Question about forces due to magnetism

In summary, the force on a particle moving through a magnetic field is given by the equation $$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over F} = q\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over v} \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over B} $$The velocity is relative to any inertial reference frame and not necessarily relative to the source of the magnetic field. In order to determine the force on a
  • #1
DCN
8
0
The force on a particle moving through a magnetic field is given by:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = q\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
B} $$
Is the velocity ##\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} ## relative to the source of magnetic field or some third party observer?
 
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  • #2
The velocity is relative to any inertial reference frame. It is not necessarily relative to the source, particularly if the source is moving non inertially.
 
  • #3
If that's the case, then please explain this.

Let's say that there is a test charge ##q## whose position from the origin is given by the radial vector ##\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} ## and there is a source charge ##Q## at the origin moving to the right along the positive ##x##-axis with a constant velocity ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over v} _s}## with respect to ##q##.

Now, in the rest frame of ##q##, ##q## has a velocity of magnitude ##0##, so it will not feel any magnetic effects. The force it will feel will be solely from the electric field generated by ##Q##. The force in the rest frame of ##q## is then:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = {{Qq} \over {4\pi {\varepsilon _0}}}{{\hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}$$
If we consider a difference reference frame in which we are moving to the right with a velocity of ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _o}##, then ##q## will have a velocity of ## - {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _o}## and ##Q## will have a velocity of ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _s} - {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _o}##. The magnetic field then produced by ##Q## is:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
B} = {{{\mu _0}Q} \over {4\pi }}{{\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o}} \right) \times \hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}$$
The force on ##q## due to the magnetic field will then be:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = - q\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o} \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
B} $$
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = - {{{\mu _0}Qq} \over {4\pi {{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o} \times \left( {\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o}} \right) \times \hat r} \right)$$
So the total force the particle will experience in the moving reference frame is:
$${{Qq} \over {4\pi {\varepsilon _0}}}{{\hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}} - {{{\mu _0}Qq} \over {4\pi {{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o} \times \left( {\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o}} \right) \times \hat r} \right)$$
Which is clearly not equal to the force found in the stationary reference frame. Am I missing something here?
 
  • #4
Yes. The fields have to transform as well when you change frames.
 
  • #5
How does that work?
 
  • #6
I don't understand the question. There is a Lorentz transformation for electromagnetic fields. You change frames, you need to use it to determine how the fields transform.
 
  • #7

FAQ: Question about forces due to magnetism

1. What is magnetism?

Magnetism is a force that causes objects with magnetic properties to attract or repel each other. It is caused by the movement of electric charges in certain materials, such as iron, nickel, and cobalt.

2. How do magnets create force?

Magnets create force by exerting a magnetic field. This field is strongest at the poles of the magnet and can attract or repel other magnetic objects within its range.

3. What is the difference between a magnetic field and an electric field?

A magnetic field is created by moving electric charges, while an electric field is created by stationary electric charges. Both fields can interact with each other and with other objects.

4. How does the strength of a magnet affect its force?

The strength of a magnet is directly proportional to the force it exerts. This means that the stronger the magnet, the greater the force it can exert on other magnetic objects.

5. Can magnets attract non-magnetic objects?

Magnets can only attract objects that have magnetic properties. Non-magnetic objects, such as wood or plastic, will not be affected by the force of a magnet.

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