I'm having a hard time to grasp one simple idea. Let us say we have a moving point charge, q, with velocity v, in the x direction. Now calculating both the Magnetic and the Electric fields is easy: We know that B=0 in the charge's frame, hence we obtain B in the lab's frame.

Now, let us say we would want to know the same fields, but with a stationary charge, positioned above the moving charge (perpendicular). I could not just surmise that there is no Magnetic field in the moving charge's frame since, in turn, the other one is now moving in the other one's frame, so in both frames there is Magnetic field.

I would be more than grateful if someone could settle this, thanks.


Remember that superposition holds for the electric and magnetic fields. That is, you can calculate them individually and then add their fields together to get the field at any point.

For the moving charge, $q_1$, the magnetic field is 0 in its frame but boosting to the co-moving frame, we have $$ \mathbf{B}_{q_1}=\gamma\frac{\boldsymbol{\beta}\times\mathbf{E}}{c}=\gamma q_1\frac{\boldsymbol{\beta}\times\hat{\mathbf{r}}}{4\pi\epsilon_0cr^2} $$ where $\gamma$ is the normal Lorentz factor, and $\boldsymbol{\beta}=\mathbf{v}/c$ is the (reduced) velocity of the particle in the lab frame (note that the above equation reduces to the Biot-Savart law for $\gamma\approx1$). Since, in the lab frame, the magnetic field of the stationary charge, $q_2$, is 0, then the total magnetic field is given by $\mathbf{B}=\mathbf{B}_{q_1}$.

In the case of the co-moving frame, as stated the magnetic field is zero for $q_1$ but now the charge $q_2$ is moving (in the opposite direction) and we get the similar magnetic field: $$ \mathbf{B}=\mathbf{B}_{q_2}=\gamma q_2\frac{-\boldsymbol{\beta}\times\hat{\mathbf{r}}}{4\pi\epsilon_0cr^2} $$

So, yes, there is going to be a magnetic field in both frames because you have a moving charge in both frames.


In the first case when you have only the moving charge (lets call it $q_1$) there is no magnetic field in a frame of reference which is moving with $q_1$. This is because the velocity of the $q_1$ is zero with respect to that frame. I gather that you already understand this.

However, when you add the second charge, which is motionless in the lab frame (lets call it $q_2$) there is now a magnetic field in the frame moving with $q_1$. This field does not come from the motion of $q_1$, however. The velocity of $q_1$ is still zero. Now, however, the velocity of $q_2$ which is zero in the lab frame is not zero in the frame moving with $q_1$. Thus, the magnetic field in the frame moving with $q_1$ comes purely from the motion of $q_2$ relative to that frame.

EDIT: I apologize I believe I misread your question initially. In order to calculate the magnetic field from a frame moving with $q_1$ all you need is the velocity of $q_2$ with respect to that frame, because of what I explained above (perhaps you already knew this). Then the magnetic field at a certain position is: $$ B = {\mu _0 \over 4 \pi } *{q_2 vsin(\theta) \over r^2}$$ where $v$ is the velocity of $q_2$ relative to your frame of motion, $r$ is the distance from the point charge to the point at which you are measuring the field, and $\theta$ is the angle between the velocity vector and the position vector of the point at which you are measuring the field. Note, this only gives you the magnitude of the field. To get the direction you can employ the right hand rule.

If you go to this link and scroll down to chapter 28 there is a nice image of this situation. With the moving charge, the displacement and velocity vectors, and the magnetic field at certain points.

  • $\begingroup$ Right, so how could one calculate the magnetic field? It's easy enough when in one frame B=0, but in our situation both frames have a moving charge, hence magnetic field which is not zero. $\endgroup$ – Danny Feb 5 '14 at 8:09
  • $\begingroup$ Each frame will experience a magnetic field that corresponds to the charge moving relative to that frame. The lab frame has a magnetic field that comes from the motion of q1 and the other frame has a magnetic field that comes from the motion of q2. Each can be calculated with the Biot-Savart law: maxwell.ucdavis.edu/~electro/magnetic_field/pointcharge.html $\endgroup$ – wgrenard Feb 5 '14 at 8:21
  • 1
    $\begingroup$ Also, note that when we talk about the magnetic field in such and such frame, the strength and direction of the field at any given point not only depends on the velocity but also the distance and angle away from the point charge. $\endgroup$ – wgrenard Feb 5 '14 at 8:29
  • $\begingroup$ I appreciate your elaborated answer, I do believe, however, that we share a misunderstanding. As it happens to be, the magnetic field is B=-(V/c)*E (i.e. minus beta times E). Which I find hard to understand, as this can only be obtained in a frame that has only stationary charges. $\endgroup$ – Danny Feb 5 '14 at 11:30
  • $\begingroup$ I'll be honest and say that I am not familiar with that equation and can't seem to find mention of it anywhere online or in my EM book. The closest equation I know that looks somewhat like that is B = E/c but that is for electromagnetic waves. Perhaps another user can enlighten us both. $\endgroup$ – wgrenard Feb 5 '14 at 17:05

Not the answer you're looking for? Browse other questions tagged or ask your own question.