Force & Work in Constant Magnetic Fields: Is it True?

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SUMMARY

A constant magnetic field does not perform work on a particle due to the perpendicular nature of the force and displacement, resulting in zero work done as per the work-energy theorem. This principle is illustrated through cyclotron motion, where particles experience a force but do not gain kinetic energy. The discussion also highlights the Faraday disk, or homopolar generator, which operates under different principles, producing significant energy output. The magnetron, used in microwave ovens, demonstrates the application of cyclotron motion in a practical device, where work is done through the radial electric field rather than the magnetic field.

PREREQUISITES
  • Understanding of the work-energy theorem
  • Knowledge of cyclotron motion and its principles
  • Familiarity with magnetic fields and their effects on charged particles
  • Basic concepts of electromagnetic devices, such as the Faraday disk and magnetron
NEXT STEPS
  • Research the principles of the work-energy theorem in detail
  • Study the operation and applications of homopolar generators
  • Learn about the design and function of magnetrons in microwave technology
  • Explore the physics of cyclotron motion and its applications in particle accelerators
USEFUL FOR

Students of physics, electrical engineers, and professionals working with electromagnetic devices or particle acceleration technologies.

cragar
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My teacher told us that a constant magnetic field cannot do work on a particle , it can only deflect a particle , But we also talked about cyclotron motion and if the particle is moving around in a circle then it is experiencing a force . is this true ?
 
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Yes, it's experiencing a force. But since the force is perpendicular to the motion, there is no work done.
 
ok , thanks , could we say that since the change in kinetic energy is 0 the work is 0
 
Last edited:
Your placement of the word "since" seems to have the cause-and-effect backwards. It goes the other way.

Force and displacement are instantaneously perpendicular to each other, therefore the work done on the particle is zero. Then use the work-energy theorem, which says that the change in kinetic energy is equal to the net work done.
 
The exception to this is the Faraday disk, a conducting disk rotating in a constant uniform magnetic field. It is also called a homopolar generator. See
http://en.wikipedia.org/wiki/Homopolar_generator
The large one built at Australian National University could store over 500 MJ, and produce a dc unipolar pulse of millions of amps for hundreds of seconds. The disk is rotating in a constant magnetic field and the direction of rotation is perpendicular to the radial current in the disk.
Bob S
 
Vanadium 50 said:
Yes, it's experiencing a force. But since the force is perpendicular to the motion, there is no work done.
I am still not understanding why there is no work done in the direction of the force.
 
dW = F dot dx. If F and dx are perpendicular, no work.
 
oh i see because the cosine of 90 is 0 , thanks.
 
If particles are being deflected, then isn't work being done? The Cyclotron is similar to the Magnotron, which is the heart of every Microwave oven. My 1000watt microwave oven does a lot of work.
 
  • #10
Relay said:
If particles are being deflected, then isn't work being done? The Cyclotron is similar to the Magnotron, which is the heart of every Microwave oven. My 1000watt microwave oven does a lot of work.
The magnetron is an interesting microwave tube, developed just before and during WW II. See

http://physics.princeton.edu/~mcdonald/examples/EM/brillouin_pr_60_385_41.pdf

In cylindrical geometry, electrons are accelerated radially outward from a hot filament by a large radial electric field, and execute cyclotron motion in a (nearly) uniform axial magnetic field. Geometry of the magnetron cavity causes the electrons to bunch, and to radiate microwave energy at the bunching frequency. So the electron energy (and work) comes from the electrons being accelerated by the radial electric field.

Bob S
 
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