Particle breaks into two others

[bb.minhtri]

Homework Statement

A stable uncharged particle is break into 2 parts of mass m, one of which is charged -2Q. The two parts then go in the direction perpendicular to a uniform magnetic field B. After what time will the particles collide?

Homework Equations

The Attempt at a Solution

I suppose the time to be the period of one revolution of each particle, which leads to the result ∏m / BQ. Is that correct?

 

[vela]

Why do you think it's the period of one revolution? Or is that just a guess?

 

[bb.minhtri]

When I settle down and draw it out, it seems that they collide after just half a period, cause they both go with the same radii...

 

[gneill]

Unless there's something clever I'm overlooking (entirely possible!), this doesn't look like a problem that's going to have an easy solution. If certain assumptions are made about the initial speeds of the daughter particles (and hence the relative influence of the electric and magnetic forces), then good approximations might be had. Otherwise the Lorentz force is going to vary in magnitude throughout the particle's trajectories, and the instantaneous radii of curvature are going to vary, too, since the electric and magnetic forces will not be co-linear. It looks like the period attributed to a single charged particle moving alone in a magnetic field is not going to hold, unless I'm missing something in the symmetry or conservation laws that will preserve it.

 

[Dadface]

I think bb minhtri is basically right.The radius of each particle is given by
r=mv/B2q and the time period of each by
T=2pir/v=2pim/B2q=pim/Bq
From this they collide when each has covered half a revolution(not a full revolution)

 

[gneill]

I think bb minhtri is basically right.The radius of each particle is given by
r=mv/B2q and the time period of each by
T=2pir/v=2pim/B2q=pim/2Bq
From this they collide when each has covered half a revolution(not a full revolution)

So what happened to the electric force between the two particles?

 

[Dadface]

So what happened to the electric force between the two particles?

Nice question.As the particles separate(first quarter of cycle) they slow down due to the attraction between them but what they lose on the outward journey they regain on the return journey(second quarter cycle).Because of the change of speed the radius of the path changes but does this change the time period?At the moment I don't know.Time for me to put the thinking cap back on.


Further thought.The time period seems to be independent of the radius so does it matter if the radius changes?

 

[vela]

I suspect the intent was to solve the problem the way bb.minhtri and Dadface seem to have done, but it seems a bit sketchy for the reasons gneill has brought up. It might just be a poorly conceived homework problem, but it's hard to say without knowing what the original problem statement said. bb.minhtri, could you post the problem statement as it was given to you?

 

[bb.minhtri]

Thank you guys so much for fast replies...
This question comes in an University Entrance Exam and I am not allowed to bring the statement home :(
I can try to state it more clearly, but when I did it in the Exam, I was also worried about the Electric Force between the two: "A stable-uncharged particle is initially at rest. It is then broken into 2 particles of equal mass m, one of which is charged -2Q. Immediately after the breaking, the two particles go in opposite direction. During their movements, there is a constant magnetic field perpendicular to the plane of movements. After what time will the particles collide?"

 

[ehild]

A stable-uncharged particle is initially at rest. It is then broken into 2 particles of equal mass m, one of which is charged -2Q.

One would think that the other particle gets charge 2Q but it is not stated. In case one particle gets charge somehow and the other does not, the problem is easy. ehild

 

[bb.minhtri]

One would think that the other particle gets charge -2Q but it is not stated. In case one particle gets charge somehow and the other does not, the problem is easy.


ehild

I think one is charged -2Q, and that the mother particle is neutral, hence the other must be charged 2Q, is that right?

 

[ehild]

I think one is charged -2Q, and that the mother particle is neutral, hence the other must be charged 2Q, is that right?

It would be logical but the problem did not say that. Maybe, they left out the other part of the sentence, "and the other particle gets +2Q". When both particles get charge they interact with the Coulomb force between them. The motion is complicated and such problem can not be part of an entrance exam without indicating that the electric interaction can be ignored.

ehild

 

[bb.minhtri]

But if one particle goes uncharged, its path is not affected by any force, hence it will never collide again?

 

[ehild]

But if one particle goes uncharged, its path is not affected by any force, hence it will never collide again?

Yes, that is true. It travels along the tangent of the circle the other particle orbits.

It is worth guessing the magnitude of the forces and distances. When a heavy atom breaks, the mass of one particle can be about 10^-25 kg, the charge is 2e, B=0.1 T, v=100 m/s for example. What is the radius of the orbit? How far would get the particles without the magnetic field? What is the ratio of the electric and magnetic forces when the particles are on opposite parts of the circle?

ehild

 

[bb.minhtri]

In my exam, this comes as a Multiple Choice Question, the 5 answers are pi * m / BQ multiplied by some constants that i don't remember. I think the problem setter's answer must be that collision really happens, but how it is affected by electric force, remains so complicated :(

 

[ehild]

Sometimes the examiners do not think the problems over... . ehild

 

[bb.minhtri]

Yeah it turns out very complicated when thought of thoroughly... Since this was only one of 50 in 2 hours, I think there were some assumptions that were missed:(
Anw, thank you guys so much for help :)