R of Curvature.doc
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A beam of particles including protons, electrons, deuterons, and helium atoms all with a speed of 2.5 x 10^8 m/s passes through a magnetic field of 0.40 T perpendicular to their velocity. The radius of curvature of the path is calculated for each particle type using the formula that relates magnetic field, particle mass, charge, and velocity. The radius is found to be 6.52 x 10^-2 m for protons, 3.559 x 10^-1 m for electrons, 1.305 x 10^-1 m for deuterons, and 2.594 x 10^-1 m for helium atoms.
R of Curvature.doc
- 1. 1 2 6 A beam of charged particles consisting of protons, electrons, deuterons, and singly ionized helium atoms and molecules all pass through a velocity selector, all emerging with speeds of 2.5 10 m/s. Th H e beam then enters a region of uniform magnetic field 0.40 directed perpendicular to their velocity. Compute the radius of curvature of the path of each type of particle. B T Solution Motion of Point Charges A particle of mass m and charge q moving with speed v in a plane perpendicular to a uniform magnetic field moves in a circular orbit. The period and frequency of the circular motion are independent of the radius of
- 2. 2 the orbit and of the speed of the particle. MAGNETIC FORCE ON A MOVING CHARGED PARTICLE Cyclotron Period Cyclotron Frequency
- 3. 3 Velocity Selector A velocity selector consists of crossed electric and magnetic fields so that the electric and magnetic forces balance for a particle moving with speed . v E v B
- 4. 4 OR
- 6. 6 Proton 27 6 19 27 6 19 2 Using 1.67 10 , 2.5 10 , 1.60 10 , 0.40 in eq.(1), we get, 1.67 10 2.5 10 R= 1.60 10 0.40 6.52 10 m X kg m v X s q e X C B T m X kgX X s X CX T X m 2 2 2 2 2 : . .. . .. . . . . . . . . .. . . . . . . . . . . . . . . . . . Note kg m kg m m kg m m s C T s C Wb s C V s kg m m N m C N m s A sV s A sV A s J C m m A s
- 7. 7 Electron 31 6 19 31 6 19 6 Using 9.11 10 , 2.5 10 , 1.60 10 , 0.40 in eq.(1), we get, 9.11 10 2.5 10 R= 1.60 10 0.40 35.59 10 m X kg m v X s q e X C B T m X kgX X s X CX T X m
- 8. 8 2 2 2 2 2 : . .. . .. . . . . . . . . .. . . . . . . . . . . . . . . . . . Note kg m kg m m kg m m s C T s CWb s CV s kg m m N m C N m s A sV s A sV A s J C m m A s
- 9. 9 Deuteron 27 6 19 27 6 19 2 Using 3.34 10 , 2.5 10 , 1.60 10 , 0.40 in eq.(1), we get, 3.34 10 2.5 10 R= 1.60 10 0.40 13.05 10 0.1305 m X kg m v X s q e X C B T m X kgX X s X CX T X m m 2 2 2 2 2 : . .. . .. . . . . . . . . .. . . . . . . . . . . . . . . . . . Note kg m kg m m kg m m s C T s C Wb s C V s kg m m N m C N m s A sV s A sV A s J C m m A s
- 10. 10 Helium (Alpha Particle) 27 6 19 27 6 19 2 Using 6.64 10 , 2.5 10 , 1.60 10 , 0.40 in eq.(1), we get, 6.64 10 2.5 10 R= 1.60 10 0.40 25.94 10 0.2594 m X kg m v X s q e X C B T m X kgX X s X CX T X m m 2 2 2 2 2 : . .. . .. . . . . . . . . .. . . . . . . . . . . . . . . . . . Note kg m kg m m kg m m s C T s C Wb s C V s kg m m N m C N m s A sV s A sV A s J C m m A s