Phys 202 1nd Edition Lecture 14Outline of Last Lecture I. Magnetic fields and cyclotronsOutline of Current Lecture II. Review of magnetic forceIII. Example problemsCurrent LectureReview of Magnetic force:Magnetic force (Fb) is the charge(q) times the velocity of the charge (v) crossed with the magnetic field(B). This is represented by the equation;FB=q(VxB)The magnitude of V cross B (V x B) is equal to the magnitude of V times the magnitude of B times sin(θ);|V x B|= |V| |B|sin(θWhen solving problems involving magnetic field, it is important to know which direction FB is going. For our problems, a dot ( . ) indicates that FB is pointing out of the page, towards you, and a + indicates that FB is pointing into the page, away from you. If it’s easier, imagine FB like an arrow. If the point of the arrow is pointing at you, it would look like a dot. If the arrow is pointing away from you, you’ll just see the fletching, which would look a bit like a cross or x. If there is no indication of which direction FB is pointing, use right hand to trace the paths of the two vectors(V and B). When your right hand lays in the direction of the velocity vector and your fingers curl towards B vector, your thumb should point in the direction of FB. However, remember that if q is negative, FB will actually point in the direction opposite of the direction that your thumb points.Example problems:Problem 1:An electron with a charge of 1.602x10-19 and a mass of 9.109x10-31kg follows a circular path around a magnetic field. The radius of the path of the electron is 23cm and the kinetic energy of the electron is These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.1.42KeV. A) What is the velocity of the electron? B) What is the magnitude of the magnetic field? C) What is the frequency and period of the path of the electron?A) First thing we have to do is convert all of our known values into units we can use;E=1.602x10-19CMe=9.109x10-31KgKE=1.42KeV=1.42x103(1.602x10-19)=2.27x10-16Jr=23cm = .23mNow we can solve for the velocity of the electron by looking at its kinetic energy;KE=.5mv2  V=√(2KE/m)V= √(2(2.27x10-16J)/9.109x10-31Kg)= 2.23x107m/sB) Solve for the magnitude of the magnetic field by applying what you know about force.FB=q|V||B|Sin(θ)F=ma=mv2/rBecause, in this case, θ is 90⁰ and Sin(90⁰) is equal to 1, we can ignore this part of the equation. Therefore, we can rearrange the equation as follows;qvB=mv2/r  B=mv/rqB=[(9.109x10-31Kg)(2.23x107m/s)]/[(.23m)(1.602x10-19C)]= 5.45x10-19 TeslaC:T=2πr/v T=2π(.23m)/(2.23x107m/s)=6.48x10-8sF=1/T F=1/6.8x10-8=1.54x107 HertzProblem 2:The figure below depicts a proton with a charge of 1.602x10-19 and a mass of 1.67x10-27 shooting throughan electric and a magnetic field. The proton is accelerated by a potential of 14kV. The magnitude of the electric field acting on the proton is 1.7x105N/c. What magnitude of the magnetic field is required to keep the proton moving in this same path with no deflection? FE +B + + + + + + + + + + + + + P- - - - - - - - FBFor this problem, we know that the force of the electric field (FE) has to equal the force of the magnetic field (FB) on the proton;FE=qEFB=qvBqE=qvB  E=vBSo to solve for B, we first need to solve for the velocity;KE=qV=.5mv2V=√(2qV/m) = √([2(1.602x10-19)(14000)]/1.67x10-27)=1.64x106m/sB=E/vB=1.7x105/1.64x106=.104