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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.

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

Added Thu, 06 Aug '15

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