A question about Magnetic Fields and A Charge

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Homework Help Overview

The discussion revolves around the effects of a magnetic field on a charged particle entering at a right angle. Participants explore how this interaction influences the kinetic energy and momentum of the particle.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between work done by the magnetic force and changes in kinetic energy and momentum. They question how the perpendicular nature of the force affects these quantities.

Discussion Status

Some participants suggest that since the force does no work, the kinetic energy remains unchanged, while the momentum changes direction but not magnitude. There is an indication of understanding developing among participants, but no explicit consensus is reached.

Contextual Notes

Participants are operating under the assumption that the magnetic force does not do work on the charged particle, which is central to their reasoning about kinetic energy and momentum.

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[SOLVED] A question about Magnetic Fields and A Charge

A charged particle enters in 90 degrees to the magnetic field i.e. the force on the charge is F = qvB as sin (angle) = 1. The question is which are the effects on the kinetic energy and momentum of the particle?

Alternatives:

A. KE => Changed
Momentum => Changed

B. KE => Unchanged
Momentum => Changed

The solution is B but I do not why? Please, help me.
 
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I think...

As the force is perpendicular to the displacement no work (W=fxcos(angle)) will be done. It's kinetic energy, a scalar, remains unchanged.

However, the force does accelerate the charged particle. This acceleration is perpendicular to it's velocity. The momentum vector changes in direction, not magnitude.

Does this help?
 
Barny said:
As the force is perpendicular to the displacement no work (W=fxcos(angle)) will be done.


In other words, the particle will circle around and around again ie it is not actually progressing anywhere. It stays where it is and circles. Hence, can it be said that there is no work done ie. KE is unchanged?
 
Yes I believe so.
 
Thanks! I think I understood :)
 

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