**Text Transcription for Educational Website:**A proton, that is accelerated from rest through a potential of 60 kV enters the velocity filter, consisting of a parallel-plate capacitor and a magnetic field, shown below. What is the proton’s speed? Take \( m_p = 1.67 \times 10^{-27} \, \text{kg} \) and \( e = 1.6 \times 10^{-19} \, \text{C} \).**Diagram Explanation:**The diagram displays a velocity filter with a proton (P) entering from the left. The filter includes:- **Parallel-Plate Capacitor:** Two horizontal lines representing the plates. An electric field (E) between them is indicated by downward-pointing vertical arrows labeled 'E'.- **Magnetic Field (B):** Depicted by crosses (×) suggesting the field is directed into the page.**Problem Details:**The speed, \( v = \) [Select an answer]The E-field between the plates is \( 1.3 \times 10^6 \, \text{N/C} \). What B-field is required so that the protons are not deflected?The magnetic field, \( B = \) [Select an answer].**Instructions:**- Calculate the speed of the proton after acceleration.- Determine the magnetic field (B) necessary to prevent proton deflection in the presence of the given electric field.

Given,Mass of the proton, mp= 1.67×10-27 kgcharge on proton, e = 1.6×10-19 Cpotential, V = 60…

A proton, that is accelerated from rest through a potential of 60 kV enters the velocity filter, consisting of a parallel-plate capacitor and a magnetic field, shown below. What is the proton's speed? Take m, = 1.67 x 10-7 kg and e = 1.6 x 10-19 c. B The speed, v = Units Select an answer The E-field between the plates is 1.3 x 10° N/C. What B-field is required so that the protons are not deflected? The magnetic field, B = Units Select an answer

A proton, that is accelerated from rest through a potential of 60 kV enters the velocity filter, consisting of a parallel-plate capacitor and a magnetic field, shown below. What is the proton's speed? Take m, = 1.67 x 10-7 kg and e = 1.6 x 10-19 c. B The speed, v = Units Select an answer The E-field between the plates is 1.3 x 10° N/C. What B-field is required so that the protons are not deflected? The magnetic field, B = Units Select an answer

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A 68.0-cm-diameter cyclotron uses a 450 V oscillating potential difference between the dees. Part A What is the maximum kinetic energy of a proton if the magnetic field strength is 0.740 T? Express your answer with the appropriate units. Value Submit Part B μA Submit Units Request Answer IV—| ΑΣΦ 3 How many revolutions does the proton make before leaving the cyclotron? ? Request Answer ?
In some facilities, real-time MRI is used to monitor a treatment. This imaging allows the treatment to be adjusted to achieve optimum results. However, the devices that deliver the treatment can themselves produce magnetic fields that interfere with the imaging. In one case, a device produces a 65 T static field parallel to the 1.5 T field in the MRI machine Part A By how much does this additional field change the proton's procession frequency? Express your answer with the appropriate units. A/= Submit 309 10 HA x₂ x+10 Hz Previous Answers Request Answe 431 12
Hey I am doing some physics homework and am stumped on this question. Any help will be great! Thanks!
Magnetic and electric field lines are in many ways similar, with few crucial differences. Which of the properties listed below does NOT apply to both fields? O All of these properties apply to both E and B field lines. The density of lines is proportional to the field strength in that area. Field lines never intersect. Tangent to the line at a point gives you the direction of the field as that point. O Field lines form closed loops.
Part D A particle with charge – 5.10 nC is moving in a uniform magnetic field B= -(1.26 T )k. The magnetic force on the particle is measured to be Calculate the scalar product v· F. Assume the velocity has only components found in the previous parts. F = -( 3.10x10-7 N )ê+(7.60x10-7 N )3. Express your answer in meters per second times newtons. Nνα ΑΣφ ? m/s N Submit Request Answer Part E What is the angle between v and F? Give your answer in degrees. Express your answer in degrees. ? = Submit Request Answer Provide Feedback
▸ Part A What value of magnetic field would make a beam of electrons, traveling to the right at a speed of 5.4×100 m/s. go undeflected through a region where there is a uniform electric field of 9400 V/m pointing vertically up? Express your answer to two significant figures and include the appropriate units. B = 1.7x10-3 T Submit ✓ Correct Part B Previous Answers Part C What is the frequency of the circular orbit of the electrons if the electric field is turned off? Express your answer to two significant figures and include the appropriate units. HA f- Value Submit Request Answer Units ?
L. B A positively charged particle is tied by a massless string and allowed to move on a smooth plane, as shown in the picture. A uniform magnetic field B = 1 T is applied perpendicular to the plane pointing down. The length of the string is L = 25 cm. The mass of the particle is m = 35 kg. The charge on this particle is q = 0.65 C. The velocity of the motion is a constant v = observing from above the plane. You can ignore friction. %3D 0.1392 m/s, in the clockwise direction when At a speed of v = 0.1392 m/s, how much is the Lorentz force (i.e. magnetic force) on the particle? (in units of Newton) Submit Answer fries 0/2 To maitain the circular motion at this speed with a radius L, the net force Fnet on the particle is pointing to the center O. How much is Fnet? (in units of Newton) Submit Answer Tries 0/2 It is strongly recommended that you draw a free body diagram to answer the next two parts of the problem. How much is the tension in the string? (in units of Newton) Submit Answer…
b
A charged particle moves through a velocity selector at constant velocity. The velocity selector is configured with "crossed" electric and magnetic fields of magnitude E = 1.50 x 104 N/C and B = 0.4 T. %3D Hint a. What is the velocity of the charged particle? Velocity of the charged particle is x 104 m/s. b. When the electric field is turned off, the charged particle travels in a circular path of radius 6 mm, as it travels through the magnetic field (still at B = 0.4 T). What is the mass-to-charge ratio of the particle? Hint for (b) Mass-to-charge ratio of the part is kg/C. (Use the "E" notation to enter your answer in scientific notation. For example, to enter 3.14 × 10´ 12".) - 12 enter "3.14E-
A positively charged particle with charge +Q passes straight through a velocity selector when its velocity is v = 2 × 10* m/s. What must be changed in order for this velocity selector to allow a particle that has a charge –2Q to pass through undeflected with this same speed? O Nothing needs to be changed The direction of the electric field (or magnetic field) It is impossible to allow this negatively charged particle to pass through undeflected The direction of both the electric and the magnetic field The strength of the electric field (or magnetic field)
e In an experiment of measure by m Thomson's method, electron travelling at a speed of 2.25 x 107 m/s are made to enter a traverse magnetic field of induction 2.32 × 10-3 Wb/m². If the radius of circular path traced out by the electron beam is 5.5 cm, find the value e of m
A 50 g horizontal metal bar, 15-cm-long, is free to slide up and down between two tall, vertical metal rods that are 15 cm apart. A 5.5×10-2 T magnetic field is directed perpendicular to the plane of the rods. The bar is raised to near the top of the rods, and a 1.2 resistor is connected across the two rods at the top. Then the bar is dropped. Part A What is the terminal speed at which the bar falls? Assume the bar remains horizontal and in contact with the rods at all times. Express your answer in meters per second. 跖 ΑΣΦ ? v = Submit Request Answer m/s