# Can electricity be induced only by a magnetic field?

I just learned that Magnetism is just electrostatics observed from a different reference point.

Also recently learned that a emf is developed on the ends of a rod which is moving in a magnetic field . Also known as motional emf

Now , According to the Theory of Motional Emf , the potential difference generated by a Magnetic field on a moving rod is given by

$$\mathscr{E}=Blv$$

when $$l$$ is length of the rod and $$v$$ is velocity of the rod.

Now Imagine being in the frame where $$B = 0$$ (or $$\vec{E}=-v×\vec{B}$$).

According to this frame there should be no emf generated on the rod since Emf produced depends only upon $$B$$.(But thats not true).

So am I missing something in this ? How will someone explain the emf induced in the rod from a reference where $$B =0$$ ?

• Is there a frame where $B=0$? $E^2-B^2$ is Lorentz invariant.
– JEB
Aug 9 '20 at 14:19
• i think here is the point where Special Relativity enters in the field. that's how things may differ from frames to frames. Aug 9 '20 at 14:21
• "Frames" should refer to the velocity being a certain amount. So you could go to the frame where velocity is 0, which would also have zero emf, but in this case special relativity would save the day. This is explained well in Purcell and Morin. Aug 9 '20 at 15:49
• There are situations in which special relativity can be used to show the relationship between $E$ and $B$ in different frames. But that does not mean that electrostatics is magnetism from another view point. I can't see how the Coulomb field arises from magnetism, but maybe there is a way. I'd like to see it. Aug 10 '20 at 1:10
• @garyp youtu.be/1TKSfAkWWN0 Aug 10 '20 at 2:38

In order to avoid special relativity (explicitly), picture a coil with a current running through it. It generates a magnetic field perpendicular to the direction of the current (say, in parallel to the opening of the coil). Think of a very big coil with the rod moving towards one of its walls. Now, a moving rod that interacts with such magnetic field will get an emf induced because the charges inside it are moving perpendicular to the field and so they are forced to move perpendicular to those directions.

Here comes the fun part. A frame of reference where the magnetic emf is zero, is one that is moving with the charges in the coil that generates the magnetic field. So basically, rotate the plane in which the rod exists at an angular velocity that matches the one of the charges in the coil (the current). In this frame, the rod is moving in an outward spiral. The electric interactions are then towards the wall and won't contribute to the emf. Why would there be an emf induced then? The charges on one side of the rod will experience more force as what used to be a constant velocity is now an always changing acceleration (since the rod is rotating around the origin, moving outwards, and changing the direction it faces). Since both ends of the rod experience different accelerations (one always going in, the other out), because they have to constantly face the wall, a force gradient (difference) is induced. More importantly, since the angular velocity of the rod is constant, the orbital velocity of the rod must increase with the radius. Therefore the rod is being accelerated in a direction that's more aligned with one side than the other, this force acts on the individual charges, creating the flow of current. Space generates the force on the charges in this frame of reference.

I hope this helps!

I'm not sure if there is a frame where $$\mathbf{B}'=0$$. $$E^2-B^2$$ is negative in your original frame (assuming $$E=0$$), but it is positive in the new frame if $$B=0$$. $$E^2-B^2$$ is Lorentz-invariant, so this cannot happen.

However, to address the more general point, motional EMF exists to solve the types of problems you are posing. If we have an electric field and no magnetic field, then there will be potential differences between different points. When we switch to frames where the magnetic field is not zero, all the equations of motions have to work out correctly in the new frame. This requires motional EMF (among other things).

• And how would the Emf generated be described from a reference frame where Velocity of the rod is 0? Aug 10 '20 at 2:36
• In that frame, the magnetic field would be moving, so there would still be a magnetic emf. In addition, the electric field in that frame may contribute to the emf across the rod. Aug 11 '20 at 3:05

I just learned that Magnetism is just electrostatics observed from a different reference point.

No: that is not quite true. It is true that there are some simple situations in which there is a magnetic field in one frame of reference, and there is another frame of reference in which there is no magnetic field and only an electric field. So in that case one could say the magnetic field is the result of the moving electric field. But more generally the relationship between magnetic and electric fields is that they are two parts of one thing: the electromagnetic field. When you examine that one thing from different reference frames, the sizes and directions of the electric and magnetic contributions can vary.

Now for your question about the emf. In the frame where the magnetic field is zero there is an electric field. This electric field provides the emf you are asking about.

• So that means Emf can be induced by electric field also? Aug 14 '20 at 6:03
• @NoahJ.Standerson emf = a voltage difference which can cause a current to flow = $E l$ for a uniform electric field $E$ and length $l$ (and more generally you would do an integral) Aug 14 '20 at 8:01

Let me first address the point about a frame with $$\pmb{B}=0$$. As already mentioned by comments and other answers, this contradicts $$\pmb{E}^2 = \pmb{B}^2$$ ($$c=1$$) being a Lorentz invariant. Let me write down the explicit transformations for the fields: $$\pmb{E}^\prime_{||} = \pmb{E}_{||}, \ \ \ \ \ \ \ \pmb{B}^\prime_{||} = \pmb{B}_{||} \\ \pmb{E}^\prime_{\perp} = \frac{\pmb{E}_{\perp} + \pmb{v}\times\pmb{B}}{1-v^2}, \ \ \ \ \ \ \ \ \ \pmb{B}^\prime_{\perp} = \frac{\pmb{B}_{\perp} - \pmb{v}\times\pmb{E}}{1-v^2}.$$
Here, the subscripts parallel and perpendicular mean parallel and perpendicular to the velocity vector $$\pmb{v}$$. For the non-relativistic speeds of the wire we can safely only keep first order of the velocity, that is, put $$1/(1-\pmb{v}^2) = 1$$.

Looking at fields at the rod, at the beginning of the process (before any electric field start piling up) we have, $$\pmb{E}_{||} = \pmb{B}_{||} = \pmb{E}_{\perp} = 0$$, and we also have the magnetic field to be completely perpendicular to the velocity. So you see the transformed fields are, $$\pmb{E}^\prime_{\perp} = |\pmb{v}| |\pmb{B}| e_{up}, \ \ \ \ \ \ \ \ \ \pmb{B}^\prime_{\perp} = \pmb{B}_{\perp}.$$ Here $$e_{up}$$ is a unit vector pointing up, with reference to the figure you provided. We see the magnetic field doesn't change much. But now you can ask, if the electrons are not moving in this frame, what is causing the force. And the answer is of course now we have electric field. Indeed you see the strength of the field is $$|\pmb{v}||\pmb{B}|$$, so the emf induced by this field is $$|\pmb{v}||\pmb{B}|l$$ as expected.

• So that means electric field can also induce current ? So why dont we talk about current induction in electrostatics ? Aug 15 '20 at 3:34
• Yeah, of course an electric field can cause a current to move. Connect any conductive wire to a battery and the electric field will cause current inside that wire. In a way current caused by electric fields are easier to understand, since the force caused by them are more straightforward. Aug 15 '20 at 12:13
• But by that logic if we put a circular loop with a bulb in a electric field , the bulb should light up. Does it? Aug 15 '20 at 12:15
• Is it a uniform electric field? If yes, then no it will not light up. You basically wanna calculate the emf around the loop, that is the line integral of the electric field inside the wire. If this integral is non-zero, a current will flow. Aug 15 '20 at 12:18
• How is “that logic” implying it would? Aug 15 '20 at 12:19