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An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.255 T. If the kinetic energy of the electron is 3.50 10-19 J, find the speed of the electron and the radius of the circular path.
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Added Thu, 06 Aug '15
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well first we can find the speed of the electron since we have the kinetic energy
so
0.5mv�=KE
0.5(9.11x10^-31)v�=3.5x10^-19
v=876576.5 m/s
well we know that the electron is moving in circles meaning there's a centripetal force
F=mv�/r
but what is providing that centripetal force, well its the magentic force right
and the magnetic force is given by the formula
F=Bqv
so
Bqv=mv�/r
Bq=mv/r
r=mv/Bq
r=(9.11x10^-31)(876576.5 ) / 0.255 (1.6x10^-19)
r=0.0000195m
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Added Thu, 06 Aug '15
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