# Similarity between electromagnetic waves $E/B=c$ compared with the equality between Lorentz force and electric force

I am very tired and in this moment I have not think very well. Using the vectorial expressions of the magnetic and electric field if an electromagnetic wave propagates in a certain direction the electric field and magnetic field are always perpendicular to each other. They arise from Maxwell's equations in the absence of sources. Thus we have,

$$\boxed{\frac{E}{B}\equiv c} \tag 1$$

When it happens, physically or experimentally, that the Lorentz force $$\mathbf F_L=q\mathbf v\times \mathbf B$$ equals the electric force $$\mathbf F_e$$?

If $$\alpha=\angle\mathbf v\mathbf B=\pi/2$$ I obtain an expression similar to the $$(1)$$. In fact I have: $$qE=qvB$$ hence $$\boxed{\frac{E}{B}=v} \tag 2$$

What is the substantial difference between the two relationships $$(1)$$ and $$(2)$$? That the first formula refers to an electromagnetic wave and the second at a speed of any charge, according to a given direction of propagation, which does not reach speeds equal to those of the light?

The first boxed equation refers specifically to the ratio of the magnitudes of electric to magnetic fields in a electromagnetic wave propagating through a vacuum.

The second equation refers to the velocity at which a particle must travel for it to experience no force given a specific electric to magnetic field ratio. If you consider uniform fields (ignoring the impossibility of this from Maxwell's equations) then it is conceivable that a moving charged particle would experience no instantaneous Lorentz force at the velocity $$\frac{E}{B} = v$$

It is worth noting that many of these values for velocity even in every day settings could require near relativistic speeds, at which point this analysis becomes significantly more complex.

• Thank you very much for you answer. May 19 '20 at 19:24

I am supplying my answer because I don't think the answer accepted by the OP is detailed enough.

To arrive at the boxed expression one needs to realise that the electric field associated with an electromagnetic wave propagating in vacuum, i.e., empty space, can be written in the form

$${\bf E} = E\cos({\bf k\cdot r} - \omega t)\,\hat{\bf e}\,,$$ where $$E$$ is the maximum amplitude of the electric field, $${\bf k}$$ is the wave vector (which points in the direction that the wave propagates) and $$\hat{\bf e}$$ is unit vector that gives the polarisation of the electric field. The vectors $${\bf k}$$ and $$\hat{\bf e}$$ are perpendicular to each other.

In the case of the magnetic field associated with the electromagnetic wave, it can be expressed as $${\bf B} = B\cos({\bf k\cdot r} - \omega t)\,(\hat{\bf n}\,{\bf \times}\hat{\bf e})\,,$$ where $$B$$ is the maximum amplitude of the magnetic field, $$\hat{\bf n}$$ is a unit vector in the direction of $${\bf k}$$ and $$\delta$$ is the phase difference between $${\bf E}$$ and $${\bf B}\,.$$

For electromagnetic waves the relationship between $${\bf B}$$ and $${\bf E}$$ is $${\bf B} = \frac{1}{c}\,{\hat{\bf n}{\bf \times E}}\,.$$

From this is obtained $${\bf B} = \frac{E}{c}\,\cos({\bf k\cdot r} - \omega t)(\hat{\bf n}{\bf \times e})\,.$$ Comparing the two expressions for $${\bf B}$$ it can be seen that $$B = \frac{E}{c}\,\Rightarrow \frac{E}{B} = c\,.$$

In the case of the second formula, consider the Lorentz force acting on a charge $$q$$ moving with velocity vector $${\bf v}$$ in a region of space where there is an $${\bf E}$$ and $${\bf B}$$ present. The force is $${\bf F} = q({\bf E} + {\bf v}{\bf \times B})\,.$$ When the force on the charge is zero it follows the electric field satisfies $${\bf E} = - {\bf v}{\bf \times B}\,.$$ This happens when conductors carrying a steady electric current are placed in a uniform magnetic field. The uniform electric field established inside the conductor is called the Hall electric field. The principle is also used in velocity selectors, which are employed in the Bainbridge mass spectrometer for example. If $${\bf E}$$ and $${\bf B}$$ are uniform, then we can let, for example, the electric field point in the positive z-direction and the magnetic field point in the positive x-direction. In such a scenario $${\bf E} = E\,\hat{\bf z}$$ and $${\bf B} = B\,\hat{\bf x}\,,$$ where $$\hat{\bf z}$$ and $$\hat{\bf x}$$ are unit vectors in the $$z$$ and $$x$$ directions respectively. This means that $${\bf v} = v\,\hat{\bf y}\,,$$ where $$\hat{\bf y}$$ is a unit vector in the positive y-direction. The speed of the charge is constant because the force on the charge is zero. From this it follows that $$E\,\hat{\bf z} = vB\,\hat{\bf z}\,\Rightarrow v = \frac{E}{B}\,.$$

This brings me to my last point, which is to address the claim by author of the accepted answer that Maxwell's equations don't allow uniform $${\bf E}$$ and $${\bf B}$$ fields to be constructed. I disagree with this statement. Maxwell's equations are a set of first order partial differential equations and to find solutions to them one needs to supply boundary conditions, and as mentioned above uniform $${\bf E}$$ and $${\bf B}$$ fields appear in different scenarios in physics. An example of the use of a uniform $${\bf B}$$ field is in a cyclotron. In the case of the electric field, the electric field deep in the interior and away from the edges of a parallel plate capacitor will be very close to being uniform if the distance between the plates is much smaller than the length and the width of the plates. Of course, there are technical difficulties involved in producing such fields but over small enough regions of space, $${\bf E}$$ and $${\bf B}$$ can be considered uniform.

• In the meantime, I thank you infinitely and, as is my custom, I thank you further after this vote. I had not seen your answer and I accept it very gladly. Thank you. Jun 9 '20 at 20:02
• Thank you @Sebastiano. Jun 9 '20 at 21:52
• No no no don't thank me it's my pleasure :-) Jun 9 '20 at 21:55