_{acc} \) is 1.75 kV. Rearrange to find \( v = \sqrt{\frac{2qV_{acc}}{m}} \).

Finally, solve for the magnetic field \( B \) using the equation \( B = \frac{E}{v} \). Substitute the expressions for \( E \) and \( v \) to find the magnitude of \( B \). The direction of \( B \) should be perpendicular to both the velocity of the particles and the electric field, following the right-hand rule.

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Key Concepts

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

Electric Field

An electric field is a region around a charged object where other charged objects experience a force. In this scenario, the electric field is created by the parallel plates connected to a 150-V battery, influencing the motion of charged particles like alpha particles. The field's strength is determined by the voltage and separation between the plates.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through electric and magnetic fields. It is the sum of the electric force and the magnetic force acting on the particle. For the alpha particles to remain undeflected, the magnetic force must counterbalance the electric force, requiring precise calculation of the magnetic field's magnitude and direction.

Magnetic Field

A magnetic field exerts a force on moving charged particles, influencing their trajectory. In this problem, the magnetic field must be oriented such that it cancels the deflection caused by the electric field on the alpha particles. The direction and magnitude of the magnetic field are crucial for ensuring the particles pass through the plates without deviation.

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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

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 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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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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