/*! 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 13 ssm In the operating room, anest... [FREE SOLUTION] | 91Ó°ÊÓ

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ssm In the operating room, anesthesiologists use mass spectrometers to monitor the respiratory gases of patients undergoing surgery. One gas that is often monitored is the anesthetic isoflurane (molecular mass \(=3.06 \times 10^{-25} \mathrm{kg} ) .\) In a spectrometer, a singly ionized molecule of isoflurane (charge \(=+e )\) moves at a speed of \(7.2 \times 10^{3} \mathrm{m} / \mathrm{s}\) /s on a circular path that hat has a radius of 0.10 \(\mathrm{m}\) . What is the magnitude of the magnetic field that the spectrometer uses?

Short Answer

Expert verified
The magnetic field is approximately 0.1377 T.

Step by step solution

01

Understand the problem

We need to find the magnitude of the magnetic field that causes a singly ionized isoflurane molecule to move in a circular path within a mass spectrometer.
02

Relevant Formula

The formula that relates the charged particle's motion, magnetic field, and radius of the path is given by the centripetal force equation:\[ F = qvB = \frac{mv^2}{r}, \]where:- \( q \) is the charge of the particle,- \( v \) is the velocity of the particle,- \( B \) is the magnetic field,- \( m \) is the mass of the particle,- \( r \) is the radius of the circular path.
03

Solve the equation for B

Rearrange the formula to solve for the magnetic field \( B \):\[ qvB = \frac{mv^2}{r} \Rightarrow B = \frac{mv}{qr}. \]
04

Substitute given values

Substitute the values given in the problem into the equation:- Charge \( q = e = 1.6 \times 10^{-19} \text{ C} \),- Mass \( m = 3.06 \times 10^{-25} \text{ kg} \),- Velocity \( v = 7.2 \times 10^{3} \text{ m/s} \),- Radius \( r = 0.10 \text{ m} \).
05

Calculate the magnetic field B

Now substitute these values into the equation:\[ B = \frac{3.06 \times 10^{-25} \times 7.2 \times 10^{3}}{1.6 \times 10^{-19} \times 0.10} \] Calculate step-by-step:1. Compute the numerator: \( 3.06 \times 10^{-25} \times 7.2 \times 10^{3} = 2.2032 \times 10^{-21} \).2. Compute the denominator: \( 1.6 \times 10^{-19} \times 0.10 = 1.6 \times 10^{-20} \).3. Divide to find \( B \):\[ B = \frac{2.2032 \times 10^{-21}}{1.6 \times 10^{-20}} = 0.1377 \text{ T} \].
06

Conclusion

The magnitude of the magnetic field required is approximately \( 0.1377 \text{ T} \).

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Key Concepts

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

Understanding Anesthetic Isoflurane in a Mass Spectrometer
Isoflurane is a commonly used anesthetic during surgeries, particularly due to its properties as a halogenated ether. In the context of a mass spectrometer, isoflurane's molecular structure and mass become crucial. Its molecular mass is approximately \(3.06 \times 10^{-25} \text{ kg}\). This value is important when analyzing how the gas behaves within the spectrometer.
In mass spectrometry, gases like isoflurane are ionized—meaning they gain or lose electrons and carry charge. A singly ionized isoflurane molecule has a charge equivalent to the elementary charge, denoted as \(+e\). Once ionized, these molecules can be manipulated using magnetic fields, allowing scientists to monitor various physical properties.
A spectrometer takes advantage of the charged state of isoflurane to create visible paths that aid in identifying the gas components. Monitoring such gases ensures safe levels of anesthetics during surgeries.
  • Isoflurane's chemical properties make it a suitable candidate for mass spectrometry analysis.
  • Understanding the mass and charge is essential for accurate calculations in a spectrometer.
  • Ensures safety and efficacy when administering anesthesia.
Unveiling the Role of Centripetal Force in Circular Motion
Centripetal force is a key concept in understanding how a particle moves in a circular path. In a mass spectrometer, this force is what keeps a charged particle, like an ionized gas molecule, on its curved trajectory.
The centripetal force is given by the equation \( F = \frac{mv^2}{r} \), where:
  • \( m \) is the mass of the particle
  • \( v \) is the velocity of the particle
  • \( r \) is the radius of the circular path

