Magnetic field of a moving point charge

Question

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.

Answer 1

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.

Answer 2

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.