Protons having a kinetic energy of 2.90 MeV are moving in the positive x-direction and enter a magnetic field of 0.0370 T in the z-direction, out of the plane of the page, and extending from x = X = 1.00 m as in the figure below. x= 1.00 m (a) Calculate the y-component of the protons' momentum as they leave the magnetic field. (Indicate the direction with the sign of your answer.) kg · m/s (b) Find the angle a between the initial velocity vector of the proton beam and the velocity vector after the beam emerges from the field. Hint: Neglect relativistic effects and note that 1 ev = 1.60 x 10-19 ). a =

Protons having a kinetic energy of 2.90 MeV are moving in the positive x-direction and enter a magnetic field of 0.0370 T in the z-direction, out of the plane of the page, and extending from x = X = 1.00 m as in the figure below. x= 1.00 m (a) Calculate the y-component of the protons' momentum as they leave the magnetic field. (Indicate the direction with the sign of your answer.) kg · m/s (b) Find the angle a between the initial velocity vector of the proton beam and the velocity vector after the beam emerges from the field. Hint: Neglect relativistic effects and note that 1 ev = 1.60 x 10-19 ). a =

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter22: Magnetic Forces And Magnetic Fields

Section: Chapter Questions

Problem 75P

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A proton moving in the plane of the page has a kinetic energy of 6.00 MeV. A magnetic field of magnitude H = 1.00 T is directed into the page. The proton enters the magnetic field with its velocity vector at an angle = 45.0 to the linear boundary of' the field as shown in Figure P29.80. (a) Find x, the distance from the point of entry to where the proton will leave the field. (b) Determine . the angle between the boundary and the protons velocity vector as it leaves the field.
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
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