A recent Veritasium video discusses the following circuit:

,_____|L|_____,    _
|             |    | 1m
\_____|B|_/ __/    -

  1 lightyear

L = Light bulb
B = Battery

Diagram of light bulb battery circuit

His claim is that when the switch is closed it takes $\sim1/c$ seconds for the light bulb to turn on (since the light is turned on by the magnetic field whose origin is only $1$ m away from the light bulb). My question is, doesn't that violate the transfer of information at the speed of light? If you close the circuit, the light bulb would turn on, but if you move the switch so it's an equal distance from the light source and battery, then presumably you would be able to transfer information (switch is opened vs. closed) to points L/B at faster than the speed of light. What am I not understanding?

  • $\begingroup$ There's a better image (native to the veritasium material) at youtube.com/channel/UCHnyfMqiRRG1u-2MsSQLbXA/… - someone with a better connection than me can probably edit it in (and hopefully also fix the link) $\endgroup$ Commented Nov 19, 2021 at 21:13
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    $\begingroup$ Nowhere in the video is the claim you're asking about here made - the switch is always next to the voltage source there. It is not clear why you think the video implies this would work the same way when the switch is located far away. (Think about where the magnetic field comes from when trying to figure out why the location of the switch still matters) $\endgroup$
    – ACuriousMind
    Commented Nov 19, 2021 at 21:32
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    $\begingroup$ @lalala Suppose the long cable is open at the end. When the switch-closed signal reaches the end, a reflected signal will travel backwards from the open part of the cable, in such a way that the zero-current condition at the broken cable is enforced. But for that matter, if the long cable is shorted at the end, the end of the cable will generate a reflected signal that enforces the zero-voltage condition at the short. To have the circuit reach equilibrium immediately when the signal reaches the end of the long cable, it would need some impedance-matching resistor to swallow the reflection. $\endgroup$
    – rob
    Commented Nov 21, 2021 at 12:31
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    $\begingroup$ @Mauricio If we were respecting units, we could not say the "time is (1m)/c seconds" (as that has units of s²). We would have to say "the time is (1m/s)/c seconds", or "the time is (1m)/c". $\endgroup$ Commented Dec 1, 2021 at 18:28
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    $\begingroup$ @electronpusher ok let’s say 1/c meters $\endgroup$
    – Mauricio
    Commented Dec 1, 2021 at 20:41

5 Answers 5


This is a transmission line problem. Transmission-line problems can be counterintuitive for people who are used to thinking about the moving charges in circuit but are used to ignoring the fields. (One example here).

Let’s imagine that the long cable to nowhere is not pair of parallel wires, but instead that the "outgoing" and "return" wires are wrapped around an iron core, like a transformer. In the approximation that the freshly-connected battery is only pushing current along the outgoing wire, this high-inductance makes it a little more obvious that the current cannot instantaneously start moving down the entire outgoing wire. The basic principle of an inductor is that the magnetic field has some intrinsic "inertia." If the magnetic field through some loop isn't constant, an electromotive force develops around the boundary of the loop; charges which are free to move under the influence of this e.m.f. will generate a current which tries to prevent the magnetic field from changing. If you treat the outgoing wire, wrapped around an iron core, as a single solenoid, each coil in the solenoid will be fighting the changing magnetic fields as the current changes in the neighboring coils.

But if the wires for the "outgoing" and "returning" currents are wrapped around the same magnetizable core, there's a much easier way for the entire system to fight the changes in the magnetic field: the outgoing current can induce a current in the return wire. (Of course I already gave away the game by talking about transformers.) Note that the direction of the return current, in this transformer-transmission-line model, depends on whether the two coils are wound with the same handedness or not. But the current in the outgoing wire will unquestionably induce some nonzero current in the return wire.

Now let's unwind our two wires and return to the parallel-cable setup from the video. Unwinding the wires from the imaginary transformer, and removing the imaginary magnetizable core, dramatically reduces the mutual inductance between them —— but that inductance is still nonzero. Pick any two points along your lightyear-long cable, and the gap between the two wires makes an Amperian loop:

                           A        B
,_____|L|__________________⋮________⋮___________,    _
|                          ⋮        ⋮           |    | 1m
\_____|B|_/ _______________⋮________⋮___________/    -
                           ⋮        ⋮
                           C        D

If the battery changes the current along the segment $CD$ from zero to not-zero, the magnetic flux changes in the loop $\mathit{ACDB}$, and the electromotive force associated with this changing flux will induce an opposing current in the segment $BA$. This is happening in both arms, and the induced current flows through the lamp, which illuminates.

