# How antenna radiates(how currents flows through wire)

I don't understand how antennas radiate a signal.

I understand basics antenna(wavelength, electron E field, ...), but I simply don't understand how can current go through a wire that doesn't have negative pole.

Can you please explain that to me.

• @Ignac: It's a lot more than just a capacitor. That's really not a good way to describe a antenna, at least anywhere near its optimal frequency. Jun 25 '13 at 13:23
• Current is simply the movement of charge. Alternating voltage pushes and pulls the charge backwards and forwards in the 'wire'. it is both the positive and negative pole at different times. This movement of charge creates a changing electric and magnetic field which can create an electromagnetic wave capable of radiating energy from the aerial. (see Maxwell equations and Hertz) Jun 25 '13 at 13:30
• Is your question, how it radiates, or how current flows. Jun 25 '13 at 14:28
• "how can current go through a wire that doesn't have negative pole". Antennas have two poles, and can be considered as capacitors (current can move even if there is a gap, at least the time required to charge the capacitor, then current must be inverted). When an antenna has only one pole (monopole, whip), the ground (or the artificial ground plane) is the second pole. Now the current flows, it must be converted into a wave, this question is still a good one.
– mins
Aug 22 '20 at 0:32

I'm guessing you don't understand how current can flow if there is no complete circuit. Let's take a simple quarter-wave dipole as an example: simulate this circuit – Schematic created using CircuitLab

How can any current flow, since there is no complete circuit from "-" to "+" of V1?

Consider this: relative to the speed at which the waves in the electromagnetic fields propagate, the dipole is long. It's true that current can't flow, but it doesn't know that until it gets to the end of the wire. As the current approaches the end of the wire but has no place to go, the charges pile up until they are pushed back in the other direction. By the time it's back, it's travelled $\lambda/2$ or experienced a $180^\circ$ phase shift. The voltage at V1 has also changed by this point, and so the current is constructively adding to the new currents being produced by V1. If it were not for some of this energy being lost as radiation, the energy in this antenna would grow without bound.

Why the energy radiates is complicated. The long answer is "Maxwell's equations". If you don't want to understand all the gritty details of that math, then here's a simple, incomplete understanding: the current in an antenna is associated with a magnetic field, and the voltage is associated with an electric field. An antenna is an arrangement such that at some distance away from the antenna (the far field) these two fields are mutually perpendicular and in phase, and what you get is a self-propagating wave like this: Red is the electric (E) field, and blue is the magnetic (B) field. This is the sort of wave that would be emitted by a dipole aligned with the Z axis.

• Sorry Phil, can't agree with the idea of current bouncing off the end of the wire. Jun 25 '13 at 15:24
• @JImDearden so where does it go, then? Jun 25 '13 at 16:38
• The alternating voltage is moving (accelerating) the charge backwards and forwards. An observer looking at one point along the wire would 'see' this as an alternating current. At the end of the aerial the electric and magnetic fields (not the current) have nowhere to go and so are reflected back (just like light reflecting from a mirror). Depending on the relationship of length of wire to wavelength this will produce a standing wave pattern. Jun 25 '13 at 17:20
• I don't think "An antenna is an arrangement such that at some distance away from the antenna (the far field) these two fields are mutually perpendicular and in phase, and what you get is a self-propagating wave like this" is an antenna but rather what an EM wave is. You can have this without an antenna. Jun 25 '13 at 18:53
• @JImDearden sure, but that voltage at the end of the wire is the result of trying to cram charge into it, and that voltage exerts a force which, if unopposed, pushes the charge back out. That feels like charge bouncing off the end, to me. I have a harder time imagining voltage "bouncing", as forces, not being made of any "stuff", can't "bounce". Jun 25 '13 at 22:23

Here's an oversimplified version that helped me get past my own noob ignorance.

There are basically two types of small antennas: the small loop antenna, and the short dipole antenna. The small loop antenna is just a ring of wire, and any current in the wire produces a magnetic field surrounding the antenna. The device is an inductor, but one that has a large space-filling magnetic field.

On the other hand, the short dipole antenna is just a pair of metal "capacitor plates" sticking out into the air, and if a voltage is applied across them, there will be an e-field in the surrounding space. The device is just a capacitor, but again, it has a large space-filling field in the surrounding region.

Apply a sine wave instead of constant volts or current, and the fields around the "antennas" will expand, then contract to zero, then expand again but pointing backwards ...then repeat. No waves are generated, so they really aren't radio antennas at all. But they are creating some local EM fields in space.

Here's the "TEAL" video project at MIT with a visual version of the process:

Expanding/contracting b-field or e-field

OK so far? The loop antenna generates a magnetic field, and the dipole antenna generates an electric field. The weird stuff starts happening when we drive either antenna with very high frequency. That, or we can build a version of either antenna with such a large size that even 60Hz will be a type of "radio signal" as far as the antenna is concerned.