This equation tells us that a faster velocity or a smaller radius requires a greater centripetal force.
In a mass spectrometer, the centripetal force is provided by the magnetic field, such that \( F = qvB \). This demonstrates that the strength of the magnetic field, the particle's velocity, and its charge directly influence the path.
Thus, centripetal force ensures that the path a particle takes is stable and predictable. Its calculation is a critical step in configuring and using a mass spectrometer effectively.
Magnetic Field Calculation in a Mass Spectrometer
The goal of a mass spectrometer is to measure various properties of a molecule by using magnetic fields to affect its path. For isoflurane, we need to calculate the magnetic field required to keep a singly ionized molecule moving in a circular path of known radius and speed.
The relevant rearranged formula to find the magnetic field \( B \) is:\[ B = \frac{mv}{qr}\]
Where:
  • \( m \) is the mass of the particle – \(3.06 \times 10^{-25} \text{ kg}\)
  • \( v \) is the velocity – \(7.2 \times 10^{3} \text{ m/s}\)
  • \( q \) is the charge of the particle – \(1.6 \times 10^{-19} \text{ C}\)
  • \( r \) is the radius of the path – \(0.10 \text{ m}\)

By substituting these values into the formula, the calculation becomes manageable:\[ B = \frac{3.06 \times 10^{-25} \times 7.2 \times 10^{3}}{1.6 \times 10^{-19} \times 0.10} = 0.1377 \text{ T}\]
This result shows the magnetic field strength needed to keep the molecule on the desired path, highlighting the precision required in scientific instruments.

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Most popular questions from this chapter

A square coil and a rectangular coil are each made from the same length of wire. Each contains a single turn. The long sides of the rectangle are twice as long as the short sides. Find the ratio \(\tau_{\text { square }} / \tau_{\text { rectangle }}\) of the maximum torques that these coils experience in the same magnetic field when they contain the same current.

A magnetic field has a magnitude of \(1.2 \times 10^{-3} \mathrm{T}\) , and an electric field has a magnitude of \(4.6 \times 10^{3} \mathrm{N} / \mathrm{C}\) . Both fields point in the same direction. A positive 1.8\(\mu \mathrm{C}\) charge moves at a speed of \(3.1 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a direction that is perpendicular to both fields. Determine the magnitude of the net force that acts on the charge.

ssm A long, cylindrical conductor is solid throughout and has a radius \(R\) . Electric charges flow parallel to the axis of the cylinder and pass uniformly through the entire cross section. The arrangement is, in effect, a solid tube of current \(I_{0} .\) The current per unit cross-sectional area (i.e., the current density ) is \(I_{0} /\left(\pi R^{2}\right) .\) Use Ampere's law to show that the magnetic field inside the conductor at a distance \(r\) from the axis is \(\mu_{0} I_{0} r /\left(2 \pi R^{2}\right)\) . Hint: For a closed path, use \(a\) circle of radius \(r\) perpendicular to and centered on the axis. Note that the current through any surface is the area of the surface times the current density.)

ssm An electron is moving through a magnetic field whose magniude is \(8.70 \times 10^{-4} \mathrm{T}\) . The clectron expericnces only a magnetic force and has an acceleration of magnitude \(3.50 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}\) . At a certain instant, it has a speed of \(6.80 \times 10^{6} \mathrm{m} / \mathrm{s}\) . Determine the angle \(\theta\) (less than \(90^{\circ}\) ) between the electron's velocity and the magnetic field.

ssm The electrons in the beam of a television tube have a kinetic energy of \(2.40 \times 10^{-15} \mathrm{J}\) . Initially, the clectrons move horizontally from west to east. The vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of \(2.00 \times 10^{-5} \mathrm{T} .\) (a) In what direction are the electrons deflected by this field component? (b) What is the acceleration of an electron in part (a)?

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