Now the amount of current that flows through the lamp, and the time it takes to reach a steady state, will depend in a complicated way on the transmission-line characteristics of the cable: its inductance, capacitance, series resistance, and shunt conductance (all traditionally measured per unit length). The point of the linked video is to remind you (or teach you) to think about these circuit properties in terms of their fields, rather than only in terms of the motions of charges.

but if you move the switch so it's an equal distance from the light source and battery?

I'm not 100% sure I understand your proposal here. But suppose there is a second switch, which is also initially open, located many light-seconds down the transmission line:

,_____|L|_____________________________________,    _
|                                             |    | 1m
\_____|B|_/ _________________________/ _______/    -
          S1                         S2 

If we close $S_1$ and leave $S_2$ open, the signal will propagate down the cable as before until it reaches $S_2$. The open switch $S_2$ will effectively function as a series capacitance, which will change the impedance of that section of cable. At an impedance mismatch in a transmission line, part of the signal is transmitted and part of it is reflected. However, the signal that "$S_1$ has been opened" has already been propagating down the cable, with current in both wires, and the lamp has already illuminated.

Likewise if we close $S_2$ but leave $S_1$ open, that signal will propagate to the left down the cable from the location of $S_2$. In this geometry, the lamp will be illuminated by the part of the signal transmitted through the impedance mismatch at $S_1$.

Electromagnetism is local. Think about the fields.

For what it’s worth, I did nearly this exact experiment (mostly by accident) back when I was a postdoctoral researcher. The context was sending pulses with a width of only a few nanoseconds down a cable with a propagation time of a few microseconds, to an LED next to a photomultiplier tube, then collecting a few-photon response from the detector a few microseconds later. My function generator output was connected by a “tee” to an oscilloscope; the cable to the scope was short, but still longer than the nanosecond rise/fall time of the brief pulse. So my setup was something like

|scope|                                       |LED|

One of the things I did as I was building this setup was to send long (millisecond) pulses, so that microsecond-scale transients had time to dissipate and the system entered a steady state. That’s roughly analogous to the Veritasium scenario of connecting the battery and waiting a year for the current to get to the end of the lightyear-long cable. If a millisecond-long pulse has a nanosecond rise (or fall) time, that leading (or trailing) edge and its reflections behave in the same way as the leading (or trailing) edge of the few-nanosecond pulse.

I might reproduce the thought experiment here by making the following changes:

  1. Introduce the battery in series with the center conductors of the coaxial cables, using a box like this one. A ground-isolated bipolar d.c. power supply might be easier to connect, but its complexity would distract a skeptical observer.

  2. In another box (or perhaps even the same box), put a small series resistance in series with the sheath conductors of the coaxial cables. Connect an oscilloscope (in its high-input-impedance mode) in parallel with this resistor. Ta-da, a fast ammeter.

  3. In a third box, perhaps with three inputs, an isolated transistor switch driven by the fast function generator between “conducting” and “insulating.”

  4. Complete the circuit by putting a zero-resistance terminator at the ends of the long cables. If I were actually doing the experiment, I’d also try swallowing the echos using an impedance-matched terminator, as well as leaving the cable ends open to demonstrate the current flows in both conductors until the reflection arrives from the open end.

The prediction in the video, as clarified by this question, is that if the resistor in series with the sheath conductor is near the switch, the current in that resistor will change when the state of the switch changes, in addition to when echos of the state change arrive from the ends of the long cables. Also, the timing of that current is essentially the same whether the switch interrupts the center conductor or the sheath conductor of the cable.

If I still worked at a place where I had all of these fast electronics lying around, I could probably put together a demo in an afternoon or two. (There would be some tedious issues related to the practice of fast electronics sharing a common ground.) Unfortunately I disassembled this setup a decade ago.

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    $\begingroup$ Hi rob, I just discovered your answer after I started writing an answer in the transmission line context here. I've linked to your answer in a comment. $\endgroup$ Commented Nov 28, 2021 at 22:15
  • $\begingroup$ What is the magnitude of the induced EMF along the segment BA when you consider Faraday’s law? $\endgroup$
    – Dave
    Commented Nov 30, 2021 at 12:16
  • $\begingroup$ @Dave See the excellent answer by Alfred Centauri, linked in the comment above yours, which computes the impedance of the transmission lines and sets up the voltage-divider equation. You could slice the transmission line at any point and do the same calculation. $\endgroup$
    – rob
    Commented Nov 30, 2021 at 16:36

Let me try to answer this without referencing the actual details of electromagnetism. When you have to worry about the finite speed of information propagation in the system (in this case, the finite speed of light), you always measure time delays due to a perturbation from the point of perturbation. Here, the perturbation to the system (circuit) is the switch closing. The location of the battery is really a red herring.

