An electron moving with Kinetic Energy 25keV moves perpendicular to a uniform magnetic field of 0.2mT. Calculate the time period of rotation of electron in the magnetic field. plz answer it fast!!!!!!!!!!!
Answers
Answered by
115
The time period is given as
T = 2πm / Bq
where
mass of the electron, m = 9.1 X 10-31 kg
magnetic field strength, B = 0.2 mT = 0.2 X10-3 T
charge on the electron, q = 1.6 X10-19 C
so,
T = (2 X3.14 X 9.1 X 10-31) / (0.2 X10-3 X 1.6 X10-19)
or
T = (57.148/0.32) X 10-9
thus, the time period will be
T = 1.785 X 10-7 secs.
T = 2πm / Bq
where
mass of the electron, m = 9.1 X 10-31 kg
magnetic field strength, B = 0.2 mT = 0.2 X10-3 T
charge on the electron, q = 1.6 X10-19 C
so,
T = (2 X3.14 X 9.1 X 10-31) / (0.2 X10-3 X 1.6 X10-19)
or
T = (57.148/0.32) X 10-9
thus, the time period will be
T = 1.785 X 10-7 secs.
Similar questions