1
$\begingroup$

I recently came across a problem where an electrically neutral particle is at rest in a uniform magnetic field. The particle now splits into two charged particles of equal mass $m$; charge and mass being conserved. Now, we know that if one particle has charge $q$, the other must have equal and opposite charge $-q$. It is said that the two particles collide after some time and we are asked to find that time.

My reasoning is that since the center of mass(com) of the system remains stationary, the particles will split in opposite directions to each other and thus collide just when each particle completes a semi-circle due to the magnetic force. With this approach, I get the answer right. Now let's forget the time period part and analyze the motion of the particles.

Second case: In analyzing the motion of the particles from a different perspective, we can say that throughout the motion, the center of mass must remain at rest as its charge is zero (q+(-q)=0) and so magnetic force is zero. So the charged particles must continue to move opposite to each other just so that the com remains stationary and not in a semi-circle. Perhaps the forces exerted on each other do the job. I am actually confused about the properties and motion of the com in situations involving charged particles in magnetic fields. I wonder whether there is a center-of-charge kind of concept where that point remains stationary when a neutral particle splits into two or more charged particles ? . Such a concept will simplify problem solving to a great extent in the same way the concept of com does.

Can someone please clarify these to me:

  1. Why is the second case incorrect ?
  2. Is there such a center-of-charge concept ?

Please feel free to edit and add relevant tags to the question.

Improve this question
$\endgroup$
2
  • $\begingroup$ You draw not the full picture of what happens. What or who brakes the body in two pices? And what gives to them a motion? Without motion no influence of a magnetic field to charged particles. $\endgroup$
    – HolgerFiedler
    Commented Oct 4, 2014 at 6:27
  • $\begingroup$ I don't think you need the specifics of the splitting. I have written what happens after the splitting. The energy released from the splitting is used as kinetic energy by the particles and that causes the motion. $\endgroup$
    – Gaurav
    Commented Oct 4, 2014 at 8:20

2 Answers 2

Reset to default
2
$\begingroup$

Is there a center-of-charge kind of concept where that point remains stationary when a neutral particle splits into two or more charged particles ?

Not when there are external forces acting on the system. As soon as the particle breaks down, the two charged particles experience magnetic force in the same direction, so their sum momentum changes and the center of mass moves as well.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Yes, agreed. But even if an external force does act on the system, the center of mass must move in such a way that all of the external forces on the system were to act at that point (according to the standard textbook definition). But here it is not clear how to interpret the situation to fit into that definition since the charge on the center of mass is zero and thus the net magnetic force is zero. $\endgroup$
    – Gaurav
    Commented Oct 4, 2014 at 8:13
  • $\begingroup$ Net magnetic force is zero only when the particles do not move. As soon as they begin to move in opposite directions, both experience magnetic force in the same direction, curving their trajectories into a circle. $\endgroup$
    – Ján Lalinský
    Commented Oct 5, 2014 at 8:06
  • $\begingroup$ But wouldn't the particles exert forces on each other, and thus wouldn't those forces affect their circular motion ? $\endgroup$
    – Gaurav
    Commented Oct 5, 2014 at 8:19
  • $\begingroup$ Yes, they would. Still, if the velocities are opposite, the magnetic forces point along the same direction. $\endgroup$
    – Ján Lalinský
    Commented Oct 5, 2014 at 8:25
0
$\begingroup$

Yes,the particles will have attractive forces between each other ,which can and will disturb their trajectories . But as far as the base question is concerned, I think we are supposed to neglect that interaction (as I saw another version of this problem which said so) . I think your first view point is better .Basic reason being that the initial momentum(net) of the system is zero, and momentum (net) should be conserved so after breaking they move antiparallel to each other. And as they move in the B-field they are both get curved in a good circle and collide (circle only after neglecting the Electro-forces)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.