Inside a cyclotron, a gold ion with a speed of v is held on a trajectory with a
ID: 2059527 • Letter: I
Question
Inside a cyclotron, a gold ion with a speed of v is held on a trajectory with a radius r by a homogeneous magnetic field pointing out of the screen.
a. What will be its radius after it doubled its velocity?
r/4
r
r/2
2r
4r
b. Gold atoms are ionized by removing some of their electrons. In which direction does the gold ion fly?
Clockwise
Counter-clockwise
c. Gold has an atomic number of 79 and a mass of (3.3·10^25 kg). If 25 electrons (charge -1.6·10^19 C) are removed in the ionization chamber, and the atom is accelerated to a speed of (95*10^6 m/s), and if the magnetic field has a strength of 4.6 T, what is the radius of the trajectory?
Image of Exercise: http://i40.tinypic.com/25s95li.jpg
If you could post instructions on how you analyzed the problem and how you reached the answer, I will appreciate it.
Thanks!
Explanation / Answer
a) mv2/r = qvB => V=rqB/m
r is proportional to V therefore on doubling velocity radius will be doubled
r' =2r
b) CLOCKWISE
for circular motion force should be in inward dirction and B is out of the plane
so from F=qV X B, V will be in clockwise direction.
c) from the formula r = mV/qB
r = 1.7038 m