/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 A singly charged ion of \(^{7} \... [FREE SOLUTION] | 91影视

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Chapter 27: Problem 26

A singly charged ion of \(^{7} \mathrm{Li}\) (an isotope of lithium) has a mass of \(1.16 \times 10^{-26} \mathrm{kg}\) . It is accelerated through a potential dif- ference of 220 \(\mathrm{V}\) and then enters a magnetic field with magnitude 0.723 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

Short Answer

Expert verified
The radius of the ion's path is 2.47 cm.

Step by step solution

01

Understand the Concepts and Equations

The problem involves the motion of a charged ion in a magnetic field. The radius of the ion's path in the magnetic field can be determined using the formula for the radius of the circular motion, given by \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.
02

Calculate the Ion's Velocity

When the ion is accelerated through the potential difference \( V \), it gains kinetic energy equal to the electrical energy: \( qV = \frac{1}{2}mv^2 \). Rearranging gives \( v = \sqrt{\frac{2qV}{m}} \). For a singly charged ion, \( q = 1.6 \times 10^{-19} \) C. Substitute the given values: \( V = 220 \) V, \( m = 1.16 \times 10^{-26} \) kg, and \( q = 1.6 \times 10^{-19} \) C into the formula.
03

Substitute Values to Find Velocity

\[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 220}{1.16 \times 10^{-26}}} \] Calculate to get \( v = \sqrt{\frac{7.04 \times 10^{-17}}{1.16 \times 10^{-26}}} = \sqrt{6.068 \times 10^{9}} \approx 2.46 \times 10^5 \) m/s.
04

Calculate the Radius of the Ion's Path

Substitute the values of \( m, v, q, \) and \( B \) into the formula for the radius: \( r = \frac{mv}{qB} \). \( m = 1.16 \times 10^{-26} \) kg, \( v = 2.46 \times 10^5 \) m/s, \( q = 1.6 \times 10^{-19} \) C, and \( B = 0.723 \) T.
05

Substitute Values and Simplify

\[ r = \frac{1.16 \times 10^{-26} \times 2.46 \times 10^5}{1.6 \times 10^{-19} \times 0.723} \] Calculate to get \( r = \frac{2.8536 \times 10^{-21}}{1.1568 \times 10^{-19}} \approx 2.47 \times 10^{-2} \) m or 2.47 cm.

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Key Concepts

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

Circular Motion
When particles like ions travel in a magnetic field, they usually follow a circular path. This happens because the magnetic force acts as a centripetal force, pulling the ion towards the center of its circular path.
The formula for the radius of this circular motion is:
  • \( r = \frac{mv}{qB} \)
Here, \( r \) is the radius of the circle, \( m \) is the mass of the ion, \( v \) is its velocity, \( q \) is its charge, and \( B \) is the magnetic field strength.
This equation comes from equating magnetic force with centripetal force \( F = \frac{mv^2}{r} \). Solving for \( r \) shows how mass, velocity, and field strength affect the size of the ion's path.
Kinetic Energy
Kinetic energy is a form of energy that an object has due to its motion. When an ion is accelerated by a potential difference \( V \), it gains kinetic energy equal to the energy it absorbs from the electric field.
The change in kinetic energy can be calculated using the equation:
  • \( KE = \frac{1}{2}mv^2 \)
But in this scenario, the energy gain from a potential difference \( V \) is given by:
  • \( qV = \frac{1}{2}mv^2 \).
By rearranging this equation to find \( v \), the ion's velocity, we get:
  • \( v = \sqrt{\frac{2qV}{m}} \)

Understanding how kinetic energy transforms and determines velocity helps explain the ion's behavior in magnetic fields.
Potential Difference
Potential difference, often measured in volts, is the amount of electrical energy provided per charge. It's the driving force that accelerates ions, giving them energy to change speed and direction.
In this exercise, the ion is accelerated through a potential difference of \( 220 \) volts.
  • The potential difference transforms into kinetic energy: \( qV \).
  • For a singly charged ion: \( q = 1.6 \times 10^{-19} \) C, thus \( qV = 1.6 \times 10^{-19} \times 220 \).