Here's the thing: the magnetic or electric fields surrounding those antennas cannot expand or contract faster than the speed of light. So, what happens if the AC pulses applied to these devices are "too fast?" The fields around inductors or capacitors have to balloon outwards and then get sucked back in again, but what if the speeds are nearly the speed of light? That's when the fields stop acting like inflating or contracting invisible balloons. Instead the fields start behaving as waves.

So, when we reverse the polarity during the AC sine wave, the e-field or the b-field doesn't get entirely sucked back in as usual. Instead it peels loose from the antenna and just keeps moving. Some of the field-energy isn't retrieved, and instead is lost into space. Our loop antenna is no longer just an inductor, and it's started making waves. And our dipole is now a wave-launcher and not just a capacitor.

YT vid: EM fields surrounding small antenna

• +1: This "oversimplified version" is an excellent step in the education process (speaking from the perspective of a senior EE taking undergrad Microwaves I) Oct 24 '13 at 3:52
• It also drives home the concept of "Nearfield region." The Nearfield is where the fields are sucked inwards, only to balloon outwards again. Outside the antenna's Nearfield, the flux-lines become closed circles, and they propagate one-way, outwards into the distance. Oct 3 '15 at 23:48
• A varying magnetic field "creates" a varying electric field and vice versa (so any antenna create both). But actually, in special relativity, one is an artifact of the other due to time contraction in the reference frame chosen. Magnetic and electric forces are only one single EM force.The EM wave is this expanding self (re)generation of EM field at light speed (we say it "propagates"). See this.
– mins
Aug 22 '20 at 0:40
• Thanks, this is a great explanation. Just to check, even in the very slow field case (with fields ballooning in and out) is there still a radio signal generated, technically? In the first TEAL video you linked I can see some field lines peeling lose just at the edge of the frame. In this case is the 'peeling' happening so far away that the field strength is really small or something?
– Matt
Sep 10 at 17:04
• @Matt heh, the TEAL vid is just a zoomed-in version of a normal antenna. When the device size is far far shorter than 1/2 wavelength, the "peeling" EM waves are gone, and all we have are the fields of coils and capacitors. Sep 14 at 1:48

Great question! Complex answer. To understand why this happens without a return path ("negative pole") you have to move beyond Ohms-Law.

All accelerated charges radiate. So everything that conducts alternating current acts as an antenna. However often they are poor antennas and don't radiate well. As a result this aspect can often be simply ignored to simplify the problem.

To make a good antenna you have to transfer power (the energy is contained in voltages and currents) into electromagnetic radiation (where the energy is contained in the E- and H-fields) travelling away from the antenna. This requires the impedance of your antenna to be roughly matched, and that the currents that cause radiation add up in-phase so they don't cancel each other out as they would in a transmission line. As Jim Dearden mentioned you can design this to get standing waves or cancel them out depending on the physical length.

The problem with your question about "not having a negative pole" is related to using a simplified circuit model that is unconcerned with the 3d aspects and fields of voltage and current. Current can flow in anything that is conductive (poles or no poles). External EM (electromagnetic) waves do this all time. However there is no ohm-law model that can predict this.

To move a step up from simple ohms law, engineers have adopted a "Radiation Resistance" model. This is used in a similar fashion as standard ohmic resistance. In ohms law the energy dissipated is turned into heat. In the radiation resistance model the energy dissipated is turned into, well, radiation.

Radiation resistance is only a simple tool to help engineers evaluate a known circuit element (i.e. usually some RF guy computed it for you) without having to use Maxwell's Equations and applying the boundary conditions to the physical circuit to understand exactly the modes of radiation.

The real key to understanding a circuit's behaviour is to understand when the radiation aspects are important to take into consideration. When the frequency of operation of a circuit has a wavelength that is physically close to the size of the circuit, then Ohm's Law starts to break down quickly. As a rule of thumb if the ratio between wavelength and circuit size is greater than 0.1 then you need to apply Maxwell's Equations to understand how that circuit is going to work. Thus the terms "quarter wave" antenna should be a clue that you need to apply EM theory to understand what the circuit does.

If you have time, try to digest this article on understanding EM Radiation. It is designed to tutor engineers in how circuits can work in ways that ohm's law fails to predict. It does have a lot of EM theory in it, but you don't need to really understand all that to appreciate there is a big difference in circuit analysis when your operating frequency gets close to the physical size of your circuit.

EDIT: I just thought of another example that might help. Capacitors have no return paths, they are just open circuits, yet somehow they work, right? This (and inductors which are just shorts) only work because of their radiation properties. Engineers have found a way to turn the EM equations into fixed elements (or lumped elements) so they can be incorporated into ohm-law models making them easier to work with. Like with antennas there can be much more going on than just a piece of metal sitting there going nowhere.

• Link is dead. Can you update it? Thanks! Mar 23 '18 at 18:36
• @robert updated link Mar 29 '18 at 19:14

I guess this approach though not completely correct may help. Try to imagine a battery and 2 wires connected at its terminals open ended. A potent exists in the battery. That means an electric field exists in the battery, now this field is through out the connected wire, causing accumulation of +ve and -ve charges at respective ends until same potential is achieved, this remains until the potential of battery is not changed. Now both the open ends are at the same magnitude of potential as that of the Battery. Now if I increase the potential of battery some more charges will move to the ends until potential is balanced. And when I decrease potential some charges will move back. Though the movement of charges is for a short period of time. This movement happens continuously when an AC voltage is applied, effectively oscillating the charges and hence producing EM waves. Hope this helps :)

This isn't really answering the Q perhaps, but unlike some throrough textual explanations, for me understanding a dipole (antenna) -and how it can radiate-, came from understanding the LC circuit Source: https://en.wikipedia.org/wiki/File:LC_parallel_simple.svg

after seeing this simple animation ("How a dipole forms"): That was really eye-opening, I hope it helps to enhance the other ton of text. • In the above (receiving antenna) animation, change "V" to "I" and I'll agree.
– Noji
Apr 21 '20 at 14:10

How currents flow thru the wire in an antenna has to do with the fact that the speed of light is finite, and the antenna has non-zero size (relative to the speed of light at the antenna’s design frequency), as well as non-zero capacitance. Basic physics.

Due to the speed of light being finite, one end of a non-zero length wire can be at different voltage and have a different charge than the other end, because the speed of light prevents them from instantly equalizing. Some time will be required (on the order of around a nanosecond for each foot of wire, or around 3 nS per meter, perhaps even a bit slower).

Say you connect a wire to a battery, current or electrons flow in one end and out the other. But what if the wire is so long it takes, say 0.25 uS for the speed of light to reach from one end to the other? Then, if current starts flowing in one end, that current won't really "know" if current is flowing out the other end of the wire into the battery until 0.25 uS later.

So, if you connect only one end of a wire to a voltage source, current starts to flow, and when it reaches the other end of the wire, charges up the far end of the wire, just like a capacitor, since it has no where else to go (no opposite battery terminal found). But if you are driving the near end with a 1 MHz oscillator instead of a DC battery, by the time the far end charges up, the near end is rapidly reversing voltage, just in time to discharge that capacitor (as it takes another 0.25 uS for that charge to travel back to the feed point).

That finite length of the wire also has inductance. That inductance will cause reverse EMF resisting the charge traveling up the wire. That resistance causes a loss of energy in the wire, and conservation of energy puts that energy into a electromagnetic field racing away from the antenna at the speed of light, and faster than any counteracting wave (caused by the charge in the wire reversing directions) can catch up and cancel it. Those alternating EM field fronts turn into standard RF waves as they radiate away from the near field of the antenna.

The negative pole of the circuit is the far end of the other half a dipole, which is getting charged and discharged in reverse. Or, in the case of a vertical monopole antenna, the planet earth (and/or the ground wire, the radio case, your hand, eventually the entire universe) ends up being the opposite plate of the capacitor.

Radio waves are invisible alternating current in the atmosphere. Light waves are visible alternating current in the atmosphere.

The antenna is a terminal of electric current; there is no current passing through an antenna, only the voltage oscillates with the input current. This oscillating voltage in the transmitter antenna induces an alternating current in the air, propagating away from the surface of the antenna at a 90-degree angle, passing through the air to reach receiver antenna and to induce oscillating voltage in it.

In the process, the antenna is like a balloon, the current is like air, and the voltage is like air pressure.

When air is pumping in and out of the balloon, the pressure in the balloon will keep changing and producing longitudinal sound waves in the air.

Similarly, when electrons are pumping in and out of the antenna, the voltage in the antenna will keep changing and producing longitudinal electrostatic waves in the air. This is, in fact, alternating current in the air.

In the vacuum space, Coulomb's force is the conductor of electric energy. The line of sight electrons on the surfaces of the antennas are constantly repelling each other with Coulomb's force. F=Ke x Q1Q2/R^2.

This repulsion force acts as a rigid rod without mass and body, and instantly transfers electric energy freely back and forth between the two antennas.

Hold a magnet in each hand, with the same poles facing each other. Do you feel the strong repulsion force? Yes. Wave one hand in and out. Feel the kinetic energy instantly transferred to the other hand? Yes. Are the two hands waving at the same frequency? Yes. Is there any magnetic wave traveling between the two hands? No.

The repulsion magnetic force is the conductor of kinetic energy between the two hands, enabling kinetic energy to freely transfer instantly. We can call this phenomenon magnetic radiation.

If we hold electrons in our hands instead of magnets, it is electrostatic radiation, a misinterpretation of electromagnetic radiation by scientists.

The alternating current direction is always perpendicular to the surface of the antenna, and it propagates in the air as a longitudinal wave.

• Welcome to Electrical Engineering! It looks like this is copy/pasted from somewhere else, which means you'll need to properly reference it. Jul 19 '19 at 6:12