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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 difference 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 approximately 0.246 m.

Step by step solution

01

Determine the ion's velocity after acceleration

The ion is accelerated through a potential difference, gaining kinetic energy. The equation to find the velocity \( v \) is derived from equating the kinetic energy \( \frac{1}{2}mv^2 \) to the electric potential energy \( qV \). For a singly charged lithium ion \( q = 1.6 \times 10^{-19} \mathrm{C} \), so:\[ \frac{1}{2} mv^2 = qV \]Rearranging for \( v \):\[ v = \sqrt{\frac{2qV}{m}} \]Substitute the given values:\[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 220}{1.16 \times 10^{-26}}} \approx 2.46 \times 10^5 \; \mathrm{m/s} \]
02

Calculate the radius of the ion's path using the velocity

Once we have the velocity, use the formula for the radius \( r \) of a charged particle's path in a magnetic field, which is given by:\[ r = \frac{mv}{qB} \]Where \( m \) is the mass of the ion, \( v \) is its velocity, \( q \) is the charge of the ion, and \( B \) is the magnetic field strength. Substituting in our known values:\[ r = \frac{1.16 \times 10^{-26} \times 2.46 \times 10^5}{1.6 \times 10^{-19} \times 0.723} \approx 0.246 \; \mathrm{m} \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is a fundamental concept in physics, describing the energy that an object possesses due to its motion. The formula for kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. When a particle like an ion is accelerated, it gains kinetic energy. This is exactly what happens in the situation described: the ion moves faster as it gains energy from the potential difference it moves through.
  • The kinetic energy depends directly on the mass of the particle and the square of its velocity.
  • Kinetic energy is always a positive value as mass and the square of velocity are positive.
  • This energy transformation is key to determining how fast an ion will move, especially when starting from rest.
Understanding kinetic energy helps explain phenomena in particle physics and is critical when calculating speeds and trajectories of charged particles in electric and magnetic fields.
Electric Potential Energy
Electric potential energy is the energy a charged particle has due to its position in an electric field. It can be thought of as stored energy that has the capacity to do work as forces act upon the charged particle.When a charged particle moves through an electric field, such as being accelerated by a potential difference, it experiences a change in electric potential energy. For a charged particle like a singly charged ion, this change can be calculated using the formula: \( U = qV \), where \( q \) is the charge of the particle and \( V \) is the potential difference.
  • As the particle accelerates, this potential energy is converted into kinetic energy.
  • The potential difference of 220 V in the exercise gives the lithium ion the energy needed to reach a certain velocity.
  • Essentially, the work done on the ion by the electric field translates entirely into kinetic energy in this ideal scenario.
This concept allows us to predict and understand how charged ions like Lithium, when accelerated, will move and change speed.
Singly Charged Ion
A singly charged ion is an atom or molecule that has lost or gained a single electron, resulting in a net positive or negative charge. For the case of the lithium ion \( \text{Li}^+ \) in the exercise, it has one extra positive charge due to losing one electron.Understanding single charge ions is crucial when dealing with their motion in electric and magnetic fields. These particles behave in predictable ways because:
  • The charge \( q \), in this case, is \( 1.6 \times 10^{-19} \; \mathrm{C} \), which is the fundamental charge of an electron or proton.
  • These ions can be influenced by potential differences and fields, changing speed and direction.
  • In a magnetic field, their paths become circular, which can be described using the radius derived from the forces acting on them.
This simplification to a single charge allows for precise calculations, such as determining the trajectory radius in a magnetic field for the singly charged lithium ion.

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Most popular questions from this chapter

A particle of mass 0.195 g carries a charge of \(-2.50 \times\) \(10^{-8} \mathrm{C}\) . The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

Determining Diet. One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon- 12 . Overneliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the \(^{10} \mathrm{C}\) and \(^{13} \mathrm{C}\) isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed 8.50 \(\mathrm{km} / \mathrm{s}\) , and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0 \(\mathrm{cm}\) for the 12 \(\mathrm{C}\) . The measured masses of these isotopes are \(1.99 \times 10^{-25} \mathrm{kg}\left(^{12} \mathrm{C}\right)\) and \(2.16 \times 10^{-26} \mathrm{kg}\left(^{13} \mathrm{C}\right) .\) (a) What strength of magnetic field is required? (b) What is the diameter of the \(^{13} \mathrm{C}\) semicircle? (c) What is the separation of the \(^{12}\mathrm{C}\) and \(^{13}\mathrm{C}\) ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

An electron in the beam of a TV picture tube is accelerated by a potential difference of 2.00 \(\mathrm{kV}\) . Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius 0.180 \(\mathrm{m}\) . What is the magnitude of the field?

The magnetic flux through one face of a cube is \(+0.120 \mathrm{Wb} .\) (a) What must the total magnetic flux through the other five faces of the cube be? (b) Why didn't you need to know the dimensions of the cube in order to answer part (a) 2 (c) Suppose the magnetic flux is due to a permanent magnet like that shown in Fig. \(27.11 .\) In a sketch, show where the cube in part (a) might be located relative to the magnet.

An Electromagnetic Rail Gun. A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\overrightarrow{\boldsymbol{B}}\) fills the region between the rails (Fig. 27.63\()\) . (a) Find the magnitude and direction of the net force on the conducting bar Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m\) , find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\) . (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{km} / \mathrm{s}) .\) Let \(B=0.50 \mathrm{T}, \quad I=2.0 \times 10^{3} \mathrm{A}\) \(m=25 \mathrm{kg},\) and \(L=50 \mathrm{cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts \((a)\) and \((b)\) , even though gravity plays an important role in an actual launch in space.
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