
Concept explainers
(a)
The sign of the charge of the ions.
(a)

Answer to Problem 94P
The sign of the charge of the ions is
Explanation of Solution
The right hand rule states that in the case of a cross product, if the index finger of the right hand points in the direction of the first vector and the second vector points in the direction of the thumb, then the cross product will point in the direction of the thumb. Magnetic Lorentz force on a moving charge is the product of the charge and the cross product of the velocity of the charge and the magnetic field.
Right hand rule for the Lorentz force can be described as if the index finger of the right hand points in the direction of velocity of charge, the middle finger in the direction of the magnetic field, then the thumb will point in the direction of the magnetic force. In the given diagram, the velocity is directed to right, the magnetic field is directed into the page and the force is directed upward. This implies the ion is positive.
(b)
The accelerating plate which is positively charged.
(b)

Answer to Problem 94P
The positively charged accelerating plate is
Explanation of Solution
In part (a), it is found that the ions are positively charged. Positive ions will be attracted by negatively charged plates and will be repelled by positively charged plates. In the given diagram, the ions are accelerated towards the right.
Since the ions are accelerated towards right, the east plate must be negatively charged. This implies the other accelerating plate must be positively charged. Thus the west plate is positively charged.
(c)
The deflection plate which is positively charged.
(c)

Answer to Problem 94P
The positively charged deflection plate is
Explanation of Solution
Right hand rule for the Lorentz force can be described as if the index finger of the right hand points in the direction of velocity of charge, the middle finger in the direction of the magnetic field, then the thumb will point in the direction of the magnetic force. In a mass spectrometer, the correct velocity is selected by making the net force which is the sum of electric and magnetic force on the charge to be zero.
According to the right hand rule, the magnetic force on the positively charged ions is northward between the deflecting plates. To make the net force zero to select the correct velocity, the electric field must be southward. Since electric field points from positive plate to the negative plate, the north plate must be positively charged.
(d)
The correct values of
(d)

Answer to Problem 94P
The value of
Explanation of Solution
Write the equation for the magnetic Lorentz force.
Here,
The angle between the magnetic field and the velocity is
Substitute
The magnetic Lorentz force provides the centripetal acceleration required for the semi-circular motion of the ions.
Write the equation for the centripetal force.
Here,
Equate equations (I) and (II) and rewrite the equation for
Write the equation for a velocity selector.
Here,
Put equation (III) in the above equation.
Write the equation for the potential difference across the deflecting plates.
Here,
Put equation (IV) in the above equation.
Apply the principle of conservation of energy for the given situation.
Here,
Rewrite the above equation for
Put equation (III) in the above equation.
Conclusion:
Given that the radius of the path is
Substitute
Substitute
Therefore, the value of
(e)
The correct values of
(e)

Answer to Problem 94P
The value of
Explanation of Solution
Equation (VI) and (V) respectively can be used to determine
Conclusion:
Given that the mass of the 14C+ ions is
Substitute
Substitute
Therefore, the value of
Want to see more full solutions like this?
Chapter 19 Solutions
Physics
- please answer this asap!!!!arrow_forwardRT = 4.7E-30 18V IT = 2.3E-3A+ 12 38Ω ли 56Ω ли r5 27Ω ли r3 28Ω r4 > 75Ω r6 600 0.343V 75.8A Now figure out how much current in going through the r4 resistor. |4 = unit And then use that current to find the voltage drop across the r resistor. V4 = unitarrow_forward7 Find the volume inside the cone z² = x²+y², above the (x, y) plane, and between the spheres x²+y²+z² = 1 and x² + y²+z² = 4. Hint: use spherical polar coordinates.arrow_forward
- ганм Two long, straight wires are oriented perpendicular to the page, as shown in the figure(Figure 1). The current in one wire is I₁ = 3.0 A, pointing into the page, and the current in the other wire is 12 4.0 A, pointing out of the page. = Find the magnitude and direction of the net magnetic field at point P. Express your answer using two significant figures. VO ΜΕ ΑΣΦ ? Figure P 5.0 cm 5.0 cm ₁ = 3.0 A 12 = 4.0 A B: μΤ You have already submitted this answer. Enter a new answer. No credit lost. Try again. Submit Previous Answers Request Answer 1 of 1 Part B X Express your answer using two significant figures. ΜΕ ΑΣΦ 0 = 0 ? below the dashed line to the right P You have already submitted this answer. Enter a new answer. No credit lost. Try again.arrow_forwardAn infinitely long conducting cylindrical rod with a positive charge λ per unit length is surrounded by a conducting cylindrical shell (which is also infinitely long) with a charge per unit length of −2λ and radius r1, as shown in the figure. What is σinner, the surface charge density (charge per unit area) on the inner surface of the conducting shell? What is σouter, the surface charge density on the outside of the conducting shell? (Recall from the problem statement that the conducting shell has a total charge per unit length given by −2λ.)arrow_forwardA small conducting spherical shell with inner radius aa and outer radius b is concentric with a larger conducting spherical shell with inner radius c and outer radius d (Figure 1). The inner shell has total charge +2q, and the outer shell has charge −2q. What's the total charge on the inner surface of the small shell? What's the total charge on the outer surface of the small shell? What's the total charge on the inner surface of the large shell? What's the total charge on the outer surface of the large shell?arrow_forward
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSONIntroduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningLecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-WesleyCollege Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON





