# Nature of Electromagnetic Waves

## Nature of Electromagnetic Waves

We know that light is an electromagnetic wave. So is a radio frequency wave. From what I understand, the 50 Hz power frequency wave in our 230 V supply at home (60Hz and a lesser voltage in countries other than India) is also an electromagnetic wave. But this power frequency wave requires a medium to travel
It does not travel in free space. But light and radio waves can travel in free space.Why is this? Or, is my assumption that the the power frequency 50 Hz/60 Hz wave is electromagnetic,incorrect? Please Help....


You are mixing up two (related) phenomena: EM waves and currents in a conductor, I guess because they both “have a frequency”.

EM waves (i.e. perturbations of EM fields that can travel in space) can be generated by time-varying currents flowing in conductors, whereas EM waves hitting conductors can induce currents in them. That’s the relationship between the two phenomena. The frequency of the generated EM wave is the same as the current’s and vice versa.

This is the general picture, in qualitative terms. Whether or not, in a given case, the generation of EM waves is negligible it depends on the specific problem at hand, and the same is true for whether or not a non-negligible current is induced in a conductor hit by an EM wave.

As a very broad rule of thumb, the radiated emission by a conducting wire becomes relevant when its length is near the wavelength of the wave that would be radiated, which depends on the frequency of the electric signal. The relationship between wavelength \$\lambda\$ and frequency f in vacuum is:

\begin{align*}
\lambda=\dfrac{c}{f}
\end{align*}

where c is the speed of light \$\approx 3\times 10^8 m/s\$.

Therefore a \$f=50\$Hz current would produce an EM wave having a wavelength

\$\lambda=\dfrac{3\times10^8\, m/s}{50\, Hz} = 6\times 10^6 m = 6000\, km\$.

Hence you would need a wire of about that size (6000 km!) to make the radiation of EM waves at that frequency non-negligible.

Electromagnetic waves are generated by oscillating currents and voltages and they can induce oscillating currents and voltages. The wavelength plays a very important role in how an electromagnetic wave behaves. Light is extremely high frequency (many THz) and as a result has an extremely short wavelength (around 1 um). Generally light will interact more with electrons in atoms and molecules than electrons in wires due to the short wavelength. Radio waves are a more reasonable wavelength of a few meters to a few cm or even mm. Antennas to transmit and receive these frequencies are around the same size as the wavelength. Oscillating electromagnetic waves will drag electrons around and produce oscillating currents in any conductors that they encounter, and this is most effective when the conductor is around the same size as the wavelength. Very low frequencies (sub 100 Hz) have extremely long wavelengths, on the order of km. You need a really, really long wire to act as an antenna at these frequencies. However, it is still possible to utilize electromagnetic energy with such a long wavelength, you just can’t really transmit it wirelessly. When your wires are much shorter than the wavelength, you still get oscillating electric and magnetic fields, however they cannot really radiate as they cancel each other out at long range.

There’s an EM field around any electrical circuit, even a DC circuit. The electric field points from high voltages to low voltages, and the magnetic field circles around currents. They don’t radiate at low frequencies without a large antenna, but they do transfer power. This web page has some good pictures. Here’s one that shows the electric and magnetic fields:

Here’s another explanation with a side view. The arrows show the direction of energy flow. (The direction is given by the cross product of the electric and magnetic fields, if that helps.)

So why doesn’t this circuit radiate well at low frequencies? Because the wires are too close together! Let’s look at the magnetic field around a DC circuit. Here’s another image showing the direction of the field around current-carrying wires. Our wires have currents flowing in opposite directions, like the pair on the right.

Here’s a top-down view:

The wires produce magnetic fields that circle in opposite directions. Between the wires, the fields point the same way. Outside of the wires, the fields point in opposite ways. Thus, outside of the wires, the fields (almost) cancel out! This happens because the current is constant everywhere in the circuit. At low frequencies (long wavelengths), this is approximately correct.

At higher frequencies (shorter wavelengths), different parts of the circuit have different currents! This is because changes in voltage and current travel at the speed of light. If the circuit is large enough or the wavelength is small enough, both wires can have current flowing in the same direction, which means they no longer cancel out.

The wavelength of an electrical signal is given by:

$$\lambda = \frac c f$$

where \$c\$ is the speed of light and \$f\$ is the frequency of the signal. For a 60 Hz sine wave, the wavelength is:

$$\lambda_{60 Hz} = \frac {3 \times 10^8 \mathrm{\frac m s}} {60 \mathrm{Hz}} \approx 5000 \mathrm km$$

So unless you have a very large current or a very large antenna, you’re not going to radiate much power at 60 Hz.

A propagating EM wave has the magnetic field (H field) in “time” phase with its E field and the ratio of E to H has to broadly match the impedance of free space (approximately 377 ohms). The magnetic field is also mechancially at right angles to the electric field.

This arrangement of E and H field naturally propagates thru free space and is called a radio wave.

An antenna creates both an electric field and magnetic field and although the near-field E and H components do not correspond with the above description of an EM wave, at some distance (usually about 1 wavelength), those fields combine to make a proper EM wave as per above.

This doesn’t happen with power transmission cables.