This equation links potential energy to the kinetic energy the ion gains, showing how potential difference influences the speed of charged particles. Understanding this concept is crucial for predicting particle motion.
Magnetic Field Strength
Magnetic field strength is represented by \( B \) and is measured in Tesla (T). It is a key factor that affects how a moving charged particle behaves in a magnetic field.
The magnetic field exerts a force on the charged particle, redirecting its path into a circular motion.
  • The formula for magnetic force is \( F = qvB \), leading to circular motion \( F = \frac{mv^2}{r} \).
  • \( B \) directly influences how tight or wide the path is; stronger fields lead to tighter circles.

In this problem, \( B \) is given as \( 0.723 \) Tesla. Understanding magnetic field strength is essential to predicting the path and radius of charged particles like ions in magnetic fields.

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Most popular questions from this chapter

In a shunt-wound dc motor with the field coils and rotor connected in parallel (Fig. E27.51), the resistance \(R _ { \text { f of the } }\) field coils is \(106 \Omega ,\) and the resistance \(R _ { r }\) of the rotor is 5.9\(\Omega .\) When a potential difference of 120\(\mathrm { V }\) is applied to the brushes and the motor is running at full speed delivering mechani- cal power, the current supplied to it is 4.82 A. (a) What is the current in the field coils? (b) What is the current in the rotor? (c) What is the induced emf developed by the motor? (d) How much mechanical power is developed by this motor?

The Electromagnetic Pump. Magnetic forces acting on conducting fluids provide a convenient means of pumping these fluids. For example, this method can be used to pump blood without the damage to the cells that can be caused by a mechanical pump. A horizontal tube with rectangular cross section (height \(h ,\) width \(w )\) is placed at right angles to a uniform magnetic field with magnitude \(B\) so that a length \(l\) is in the field (Fig. P27.90). The tube is filled with a conducting liquid, and an electric current of density \(J\) is maintained in the third mutually perpendicular direction. (a) Show that the difference of pressure between a point in the liquid on a vertical plane through \(a b\) and a point in the liquid on another vertical plane through \(c d ,\) under conditions in which the liquid is prevented from flowing, is \(\Delta p = J / B\) . (b) What current density is needed to provide a pressure difference of 1.00 atm between these two points if \(B = 2.20 \mathrm { T }\) and \(l = 35.0 \mathrm { mm } ?\)

A wire 25.0\(\mathrm { cm }\) long lies along the \(z\) -axis and carries a current of 7.40 A in the \(+ z\) -direction. The magnetic field is uniform and has components \(B _ { x } = - 0.242\) T, \(B _ { y } = - 0.985 \mathrm { T }\) , and \(B _ { z } = - 0.336 \mathrm { T }\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the magnetic force on the wire?

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 \(\mathrm{V} / \mathrm{m}\) and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 \(\mathrm{T}\) 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 \(17.5 \mathrm{cm},\) what is their mass?

Force on a Current Loop in a Nonuniform Magnetic Field. It was shown in Section 27.7 that the net force on a current loop in a uniform magnetic field is zero. But what if \(\vec { B }\) is not uniform? Figure P27.85 shows a square loop of wire that lies in the \(x y\) -plane. The loop has corners at \(( 0,0 ) , ( 0 , L ) , ( L , 0 ) ,\) and \(( L , L )\) and carries a constant current \(I\) in the clockwise direction. The magnetic field has no \(x\) -component but has both \(y\) - and z-components: \(\vec { B } = \left( B _ { 0 } z / L \right) \hat { J } + \left( B _ { 0 } y / L \right) \hat { k } ,\) where \(B _ { 0 }\) is a positive constant. (a) Sketch the magnetic field lines in the \(y z\) -plane. (b) Find the magnitude and direction of the magnetic force exerted on each of the sides of the loop by integrating Eq. (27.20). (c) Find the magnitude and direction of the net magnetic force on the loop.
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