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Magnetic Force and Circular Motion Mrs. Coyle AP Physics C Force on a Charged Particle in a Magnetic Field For a charged particle moving in an external magnetic field with its velocity perpendicular to the field: • The force is always directed toward the center of the circular path • The magnetic force causes a centripetal acceleration, changing the direction of the velocity of the particle Finding the radius of the circular path • Equate the magnetic and centripetal forces: mv FB qvB r • Solving for r: mv r 2 qB – Note: r is proportional to the momentum of the particle and inversely proportional to the magnetic field More About Motion of Charged Particle • The angular speed of the particle is v qB ω r m – The angular speed, w, is also referred to as the cyclotron frequency • The period of the motion is 2πr 2π 2πm T v ω qB Force on a charge moving in a magnetic field Radius: mv r qB qBr Velocity: v m qB Frequency: w m Period: 2m T qB If the angle between v and B is not 90o . • The path is a helix • Same equations apply, with v v v 2 y 2 z Problem #29. The magnetic field of the Earth at a certain location is directed vertically downward and has a magnitude of 50.0 μT. A proton is moving horizontally toward the west in this field with a speed of 6.20 × 106 m/s. (a) What are the direction and magnitude of the magnetic force the field exerts on this charge? (b) What is the radius of the circular arc followed by this proton? Ans: a)4.96x10-17 N South, b) 1.29km Problem #32. A proton moving freely in a circular path perpendicular to a constant magnetic field takes 1.00 μs to complete one revolution. Determine the magnitude of the magnetic field. Ans: 6.56 x10-2 T Problem #39. A singly charged positive ion moving at 4.60 × 105 m/s leaves a circular track of radius 7.94 mm along a direction perpendicular to the 1.80-T magnetic field of a bubble chamber. Compute the mass (in atomic mass units 1amu=1.66x10-27 kg) of this ion. Ans: 4.97x10-27 kg=2.99amu