How does a magnetic field perform no work?

[mahrap]

1. I understand that in order for a force to qualify as doing work it must displace an object. The magnetic field does indeed induce a "force" on a moving charge and displaces the moving charge toward a different trajectory. However, it does not change the kinetic energy of the moving charge. So for something to qualify to do work must it also change the kinetic energy of the object? I guess what I really need is a better definition of work and how it relates to force. Please guide me in any way you can. Thanks for all the help.

2. F = q (v x B)

 

[rcgldr]

If the force is always perpendicular to velocity, then no work is done, a change in direction, but not a change in speed.

 

[vela]

You should be able to find answers to your questions in your textbook. You probably paid a lot of good money for it, so you should make use of it! What is the definition of work? What does the work-energy theorem tell you?

 

[rpthomps]

I like to think of work as transferring energy via a force. In your particular case, if a charged particle is moving parallel to magnetic field lines then no work would be done because, as you said, there would be no change in energy of the particle and consequently no work done.

 

[collinsmark]

The magnetic field does indeed induce a "force" on a moving charge and displaces the moving charge toward a different trajectory.

But consider the total, instantaneous displacement of the charge. It's moving, don't forget. So you can find this infinitesimal displacement, \vec {ds} by by finding the distance traveled over the infinitesimal time, dt. Don't forget, \vec {ds} is a vector and it has a direction.

Now ask yourself, knowing what you know about magnetic fields, what is the that vector's direction with respect to the direction of the magnetic force, \vec f?

[Edit: bonus hint: how does the charge's instantaneous velocity \vec v relate to \vec {ds} and dt?]

I guess what I really need is a better definition of work and how it relates to force. Please guide me in any way you can. Thanks for all the help.

Work is defined as

W = \vec f \cdot \vec {s}
or in the case of a varying force, or force with a varying direction,
W = \int_{s=a}^b \vec f \cdot \vec {ds}

Notice that in both cases, there is a dot product involved. (It's not as simple as just multiplying the magnitudes together.) What does that dot product tell you about the work done by magnetic fields, given what you've just found about the charge's displacement and the direction of magnetic force?

 

[rude man]

I like to think of work as transferring energy via a force. In your particular case, if a charged particle is moving parallel to magnetic field lines then no work would be done because, as you said, there would be no change in energy of the particle and consequently no work done.

The relative direction of the mag field compared to the direction of the particle velocity is immaterial. A mag field produces no work on a moving charge, period. Reason? (v x B)*v = 0 always. Algebraic identity. * means dot-product.

 

[rpthomps]

The relative direction of the mag field compared to the direction of the particle velocity is immaterial. A mag field produces no work on a moving charge, period. Reason? (v x B)*v = 0 always. Algebraic identity. * means dot-product.

Can you parse what you said a little more? First, why is this equation (v x B)*v = 0 relevant? And secondly, if no work is done by the mag field then where does the energy come from?

 

[rude man]

qv x B is force. v is rate of change of displacement. So (qv x B)*v is power. Algebraically, this is always zero since v is always orthogonal to v x B.

The energy of the particle is imparted to it by some other force, usually an electric field. Never by a B field.

 

[rpthomps]

qv x B is force. v is rate of change of displacement. So (qv x B)*v is power. Algebraically, this is always zero since v is always orthogonal to v x B.

The energy of the particle is imparted to it by some other force, usually an electric field. Never by a B field.

Okay. Thanks for your response and your time.

 

[rude man]

Okay. Thanks for your response and your time.

10-4.