Magnetic Force on a Charged Particle in Constant Magnetic Field

AI Thread Summary
In a uniform magnetic field, the magnetic force on a charged particle does not change its speed or kinetic energy, as it acts perpendicular to the particle's velocity. The false statement in the discussion is that the magnetic force can do work on the particle; it cannot, since work depends on the angle between force and motion, and the magnetic force is always perpendicular. This means the magnetic force alters the direction of the particle's velocity but not its magnitude. The confusion arises from misunderstanding the nature of the magnetic force, which is defined by the equation F=q(VxB). Overall, the magnetic force on a charged particle in a constant magnetic field does not perform work.
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Homework Statement


A charged particle is moving in a uniform, constant magnetic field. Which one of the following statements concerning the magnetic force exerted on the particle is false?
It changes the velocity of the particle.
It increases the speed of the particle.
It does not change the kinetic energy of the particle.
It can act only on a particle in motion.
It does no work on the particle.


Homework Equations


Could you explain why it false?


The Attempt at a Solution


I am told that the answer is the last one.
 
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3. The Attempt at a Solution
I am told that the answer is the last one.
Not really an attempt!
 
This isn't obvious but a static magnetic field, whose force on a charge particle is given by F=q(VxB), can not do work.

Remember that work in general depends on the direction between the applied force and the direction of motion. If they're in the same direction then you get the simple W=F*d, and slightly more generally, W=F*d*cos(theta) where theta is the angle between them.(because really it's a dot product)

So now look at the equation for force, and tell me, knowing the definition of a cross product, if the force vector will ever be anything but perpendicular to the velocity of the particle? What's the result of that?
 
mgb_phys said:
Not really an attempt!

not helpful!
 
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