A magnetic field turns the velocity of a particle but does not change the speed, because the force isalways perpendicular to the velocity. Particle accelerators (like CERN,) bubble chambers (to detectand characterize particles,) and mass spectrometers (to identify ions) all rely on this circular motionof charged particles in a magnetic field. In section 26.3, we learn that the radius of the circle madeby a charged particle moving perpendicular to a magnetic field isT =mvqB(a) A particle is observed moving to the right when it enters a magnetic field. The magnetic fieldpoints into the page. When the particle enters the field, it moves in a clockwise circle. What isthe sign of the charge?(b) Explain using physics language why the radius gets larger when the mass increases, and smallerwhen the charge or field increase. Discuss also why the radius tends to infinity (ie the pathbecomes 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 thispart of the problem we will figure out the magnetic field needed to bend them into this circle. Iwill 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 (inor out of the page.) Use the right hand rule to decide which way the force points and add theforce vector to your diagram. Calculate the magnetic field, in Tesla, that is needed - if we ignore relativity - to get a radiusof r = 4.0 km and speed v equal to c = 3.0 × 108 m/s (the speed of light) for a proton (mass1.67 x 10-27 kg and charge 1.6 × 10-1⁹ C.)(d) FOR E LEVEL HOMEWORK: The magnetic field you calculated for the CERN experimentshould look ridiculously small. Why not make the field larger (CERN uses giant multi-Teslamagnets!) 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 isabout 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 actuallycontinue 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 nowbe several Tesla, which explains the ever growing size of the CERN accelerator rings.

As per guidelines, only 3 sub parts are solved here. Given, A charged particle moving perpendicular…

Answered: A magnetic field turns the velocity of…