TBR: Can Magnetic FIELDS do work?

[justadream]

I read a thread (http://forums.studentdoctor.net/threads/magnetic-field-and-force-and-work.937666/) that said that magnetic fields do no work EVER* (so both magnetic force and magnetic fields don't do work).





If this is true, then in “velocity selectors”, does that mean if a particle is deflected because magnetic force > electric force, the magnetic field is still doing no work?

I ask because TBR physics book II page 156 #23 mentions that undeflected particles experience NO WORK but doesn't say anything about deflected particles.

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[Cawolf]

A deflection is not a change in energy. So no change in energy means no work by the Work-Energy theorem.

Think of this, a magnetic field always exerts its force perpendicular.
Work = F dot D = FDcos(theta) = FDcos(90) = 0

 

[justadream]

@Cawolf

So even though as the particle deflects and the field isn't directly perpendicular anymore, it's only the component of the field that IS perpendicular that is acting?

 

[Cawolf]

Yes it is always the cross product. So it always acts perpendicular. Every small deflection causes the field to exert it's force in a different direction.

It is well defined that magnetic fields lack a capacity to do work - this is why a mass spectrometer works so well. It can deflect a particle without effecting it's speed. It changes the vector components of velocity, but the magnitude (speed) of the velocity will be constant through the field.

 

[Cawolf]

Here is my poor sketch of a charged particle accelerated into a constant magnetic field.

http://imgur.com/6exzapt

Note how you can picture the force as a bunch of small forces that constantly act in the direction of the cross product. This is similar to what you might see in uniform circular motion.
The particle is constantly accelerating towards the center while it's speed is constant.

 

[justadream]

@Cawolf

Thank you for taking the time to draw that.

I guess I was confused since the magnetic force does not work (which your diagram shows) and since the magnetic field is providing that force, by extension then, the magnetic field does no work too.

 

[Cawolf]

No prob - hope it helps.

 

[tshank]

Thanks the responses and explanations. I have a question. How does the particle in your picture change direction but not velocity? I thought velocity is a vector. If the directional aspect of velocity changes, does the velocity as a whole change? Since direction changes, velocity must change also, therefore work as seen in kinetic energy? 1/2mv^2


If the magnetic field does not cause the velocity change, doing the work, what does cause the velocity change and do work?

Maybe someone can walk through this one a little for me. Thank!

 

[Cawolf]

Changing direction while moving is by definition, a change in velocity. So as you state, the velocity is changing.

Kinetic energy is calculated using the speed, which is the magnitude of the velocity - which does not change with direction.

 

[type12]

Changing direction while moving is by definition, a change in velocity. So as you state, the velocity is changing.

Kinetic energy is calculated using the speed, which is the magnitude of the velocity - which does not change with direction.

Just want to remphasize this. Work is your system changing the amount of energy it has. The total amount of energy (or 1/2mv^2) never changes with a magnet, so there is no change in the amount of energy, so there is no work done.

 

[tshank]

Thanks, that makes sense. No change of energy within system = no change of work.

 

[AmishExpressWay]

I read a thread (http://forums.studentdoctor.net/threads/magnetic-field-and-force-and-work.937666/) that said that magnetic fields do no work EVER* (so both magnetic force and magnetic fields don't do work).





If this is true, then in “velocity selectors”, does that mean if a particle is deflected because magnetic force > electric force, the magnetic field is still doing no work?

I ask because TBR physics book II page 156 #23 mentions that undeflected particles experience NO WORK but doesn't say anything about deflected particles.

In this case, magnitude of speed is changing, and therefore there is work being done by the magnetic/electric field.

Particle moves right with a velocity, v.
Magnetic field/electric field apply a net force perpendicular to the velocity v, Fapplied.
Since one v is a speed in the x direction, and Fapplied creates a speed in the y direction, the speed additions are additive and thus:

The new velocity, which is greater than the old velocity, is {[(Fapplied)(t)/(mass particle)]^2 + v^2}^(1/2) = Vnew.

(1/2)mVnew^2 - (1/2)mV^2 = Delta change in work = Total work done by magnetic field/electric field.

 

[type12]

In this case, magnitude of speed is changing, and therefore there is work being done by the magnetic/electric field.

Particle moves right with a velocity, v.
Magnetic field/electric field apply a net force perpendicular to the velocity v, Fapplied.
Since one v is a speed in the x direction, and Fapplied creates a speed in the y direction, the speed additions are additive and thus:

The new velocity, which is greater than the old velocity, is {[(Fapplied)(t)/(mass particle)]^2 + v^2}^(1/2) = Vnew.

(1/2)mVnew^2 - (1/2)mV^2 = Delta change in work = Total work done by magnetic field/electric field.

There's a couple of things. Magents can never do work, period. In the particular example given, only the electric field can do work, but that's only if the velocity is not along the x-axis. Applying a net force is not sufficient grounds for work to have been done on an object (the force must be parallel - not necessarily the same direction, but parallel - to the displacement).

Also, vector addition is not additive: the hypotenuse of a triangle is not equal to the sum of its sides. Lastly, the velocity only increases if it has a y-component and goes up, but decreases if it has a y-component and goes down.