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Chapter 22: Q90PE (page 816)

Consider a mass separator that applies a magnetic field perpendicular to the velocity of ions and separates the ions based on the radius of curvature of their paths in the field. Construct a problem in which you calculate the magnetic field strength needed to separate two ions that differ in mass, but not charge, and have the same initial velocity. Among the things to consider are the types of ions, the velocities they can be given before entering the magnetic field, and a reasonable value for the radius of curvature of the paths they follow. In addition, calculate the separation distance between the ions at the point where they are detected.

Short Answer

Expert verified
  1. The magnetic field strength is \({\rm{0}}{\rm{.4}}\;{\rm{T}}\).
  2. The curvature radius is\({\rm{0}}{\rm{.37}}\;{\rm{m}}\).
  3. The mass is\({\rm{5}}{\rm{.92}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}\;{\rm{kg}}\).
  4. The separation distance is \({\rm{0}}{\rm{.2}}\;{\rm{m}}\).

Step by step solution

01

Define Magnetic field

Magnetic field lines are never crossed. The strength of the field is determined by the density of the field lines.

02

Constructing a problem

In a mass spectrometer, single-charged chloride\({\rm{Cl - }}\)and oxide \({\rm{O - }}\)ions are separated. The mass ratio of these ions is\({\rm{27}}\)to\({\rm{59}}\), with oxide having a mass of\({\rm{2}}{\rm{.7 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{kg}}\)Both ions travel at a speed of\({\rm{4}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}\;{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}\).

(a) Determine the magnetic field strength required to keep\({\rm{O - }}\)in a circular path with a radius of\({\rm{0}}{\rm{.17}}\)m.

(b) What is the curvature radius that\({\rm{Cl - }}\)follows?

(c) Calculate the mass of\({\rm{Cl - }}\)using the findings from (a) and (b).

(d) If the ions are identified after crossing a semicircle, what is the separation distance between them?

03

Solution- Evaluating the magnetic field strength

(a)

Solving for\({\rm{B}}\)by using the equation,

\(\begin{aligned}{c}{\rm{r}} &= \frac{{{\rm{mv}}}}{{{\rm{qB}}}}\\{\rm{B}} &= \frac{{{\rm{mv}}}}{{{\rm{qr}}}}\\{\rm{B}} &= \frac{{\left( {{\rm{2}}{\rm{.7}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}\;{\rm{kg}}} \right) \times \left( {{\rm{4}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}\;{m \mathord{\left/

{\vphantom {m s}} \right.

\kern-\nulldelimiterspace} s}} \right)}}{{\left( {{\rm{1}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}} \right) \times \left( {{\rm{0}}{\rm{.17}}\;{\rm{m}}} \right)}}\\{\rm{B}} &= {\rm{0}}{\rm{.4}}\;{\rm{T}}\end{aligned}\)

Therefore, the magnetic field strength is \({\rm{0}}{\rm{.4}}\;{\rm{T}}\).

04

Solution- Evaluating the curvature radius

(b)

Since the radius is directly proportional to mass, the radius of curvature is obtained by considering the ratio of ions masses.

\(\begin{aligned}{l}\frac{{{{\rm{r}}_{{\rm{O - }}}}}}{{{{\rm{r}}_{{\rm{Cl - }}}}}} &= \frac{{{{\rm{m}}_{{\rm{O - }}}}}}{{{{\rm{m}}_{{\rm{Cl - }}}}}}\\\frac{{{\rm{0}}{\rm{.17}}\;{\rm{m}}}}{{{{\rm{r}}_{{\rm{Cl - }}}}}} &= \frac{{{\rm{27}}}}{{{\rm{59}}}}\\{{\rm{r}}_{{\rm{Cl - }}}} &= {\rm{0}}{\rm{.17}}\;{\rm{m}} \times \left( {\frac{{{\rm{59}}}}{{{\rm{27}}}}} \right)\\{{\rm{r}}_{{\rm{Cl - }}}} &= {\rm{0}}{\rm{.37}}\;{\rm{m}}\end{aligned}\)

Therefore, the curvature radius is \({\rm{0}}{\rm{.37}}\;{\rm{m}}\).

05

Solution- Evaluating the mass

(c)

Solving for\({{\rm{m}}_{{\rm{Cl - }}}}\)and filling in the numbers using the findings from (a) and (b) provides

\(\begin{aligned}{c}{{\rm{r}}_{{\rm{Cl - }}}} &= \frac{{{{\rm{m}}_{{\rm{Cl - }}}}{\rm{v}}}}{{{\rm{qB}}}}\\{{\rm{m}}_{{\rm{Cl - }}}} &= \frac{{{{\rm{r}}_{{\rm{Cl - }}}}{\rm{qB}}}}{{\rm{v}}}\\ &= \frac{{\left( {{\rm{0}}{\rm{.37}}\;{\rm{m}}} \right) \times \left( {{\rm{1}}{\rm{.6 \times 1}}{{\rm{0}}^{{\rm{ - 19}}}}\;{\rm{C}}} \right) \times \left( {{\rm{0}}{\rm{.4}}\;{\rm{T}}} \right)}}{{{\rm{4}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}\;{m \mathord{\left/

{\vphantom {m s}} \right.

\kern-\nulldelimiterspace} s}}}\\ &= {\rm{5}}{\rm{.92}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}\;{\rm{kg}}\end{aligned}\)

Therefore, the mass is \({\rm{5}}{\rm{.92}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 27}}}}\;{\rm{kg}}\).

06

Solution- Evaluating the separation distance

(d)

When the ions are discovered, the separation distance\({\rm{d}}\)between them is just the difference in their radii.

\(\begin{aligned}{c}{\rm{d}} &= {{\rm{r}}_{{\rm{Cl - }}}} - {{\rm{r}}_{{\rm{O - }}}}\\ &= {\rm{0}}{\rm{.37}}\;{\rm{m}} - {\rm{0}}{\rm{.17}}\;{\rm{m}}\\ &= {\rm{0}}{\rm{.2}}\;{\rm{m}}\end{aligned}\)

Therefore, the separation distance is\({\rm{0}}{\rm{.2}}\;{\rm{m}}\).

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Most popular questions from this chapter

(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than1.00脳10鈥�12N. What is the greatest the charge can be if it moves at a maximum speed of30.0m/sin the Earth鈥檚 field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent.

The force on the rectangular loop of wire in the magnetic field in Figure 22.56 can be used to measure field strength. The field is uniform, and the plane of the loop is perpendicular to the field. (a) What is the direction of the magnetic force on the loop? Justify the claim that the forces on the sides of the loop are equal and opposite, independent of how much of the loop is in the field and do not affect the net force on the loop. (b) If a current of 5.00 A is used, what is the force per tesla on the 20.0-cm-wide loop?


Figure 22.56 A rectangular loop of wire carrying a current is perpendicular to a magnetic field. The field is uniform in the region shown and is zero outside that region.

Suppose two long straight wires run perpendicular to one another without touching. Does one exert a net force on the other? If so, what is its direction? Does one exert a net torque on the other? If so, what is its direction? Justify your responses by using the right hand rules.

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