# Question about forces due to magnetism

1. Jun 12, 2015

### DCN

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?

2. Jun 12, 2015

### Staff: Mentor

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. Jun 13, 2015

### DCN

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. Jun 13, 2015

Staff Emeritus
Yes. The fields have to transform as well when you change frames.

5. Jun 13, 2015

### DCN

How does that work?

6. Jun 13, 2015