# Is Magnetic force different in different frames?

Suppose a charged particle moves with a velocity $$v$$ near a wire carrying an electric current. So, a magnetic force acts on it. If the same particle is seen from a frame moving with velocity $$v$$ in the same direction, the charge will be found to be at rest. Will the magnetic force become zero in this frame? Will the magnetic field become zero in this frame?

Even though the solution says that the force shall be zero but the magnetic field will be independent of frame and henceforth exist, I am not convinced with the idea of zero force because even if I were to place myself in a frame moving with velocity $$v$$, I will see the particle executing a circular motion around a different center though. In reality I know that its the magnetic force that's causing it to move in a circle but in my frame if I say that there is no magnetic force, how would I explain the circular motion of the particle ?

As a rule of thumb, if you ever transform away a magnetic field the motion of any particle that was due to $$\vec{B}$$ in the original frame will be due to $$\vec{E}$$ in the new frame.

This is governed by the lorentz transformation of EM fields.

If a neutral wire had a current going to the right (i.e. electrons traveling left) and you then travel to the right at the speed of the current, the wire is no longer neutral. The negative charges experience greater length contraction relative to you in the new frame and thus linear charge density is nonzero. A positive charge outside the wire will be attracted to the wire by the $$\vec{E}$$ force now, where before you would have explained it with the $$\vec{B}$$ field.

If you check the direction of a B field around a wire, you'll notice they both bring positive charges moving with the current in towards the wire.

Frame 1:

$$\vec{B}_{wire} = \frac{\mu_0 I}{r} \hat{\phi}$$

$$\vec{F} = q\vec{v} \times \vec{B} = qv\hat{z} \times \mu_0 I \hat{\phi} = -\frac{qv\mu_0 I/r}{r} \hat{r}$$

Frame 2:

$$\vec{E}_{wire} = -\frac{\lambda}{\epsilon_0 r}\hat{r} ; \lambda = \frac{Iv}{c^2} = Iv \epsilon_0 \mu_0$$

$$\vec{F} = q\vec{E} = - \frac{qIv\mu_0}{r} \hat{r}$$

• This is not quite right. If you boost to a frame in which $\vec{B}$ is zero, the the magnetic force will be replaced by a solely electric force. However, the magnitude of the total force will also, in general, change. – Buzz Jun 7 at 0:48

The answer here is that you have to consider electromagnetic force as a whole. In special relativity you indeed learn that those 2 forces are actually related to the same phenomenon and when changing from a frame to another, electric field $$\vec{E}$$ and magnetic field $$\vec{B}$$ "mix" each other.

In your specific case you'll see the particle being at rest, but it will be subject to a force due to the electric field that you see because of the change of frame.

Will the magnetic force become zero in this frame?

It will, for the particle's velocity is zero in this frame.

Will the magnetic field become zero in this frame?

It will not. As long $$v \ll c$$, where $$c$$ is the speed of light, the magnetic field in this frame will be practically the same as in the original frame.

if I say that there is no magnetic force, how would I explain the circular motion of the particle ?

By the electric field. You have no electric field in the original frame. In your original frame of reference, you have only a magnetic field $$\vec{B}$$. However, your new frame of reference moves at a constant velocity $$\vec{v}$$ relative to the original frame. Hence, according to the law of electromagnetic field transformation, you will have an electric field in the new reference frame, namely $$\vec{E}=\vec{v} \times \vec{B}$$. This formula is valid as long as $$v \ll c$$.

In short, as long as $$v \ll c$$, the total of the electric and magnetic forces does not depend on the reference frame, but the choice of the reference frame affects the ratio of the electric and magnetic components of the total force.

Remember: the electric and magnetic fields depend on the choice of the reference frame.