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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 and characterize particles,) and mass spectrometers (to identify ions) all rely on this circular motion of charged particles in a magnetic field. In section 26.3, we learn that the radius of the circle made by a charged particle moving perpendicular to a magnetic field is r= mv qB (a) A particle is observed moving to the right when it enters a magnetic field. The magnetic field points into the page. When the particle enters the field, it moves in a clockwise circle. What is the sign of the charge? (b) Explain using physics language why the radius gets larger when the mass increases, and smaller

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 and characterize particles,) and mass spectrometers (to identify ions) all rely on this circular motion of charged particles in a magnetic field. In section 26.3, we learn that the radius of the circle made by a charged particle moving perpendicular to a magnetic field is r= mv qB (a) A particle is observed moving to the right when it enters a magnetic field. The magnetic field points into the page. When the particle enters the field, it moves in a clockwise circle. What is the sign of the charge? (b) Explain using physics language why the radius gets larger when the mass increases, and smaller

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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 and characterize particles,) and mass spectrometers (to identify ions) all rely on this circular motion of charged particles in a magnetic field. In section 26.3, we learn that the radius of the circle made by a charged particle moving perpendicular to a magnetic field is T = mv qB (a) A particle is observed moving to the right when it enters a magnetic field. The magnetic field points into the page. When the particle enters the field, it moves in a clockwise circle. What is the sign of the charge? (b) Explain using physics language why the radius gets larger when the mass increases, and smaller when the charge or field increase. Discuss also why the radius tends to infinity (ie the path becomes a straight line) when the charge goes to zero. (c) At CERN, protons move almost the speed of light in a circle whose radius is over 4 km. In this part of the 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.) 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.
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 and characterize particles,) and mass spectrometers (to identify ions) all rely on this circular motion of charged particles in a magnetic field. In section 26.3, we learn that the radius of the circle made by a charged particle moving perpendicular to a magnetic field is T = mv qB (a) A particle is observed moving to the right when it enters a magnetic field. The magnetic field points into the page. When the particle enters the field, it moves in a clockwise circle. What is the sign of the charge? (b) Explain using physics language why the radius gets larger when the mass increases, and smaller when the charge or field increase. Discuss also why the radius tends to infinity (ie the path becomes a straight line) when the charge goes to zero. (c) At CERN, protons move almost the speed of light in a circle whose radius is over 4 km. In this part of the 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.) 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.
Calculate the magnetic field, in Tesla, that is needed - if we ignore relativity - to get a radius of r = 4.0 km and speed v equal to c = 3.0 × 108 m/s (the speed of light) for a proton (mass 1.67 x 10-27 kg and charge 1.6 × 10-1⁹ C.) (d) FOR E LEVEL HOMEWORK: The magnetic field you calculated for the CERN experiment 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 mv² 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:Calculate the magnetic field, in Tesla, that is needed - if we ignore relativity - to get a radius of r = 4.0 km and speed v equal to c = 3.0 × 108 m/s (the speed of light) for a proton (mass 1.67 x 10-27 kg and charge 1.6 × 10-1⁹ C.) (d) FOR E LEVEL HOMEWORK: The magnetic field you calculated for the CERN experiment 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 mv² 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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