A proton with a speed of 2.7 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.15 m. As the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field, so the proton just misses colliding with the opposite plate?

A proton with a speed of 2.7 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.15 m. As the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field, so the proton just misses colliding with the opposite plate?

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter30: Magnetic Fields And Forces

Section: Chapter Questions

Problem 12PQ: Review A proton is accelerated from rest through a 5.00-V potential difference. a. What is the...

See similar textbooks

Similar questions

An alpha-particle ( m=6.641027kg , q=3.21019C ) travels in a circular path of radius 25 cm in a uniform magnetic field of magnitude 1.5 T. (a) What is the speed of the particle? (b) What is the kinetic energy in electron-volts? (c) Through what potential difference must the particle be accelerated in order to give it this kinetic energy?
An electron is accelerated through 2.40 103 V from rest and then enters a uniform 1.70-T magnetic field. What are (a) the maximum and (b) the minimum values of the magnetic force this particle experiences?
A singly charged ion of mass m is accelerated from rest by a potential difference V. It is then deflected by a uniform magnetic field (perpendicular to the ions velocity) into a semicircle of radius R. Now a doubly charged ion of mass m' is accelerated through the same potential difference and dellected by the same magnetic field into a semicircle of radius R' = 2R. What is the ratio of the masses of the ions?
Unreasonable Results A charged particle having mass 6.641027kg (that of a helium atom) moving at 8.70105m/s perpendicular to a 1.50T magnetic field travels in a circular path of radius 16.0 mm. (a) What is the charge of the particle? (b) What is unreasonable about this result? (c) Which assumptions are responsible?
A laboratory electromagnet produces a magnetic field of magnitude 1.50 T. A proton moves through this field with a speed of 6.00 106 m/s. (a) Find the magnitude of the maximum magnetic force that could be exerted on the proton. (b) What is the magnitude of the maximum acceleration of the proton? (c) Would the field exert the same magnetic force on an electron moving through the field with the same speed? (d) Would the electron undergo the same acceleration? Explain.
A particle of charge q and mass m is accelerated from rest through a potential difference V, after silica it encounters a uniform magnetic field B. If the particle moves in a plane perpendicular to B, shaft is the radius of its circular orbit?
, A proton, deuteron, and an alpha-particle ae all accelerated from rest through the same potential difference. They then enter the same magnetic field, moving perpendicular to it. Compute the ratios of the radii of their circular paths. Assume that md= 2wmp and ma= 4mp.
Electrons in Earths upper atmosphere have typical speeds near 6.00 105 m/s. (a) Calculate the magnitude of Earths magnetic field if an electrons velocity is perpendicular to the magnetic field and its circular path has a radius of 7.00 102 m. (b) Calculate the number of times per second that an electron circles around a magnetic field line.
At a particular instant an electron is traveling west to east with a kinetic energy of 10 keV. Earth's magnetic field has a horizontal component of 1.8105 T north and a vertical component of 5.0105 T down. (a) What is the path of the election? (b) What is the radius of curvature of the path?
A particle’s path is bent when it passes through a region of non-zero magnetic field although its speed remains unchanged. This is very useful for “beam steering’’ in particle accelerators. Consider a proton of speed 4106m/s entering a region of uniform magnetic field 0.2 T over a 5-cm-wide region. Magnetic field is perpendicular to the velocity of the particle. By how much angle will the path of the proton be bent? (Hint: the particle comes out tangent to a circle.
Why is the following situation impossible? Figure P28.46 shows an experimental technique for altering the direction of travel for a charged particle. A particle of charge q = 1.00 C and mass m = 2.00 1015 kg enters the bottom of the region of uniform magnetic field at speed = 2.00 105 m/s, with a velocity vector perpendicular to the field lines. The magnetic force on the particle causes its direction of travel to change so that it leaves the region of the magnetic field at the top traveling at an angle from its original direction. The magnetic field has magnitude B = 0.400 T and is directed out of the page. The length h of the magnetic field region is 0.110 m. An experimenter performs the technique and measures the angle at which the particles exit the top of the field. She finds that the angles of deviation are exactly as predicted. Figure P28.46
Review A proton is accelerated from rest through a 5.00-V potential difference. a. What is the protons speed after it has been accelerated? b. What is the maximum magnetic field that this proton produces at a point that is 1.00 m from the proton?