A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to the magnetic field. What is the ratio of the radii of their orbits? a. 2.8
b. 1.7.
c. 1.3.
d. 4.0.
e. 1.0.

ratio of the radii of their orbits = 1.3 --- C

Explanation:

1- eV = to the kinetic energy of the electrons

and kinetic energy is given as

K.E= 1/2mv2

v = √(2E/m)----- equation 1

The force on the particles  relating to the magnetic and circular motion ( centripetal force is given as

F = magnetic force = centripetal force

F= qvB = mv2/r

qvB = mv2/r

  r = mv/qB ------ equation 2

We know from equation 1 that v = √(2E/m)

Therefore,

r = √(2mE)/qB------ equation 3

We can now say that the  ratio of the two radii of their orbits can be calculated as

r1/r2 =(√(2mE1)/qB) /(√(2mE2)/qB

Where E1 = 500-eV  and E2 = 300-eV (1- eV = to the kinetic energy of the electrons)

r1/r2 = (√(2m x500)/qB) /(√(2mx 300)/qB

Cancelling out common variables, we are left with

r1/r2 =

r1/r2= 1.29 ≈ 1.3

the answer is b, have negligible mass