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An ionized deuteron (a bound proton-neutron system with a net \(+e\) charge) passes through a velocity selector whose perpendicular magnetic and electric fields have magnitudes of \(40 \mathrm{mT}\) and \(8.0 \mathrm{kV} / \mathrm{m}\), respectively. Find the speed of the ion.

Short Answer

Expert verified
The speed of the ion is \( 2.0 \times 10^5 \mathrm{m/s} \).

Step by step solution

01

Understanding the Velocity Selector

A velocity selector is a device that allows particles with a certain velocity to pass through it without deflection. It uses perpendicular electric and magnetic fields. For a charged particle moving through both fields, the electric force \( F_e \) and the magnetic force \( F_b \) must balance for it to pass undeflected.
02

Equating Electric and Magnetic Forces

The electric force \( F_e \) on a charge \( q \) is given by \( F_e = qE \), and the magnetic force \( F_b \) is given by \( F_b = qvB \), where \( v \) is the speed of the particle, \( E \) is the electric field, and \( B \) is the magnetic field. For no deflection, \( F_e = F_b \), so \( qE = qvB \).
03

Solving for the Speed of the Ion

Since the charges \( q \) cancel out from the equation \( qE = qvB \), we simplify it to \( E = vB \). Rearranging gives \( v = \frac{E}{B} \). Now substituting the given values, \( E = 8.0 \times 10^3 \mathrm{V/m} \) and \( B = 40 \times 10^{-3} \mathrm{T} \), we calculate the speed \( v = \frac{8.0 \times 10^3}{40 \times 10^{-3}} \).
04

Calculating the Speed

Perform the calculation \( v = \frac{8.0 \times 10^3}{40 \times 10^{-3}} = \frac{8.0 \times 10^3}{0.04} = 2.0 \times 10^5 \mathrm{m/s} \). Thus, the speed of the ion is \( 2.0 \times 10^5 \mathrm{m/s} \).

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Key Concepts

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

Electromagnetic Fields
Electromagnetic fields are fundamental in physics, comprising electric and magnetic components that interact with charged particles. An electric field is generated by electric charges or varying magnetic fields, while a magnetic field arises from moving electric charges or intrinsic magnetic moments. These fields are pivotal in a wide range of applications, including communication systems, motors, and more complex phenomena in space.

In settings like a velocity selector, electromagnetic fields are crucial. By manipulating these fields, scientists can select particles based on their velocities. The electric and magnetic fields work perpendicularly to precisely maneuver charged particles without deflection. This balance is achieved by synchronizing the electric force, which pushes the charge, with the magnetic force, which can change the direction of motion.
Ionized Deuteron
An ionized deuteron is an atomic nucleus consisting of one proton and one neutron. Deuterons are isotopes of hydrogen, and when they lose an electron, they become positively charged ions, referred to as ionized deuterons.

These ions are significant in fields such as nuclear physics and fusion energy research, as understanding their behavior under different conditions helps scientists advance technology. When passing through a velocity selector, ionized deuterons are subjected to finely-tuned electromagnetic fields to determine their velocities accurately. Due to their simplicity, deuterons serve as a useful model for studying nuclear interactions and are often used in experiments assessing particle behavior.
Electric Force
Electric force is the fundamental force that charged particles exert on each other. It arises from the presence of electric charges and is described by Coulomb's law. The electric force (\( F_e \)) acting on a charge (\( q \)) in an electric field (\( E \)) is given by the equation: \[ F_e = qE \].

In a velocity selector, the electric force plays a vital role in balancing the magnetic force to allow particles to pass through without deflection. By adjusting the magnitude of the electric field, scientists effectively control the degree to which charged particles deviate from their path. This precision is crucial in both research and applications such as mass spectrometry, where desired particle velocities are isolated.
Magnetic Force
Magnetic force refers to the force exerted on moving charges in a magnetic field. Unlike the electric force, which acts independently of motion, the magnetic force requires movement to exist. The magnetic force (\( F_b \)) on a particle with charge (\( q \)) moving with velocity (\( v \)) in a magnetic field (\( B \)) is given by: \[ F_b = qvB \].

This magnetic force can be perpendicular to the direction of movement, causing a change in direction rather than speed. Within a velocity selector, it is crucial for balancing with the electric force to ensure particles with a specific velocity pass unaltered. This feature is indispensable in fields ranging from molecular chemistry to particle physics, where understanding the trajectory of charged particles is key to unlocking new discoveries.

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Most popular questions from this chapter

A cylindrical solenoid \(10 \mathrm{~cm}\) long has 3000 turns of wire and carries a current of 5.0 A. A second solenoid, consisting of 2000 turns of wire and the same length as the first solenoid, surrounds it and is concentric (shares a common central axis) with it. The outer coil carries a current of 10 A in the same direction as the current in inner one. (a) Find the magnetic field near their common center. (b) What current in the second solenoid (magnitude and relative direction) would make the net field strength at the center twice that of the first solenoid alone? (c) What current in the second solenoid (magnitude and relative direction) would result in zero net magnetic field near their common center?

A positive charge moves horizontally to the right across this page and enters a magnetic field directed vertically downward in the plane of the page. (a) What is the direction of the magnetic force on the charge: (1) into the page, (2) out of the page, (3) downward in the plane of the page, or (4) upward in the plane of the page? Explain. (b) If the charge is \(0.25 \mathrm{C}\), its speed is \(2.0 \times 10^{2} \mathrm{~m} / \mathrm{s},\) and it is acted on by a force of \(20 \mathrm{~N},\) what is the magnetic field strength?

A circular wire coil consists of 100 turns and is wound tightly around a very long iron cylinder with a radius of \(2.5 \mathrm{~cm}\) and a relative permeability of \(2200 .\) The loop has a current of \(7.5 \mathrm{~A}\) in it. Determine the magnetic field strength produced by the coil (a) at the center of the coil and (b) at a location on the central axis of the iron cylinder \(5.0 \mathrm{~cm}\) above the center of the circular coil.

A 100 -turn circular coil of wire has a radius of \(20.0 \mathrm{~cm}\) and carries a current of 0.400 A. The normal to the coil area points due east. A compass, when placed at the center of the coil, does not point east, but instead makes an angle of \(60^{\circ}\) north of east. Using this data, determine (a) the magnitude of the horizontal component of the Earth's field at that location and (b) the magnitude of the Earth's field at that location if it makes an angle of \(55^{\circ}\) below the horizontal.

In a mass spectrometer, a beam of protons enters a magnetic field. Some protons make exactly a one-quarter circular arc of radius \(0.50 \mathrm{~m}\). If the field is always perpendicular to the proton's velocity, (a) what is the field's magnitude if exiting protons have a kinetic energy of \(10 \mathrm{keV} ?\) (b) How long does it take the proton to complete the quarter circle? (c) Find the net force (magnitude) on a proton while it is in the field.

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