In: Physics
An electron of kinetic energy 1.69 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 20.1 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.
Given KE is in Kev ,first
convert it into joules using 1ev=1.6×10^(-19).after that using
forumla of K.E=1/2(mv^2) , velocity can be find out.
In part b) as Ques says electron moves in a circular path ,then ,it's radius is given by r=m×v/(e×B).From this magnetic field comes out to be B=m×v/(e×r)
Similarly for part c and d formula is given from that we can find out the values that to be found.