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A magnetic field turns the velocity of a particle but does not change the speed, because the force is always perpendicular to the velocity. Particle accelerators (like CERN,) bubble chambers (to detect 4 and characterize particles,) and mass spectrometers (to identify ions) all rely on this circular motion of charged particles in a magnetic field. At CERN, protons move almost the speed of light in a circle whose radius is over 4 km. In this problem we will figure out the magnetic field needed to bend them into this circle. I will walk you through the analysis. (The text also goes over a similar example in section 26.3) (a) Draw a diagram of a proton moving in the page, and the B field pointing perpendicular to it (in or out of the page.) Use the right hand rule to decide which way the force points and add the force vector to your diagram. (b) Write the magnitude of the force in terms of q, v and B. (c) Recall that a force perpendicular to the motion will result in uniform circular motion with centripetal acceleration a. = already wrote, and the second law (F=ma) to solve for the magnetic field, in Tesla, that is needed – if we ignore relativity - to get a radius r = 4 km and speed v equal to c = 3X10$ (the speed of light) for a proton (mass 1.67 x 10-27 kg and charge 1.6 × 10-19 C.) v2 /r. Use this, the expression for the magnetic force that you (d) This should look ridiculously small. Why not make the field larger (CERN uses giant multi-Tesla magnets!) and the radius smaller? The answer is relativity. Without getting into the details, the error is that at speeds near c, the effective mass is almost all energy! To account for this,

A magnetic field turns the velocity of a particle but does not change the speed, because the force is always perpendicular to the velocity. Particle accelerators (like CERN,) bubble chambers (to detect 4 and characterize particles,) and mass spectrometers (to identify ions) all rely on this circular motion of charged particles in a magnetic field. At CERN, protons move almost the speed of light in a circle whose radius is over 4 km. In this problem we will figure out the magnetic field needed to bend them into this circle. I will walk you through the analysis. (The text also goes over a similar example in section 26.3) (a) Draw a diagram of a proton moving in the page, and the B field pointing perpendicular to it (in or out of the page.) Use the right hand rule to decide which way the force points and add the force vector to your diagram. (b) Write the magnitude of the force in terms of q, v and B. (c) Recall that a force perpendicular to the motion will result in uniform circular motion with centripetal acceleration a. = already wrote, and the second law (F=ma) to solve for the magnetic field, in Tesla, that is needed – if we ignore relativity - to get a radius r = 4 km and speed v equal to c = 3X10$ (the speed of light) for a proton (mass 1.67 x 10-27 kg and charge 1.6 × 10-19 C.) v2 /r. Use this, the expression for the magnetic force that you (d) This should look ridiculously small. Why not make the field larger (CERN uses giant multi-Tesla magnets!) and the radius smaller? The answer is relativity. Without getting into the details, the error is that at speeds near c, the effective mass is almost all energy! To account for this,

College Physics
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)...
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A magnetic field turns the velocity of a particle but does not change the speed, because the force is always perpendicular to the velocity. Particle accelerators (like CERN,) bubble chambers (to detect 4 and characterize particles,) and mass spectrometers (to identify ions) all rely on this circular motion of charged particles in a magnetic field. At CERN, protons move almost the speed of light in a circle whose radius is over 4 km. In this problem we will figure out the magnetic field needed to bend them into this circle. I will walk you through the analysis. (The text also goes over a similar example in section 26.3) (a) Draw a diagram of a proton moving in the page, and the B field pointing perpendicular to it (in or out of the page.) Use the right hand rule to decide which way the force points and add the force vector to your diagram. (b) Write the magnitude of the force in terms of q, v and B. (c) Recall that a force perpendicular to the motion will result in uniform circular motion with v2 /r. Use this, the expression for the magnetic force that you centripetal acceleration a. = already wrote, and the second law (F=ma) to solve for the magnetic field, in Tesla, that is needed – if we ignore relativity - to get a radius r = 4 km and speed v equal to c = 3X10$ (the speed of light) for a proton (mass 1.67 x 10-27 kg and charge 1.6 × 10-19 C.) (d) This should look ridiculously small. Why not make the field larger (CERN uses giant multi-Tesla magnets!) and the radius smaller? The answer is relativity. Without getting into the details, the error is that at speeds near c, the effective mass is almost all energy! To account for this, simply replace the kinetic energy we used earlier mv? with the relativistic energy, which is about 7 TeV (= 7 × 1012 eV. ) Convert the energy to SI (J) units, noting that J=(charge of proton in C)(energy in eV.) Notice that the expression mv² shows up in our analysis via the net force so you can actually continue to work either in eV or use the converted energy in J. Either way, be careful of units! Revise the analysis to include this correction and recalculate the needed B field. It should now be several Tesla, which explains the ever growing size of the CERN accelerator rings.
Transcribed Image Text:A magnetic field turns the velocity of a particle but does not change the speed, because the force is always perpendicular to the velocity. Particle accelerators (like CERN,) bubble chambers (to detect 4 and characterize particles,) and mass spectrometers (to identify ions) all rely on this circular motion of charged particles in a magnetic field. At CERN, protons move almost the speed of light in a circle whose radius is over 4 km. In this problem we will figure out the magnetic field needed to bend them into this circle. I will walk you through the analysis. (The text also goes over a similar example in section 26.3) (a) Draw a diagram of a proton moving in the page, and the B field pointing perpendicular to it (in or out of the page.) Use the right hand rule to decide which way the force points and add the force vector to your diagram. (b) Write the magnitude of the force in terms of q, v and B. (c) Recall that a force perpendicular to the motion will result in uniform circular motion with v2 /r. Use this, the expression for the magnetic force that you centripetal acceleration a. = already wrote, and the second law (F=ma) to solve for the magnetic field, in Tesla, that is needed – if we ignore relativity - to get a radius r = 4 km and speed v equal to c = 3X10$ (the speed of light) for a proton (mass 1.67 x 10-27 kg and charge 1.6 × 10-19 C.) (d) This should look ridiculously small. Why not make the field larger (CERN uses giant multi-Tesla magnets!) and the radius smaller? The answer is relativity. Without getting into the details, the error is that at speeds near c, the effective mass is almost all energy! To account for this, simply replace the kinetic energy we used earlier mv? with the relativistic energy, which is about 7 TeV (= 7 × 1012 eV. ) Convert the energy to SI (J) units, noting that J=(charge of proton in C)(energy in eV.) Notice that the expression mv² shows up in our analysis via the net force so you can actually continue to work either in eV or use the converted energy in J. Either way, be careful of units! Revise the analysis to include this correction and recalculate the needed B field. It should now be several Tesla, which explains the ever growing size of the CERN accelerator rings.
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