Question

An electron of kinetic en-
ergy 1.20 keV circles in a plane per-
pendicular to a uniform magnetic
field. The orbit radius is 32.7 cm. Find
(a) the electron’s speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

For problem 28.21 calculate the cyclotron period in nanoseconds using a cyclotron radius of 32.7 cm. 5 sig. figs.

Answer 1

Kinetic energy of the electron = 1.2 KeV =

= (J=Joule)

orbit radius = 32.7 cm =0.327 m

a) electron's speed (velocity )

we know that

(where KE is kinetic energy ,m is mass ,V is velocity)

so

substitute the values , we get

so speed of electron is m/s

(b) The magnetic field magnitude

the electron is in a magnetic field

the orbit radius is 0.327 m

we know that

(r is radius ,V is velocity,e is magnitude of charge and B is magnitude of the MF)

so

substitute the values we get

The magnetic field magnitude is  T .

(c) The circling frequency

The circling freq =the velocity of the electron over the circumference ( here circumference =)

Freq is

(d) The period of the motion

period is the reciprocal of freq

so ,

period is