How electrons get deflected in magnetic field while moving?

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Discussion Overview

The discussion centers around the behavior of electrons in a magnetic field, particularly focusing on how they are deflected while in motion. Participants explore the underlying physics concepts, including the forces acting on charged particles in magnetic fields, and the challenges of using analogies to explain these phenomena.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express confusion about why electrons do not move in the same way as light objects like crumpled paper when influenced by a magnetic field.
  • Others clarify that a magnetic field does not behave like wind or other mechanical forces, emphasizing that it exerts a force described by the equation $$\vec{F}=q \frac{\vec{v}}{c} \times \vec{B}$$.
  • Participants discuss the implications of the cross product in the force equation, noting that the force is always perpendicular to both the velocity of the particle and the magnetic field.
  • There is a suggestion to simplify the force equation to $$f = q + v \times B$$, with a request for example values for charge, velocity, and magnetic field strength.
  • Some participants repeatedly ask for specific values to substitute into the force equation to calculate the force acting on a charge in a magnetic field.
  • A later reply provides an example using $$q = 1 C$$, $$v = 1 m/s$$, and $$B = 1 T$$, resulting in a calculated force of $$1 N$$, but the context of this example remains under discussion.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical formulation of the force on a charged particle in a magnetic field, but there is disagreement regarding the appropriateness of analogies used to explain the behavior of electrons. The discussion remains unresolved regarding the best way to conceptualize these forces.

Contextual Notes

Some limitations are noted regarding the use of mechanical analogies to describe magnetic forces, as they do not behave in the same manner. Additionally, there are unresolved questions about the specific values to use in calculations, which depend on the context of the problem.

mo0nfang
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I don't understand why electron moves this way... e.g. A light object (crampled paper) going down until gets hit by the wind will go parallel (at least a few seconds) to the wind direction ... why not with electron?

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A B-field does not behave like wind.
 
Dale said:
A B-field does not behave like wind.
I know, I just want to give an example of a force.. I don't understand why electrons move that way... If I push a table forward, it will probably move forward (except with some uncommon circumstances)... I can't imagine pushing a table forward and then it moves left, right, or any other direction.
 
The force of a moving charge in an magnetic field is given by
$$\vec{F}=q \frac{\vec{v}}{c} \times \vec{B},$$
where ##q## is the charge, ##\vec{v}## the particle's velocity, ##c## the speed of light (in a vacuum), and ##\vec{B}## the magnetic field.

This is sometimes a bit complicated to figure out. The cross product is telling you that the force is always perpendicular to the velocity of the particle and the magnetic field. You get the direction by the right-hand rule, i.e., putting the thumb of your right hand in direction of the particle's velocity and the index finger in direction of the magnetic field, then your middle finger points in the direction of the magnetic force on the particle.

Since the force is always perpendicular to the force, a magnetic field doesn't change the magnetitude of the particle's velocity but only its direction. For a homogeneous magnetic field the trajectory of the particle is a helix, i.e., it is moving with a constant component of the velocity in direction of the magnetic field, and the projection of the velocity to the plane perpendicular the field is a circle.
 
vanhees71 said:
The force of a moving charge in an magnetic field is given by
$$\vec{F}=q \frac{\vec{v}}{c} \times \vec{B},$$
where ##q## is the charge, ##\vec{v}## the particle's velocity, ##c## the speed of light (in a vacuum), and ##\vec{B}## the magnetic field.

This is sometimes a bit complicated to figure out. The cross product is telling you that the force is always perpendicular to the velocity of the particle and the magnetic field. You get the direction by the right-hand rule, i.e., putting the thumb of your right hand in direction of the particle's velocity and the index finger in direction of the magnetic field, then your middle finger points in the direction of the magnetic force on the particle.

Since the force is always perpendicular to the force, a magnetic field doesn't change the magnetitude of the particle's velocity but only its direction. For a homogeneous magnetic field the trajectory of the particle is a helix, i.e., it is moving with a constant component of the velocity in direction of the magnetic field, and the projection of the velocity to the plane perpendicular the field is a circle.

Can this: f = q + v x B be used instead to make it simpler? also can you please give simple value of q, v, & B, as well as the solution...
 
mo0nfang said:
I know, I just want to give an example of a force.. I don't understand why electrons move that way... If I push a table forward, it will probably move forward (except with some uncommon circumstances)... I can't imagine pushing a table forward and then it moves left, right, or any other direction.
A B-field is not a push. It is not a wind. It is not any other mechanical analogy.

A B-field exerts a force given by ##F=qv \times B##. That is it. If you wish to make an analogy with something else then it needs to behave approximately like that.
 
Dale said:
A B-field is not a push. It is not a wind. It is not any other mechanical analogy.

A B-field exerts a force given by ##F=qv \times B##. That is it. If you wish to make an analogy with something else then it needs to behave approximately like that.
I don't know what value will I substitute to q, v, & B to get the value of Force... Can you please give simple value of q, v, & B, as well as the solution...
 
mo0nfang said:
I don't know what value will I substitute to q, v, & B to get the value of Force... Can you please give simple value of q, v, & B, as well as the solution...
What do you mean by substitute a value?
 
mo0nfang said:
I don't know what value will I substitute to q, v, & B to get the value of Force... Can you please give simple value of q, v, & B, as well as the solution...
How about q = 1 C, v = 1 m/s down, and B = 1 T to the left. Then F = 1 N forwards. That would be the force on a 1 C charge falling at 1 m/s through a 1 T field.

If you want the force from a different charge or field configuration then substitute the values you want.
 
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