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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?

Short Answer

Expert verified
(a) The ions move at approximately 4920.63 m/s. (b) Their mass is about \( 9.76 \times 10^{-26} \text{ kg} \).

Step by step solution

01

Understand the Problem - Velocity Selector

The velocity selector allows charged particles to pass through if the electric force equals the magnetic force exerted on them. Given the electric field, \( E = 155 \, \text{V/m} \), and magnetic field, \( B_1 = 0.0315 \, \text{T} \), we can find the velocity of ions.
02

Calculate Ion Velocity

For ions to pass through without deflection in a velocity selector, the electric force \( qE \) must equal the magnetic force \( qvB_1 \). Therefore, \( E = vB_1 \). Substituting the given values, solve for velocity: \[ v = \frac{E}{B_1} = \frac{155}{0.0315} \, \text{m/s}. \] Calculating, we get the velocity \( v \approx 4920.63 \, \text{m/s} \).
03

Understand the Problem - Circular Motion in Magnetic Field

After passing through the velocity selector, ions enter a magnetic field \( B_2 = 0.0175 \, \text{T} \) perpendicular to their velocity, causing them to move in a circular path with radius \( r = 0.175 \, \text{m} \). We need to find the ion's mass.
04

Apply Circular Motion Formula

In a magnetic field, a charged particle moving in a circle experiences a centering force equal to the magnetic force. The force is given by \( qvB_2 = \frac{mv^2}{r} \), where \( m \) is the mass. Solving for mass, we have \[ m = \frac{qB_2r}{v}. \] Substituting known quantities using \( q = e = 1.6 \times 10^{-19} \text{C} \), \( v = 4920.63 \, \text{m/s} \), \( B_2 = 0.0175 \, \text{T} \), and \( r = 0.175 \, \text{m} \).
05

Calculate the Mass

Using the formula, \[ m = \frac{(1.6 \times 10^{-19} \, \text{C})(0.0175 \, \text{T})(0.175 \, \text{m})}{4920.63 \, \text{m/s}}. \] Calculating gives \( m \approx 9.76 \times 10^{-26} \, \text{kg} \).

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Key Concepts

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

Electric Field
The concept of an electric field is essential in understanding how charged particles, like ions, interact with each other and with other forces. An electric field is a region of space around a charged particle where forces are exerted on other charged particles. It is measured in volts per meter (V/m).

In the context of a velocity selector, the electric field (\(E\)) is crucial because it exerts an electric force on the ions moving through it. This force can be calculated using the formula:
  • \( F_E = qE \)
where \(q\) is the charge of the ion. The velocity selector uses this field to ensure that only ions with a specific velocity will pass through without being deflected.

The electric field works in conjunction with a magnetic field to apply forces on the ions. Thus, understanding this field and its interaction with charged particles helps in controlling and directing them in various applications, such as in particle accelerators and mass spectrometers.
Magnetic Field
Magnetic fields play a crucial role in controlling the motion of charged particles. In our example, we have two different magnetic fields. The first magnetic field (\(B_1\)) works with the electric field in the velocity selector, while the second (\(B_2\)) affects the motion after the ions pass through the selector.

A magnetic field can exert a force on a moving charge, calculated using the formula:
  • \( F_B = qvB \)
where \(v\) is the velocity of the ion and \(B\) is the magnetic field strength. In the velocity selector, the magnetic field's role is to balance the force from the electric field.

The second magnetic field, \(B_2\), causes the ions to move in a circular path. This is because the magnetic force acts as a centripetal force that bends the ion's path. The magnetic field thus becomes an essential tool for studying the properties of charged particles by manipulating their paths, allowing scientists to measure properties such as mass and energy.
Circular Motion
After leaving the velocity selector, ions enter the second magnetic field that causes them to move in a circle. This motion results from the magnetic force acting as a centripetal force, steering the ions into a circular path.

Circular motion occurs when the expression for the magnetic force, \( qvB \), equals the centripetal force required to keep the ion in a circular path, \( \frac{mv^2}{r} \). Thus, we have:
  • \( qvB = \frac{mv^2}{r} \)
where \(m\) is the ion's mass and \(r\) is the radius of the circle.

Understanding circular motion in this context helps in designing instruments like cyclotrons and in calculating unknown properties of ions, such as mass or charge. It highlights how forces combine to produce motion and makes it possible to analyze the paths and behaviors of particles.
Ion Mass Calculation
Calculating the mass of an ion is essential in fields like chemistry and physics. In this scenario, the velocity of ions and their circular motion in a magnetic field provides the framework to determine their mass.

The mass can be calculated by re-arranging the equation used for circular motion:
  • \( m = \frac{qB_2r}{v} \)
Inserting known values such as the charge \(q\) (typically the elementary charge \(1.6 \times 10^{-19}\) C), the magnetic field \(B_2\), the radius \(r\), and the velocity \(v\), one can calculate the ion's mass.

This calculation is significant because it provides a method to measure unknown mass quantities of ions. Such precise measurements are critical in research and development in fields such as biology, materials science, and astrophysics. Accurately determining the mass of ions allows for better understanding and analysis of various elements or compounds.

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Most popular questions from this chapter

A particle with charge -5.60 nC is moving in a uniform magnetic field \(\overrightarrow{B} =\) -(1.25 T)\(\hat{k}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) -(3.40 \(\times\) 10\(^{-7}\)N)\(\hat{\imath}\) + (7.40 \(\times\) 10\(^{-7}\)N)\(\hat{\jmath}\). (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{v}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\vec{v}\) and \(\overrightarrow{F}\)?

A circular area with a radius of 6.50 cm lies in the \(xy\)-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field \(B =\) 0.230 T (a) in the \(+z\)-direction; (b) at an angle of 53.1\(^\circ\) from the \(+z\)-direction; (c) in the \(+y\)-direction?

\(\textbf{Determining Diet.}\) 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. Overreliance 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 \(^{12}\)C and \(^{13}\)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 km /s, and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0 cm for the \(^{12}\)C. The measured masses of these isotopes are 1.99 \(\times\) 10\(^{-26}\) kg (\(^{12}\)C) and 2.16 \(\times\) 10\(^{-26}\) kg (\(^{13}\)C). (a) What strength of magnetic field is required? (b) What is the diameter of the \(^{13}\)C semicircle? (c) What is the separation of the \(^{12}\)C and \(^{13}\)C ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64 \(\times\) 10\(^{-27}\) kg) traveling horizontally at 35.6 km>s enters a uniform, vertical, 1.80-T magnetic field. (a) What is the diameter of the path followed by this alpha particle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it.

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the \(+x\)-direction experiences a force of 2.25 \(\times\) 10\(^{-16}\) N in the \(+y\)-direction, and an electron moving at 4.75 km/s in the \(-z\)-direction experiences a force of 8.50 \(\times\) 10-16 N in the \(+y\)-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\)-direction at 3.20 km/s?

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