So, if the switch is close to the battery (1 m from the bulb) then the time to light up is $1/c$ m as advertised. If the switch is far away, then the EM fields are generated from the closed switch and take longer to propagate to the bulb/battery. So causality is intact :)

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    $\begingroup$ So is it the position of the switch or the position of the bulb and battery? // What if the bulb and battery were at the "far ends" of the wires (i.e. ~1 light year from each other)? Would the bulb still turn on in ~1/c seconds? Or is it the same and the only way to increase the turn on time the distance of the switch from the battery? That is, the longest possible turn on time is when the switch is ~1/2 light year from either the bulb or battery. $\endgroup$
    – KDecker
    Commented Nov 19, 2021 at 21:40
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    $\begingroup$ I think like @dkarkada said that the position of the battery is a red herring. Assuming the battery has been hooked up to the switch for long enough, there is a potential difference across the switch when it is open. So when the switch is closed, current will start to flow, and a signal from the switch will propagate to the light bulb to generate a current due to induction. $\endgroup$
    – Andrew
    Commented Nov 19, 2021 at 21:43
  • $\begingroup$ Regardless of what the video said, 1/c is not a measurement of time. It has units of seconds per meter. $\endgroup$
    – Matt
    Commented Dec 1, 2021 at 12:48

I hope this answer is not overlooked or taken as naive. I would be sad if it is downvoted without comments.

There are two ways to see why the answer in the video is wrong,

  • From the causal point of view, if the light bulb lights up after $1/c$ seconds then, all what matters for the bulb in the damn universe (imagine you are the light bulb) is a battery with a switch (on) beside to be detected in a radius of $1$ m. It doesn't matter if the wires are connencted to this bulb or to another light bulb. That implication is absurd since if we wait enough, in case it is connected to another bulb, that other would be the one that lights up not the first.

  • From the point of view of fields, it has to take time for the fields to take the shape of the steady-state in which the poynting vector converges into the bulb, and of course that doesn't happen before all parts of the circuit have interacted with other for a while. And of course considering your self the bulb again, will tell you that, for the configuration in the video, you are the first part on the other side of the circuit that recieves the signal and what ever the shape of the field is it will not be with poynting field converging into the bulb; why would it converge to the bulb not to any other point (in a ball of radius $1$ m, around the battery) if nothing would have had time (within $1/c$ seconds) to send light back to the battery conveying the info that there where it should converge ? that happens only after all parts interacted with each other, i.e. after sufficiently long time.

  • $\begingroup$ Veritasium is only slightly wrong. In theory a switch closing anywhere in the universe is enough to "light up" the bulb. A closing switch causes an electromagnetic disturbance that travels outwards. That can cause a non-zero Poytning vector in some other nearby wire and in turn that can create a small current . See also this video youtube.com/… . Every realistic wire has some inductance so you can see this setup as a transformer with very large spacing between the two coils. $\endgroup$ Commented Dec 16, 2021 at 12:30
  • $\begingroup$ What about the first point ? $\endgroup$
    – Physor
    Commented Dec 16, 2021 at 12:31
  • $\begingroup$ Veritasium is wrong in the sense that this current will be too tiny to light up any physical light bulb but in theory a small current will go through the bulb. $\endgroup$ Commented Dec 16, 2021 at 12:31
  • $\begingroup$ should I remove my answer ? $\endgroup$
    – Physor
    Commented Dec 16, 2021 at 12:32
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    $\begingroup$ Yes I thougth about that tiny short current because of induction, but I was talking about the man current that flows in the steady state. Thanks for the comment by the way $\endgroup$
    – Physor
    Commented Dec 16, 2021 at 12:36

I like the simplification of taking 1/2 of circuit out. Another simplification is take out the load replacing with wire. Now suppose we give a range of behavior for unknown C and unknown L and based on this unknown fr=12π√LC Where: fr = resonant frequency (Hz) L = circuit inductance (H) C = circuit capacitance (F) V is unknown but battery so DC

This is a tank circuit when load is replaced with a wire. R = 0 so tank circuit is going to oscillate on start of any current. Tank will keep the ramp up to a DC going storing charge initially in the magnetic field full stop current collapses the field to the capacitor. This goes back and forth where energy transfers from flux to charge and flux and charge... Because we have perfect circuit and zero resistance. Point of experiment is to say flux transfers energy. True, but false in how. The DC well takes it's time and moves like voltage does in a wire. So you got the AC oscillation riding on the DC rising voltage. W/out load is the simplification. This is called Norton's current. Simplified current source and load will with unknown's disparate the energy going back and forth. So depends may have just a glitch and may have a little second glitch. And tiny load with big current may have a long ringing AC voltage that DC settles after time for 1 second. The AC doesn't last long as it is consumed quickly if load is applied to a light bulb.


What Veritasium's video is describing is electromagnetic induction and EM power and signal wireless transmission. But the circuit he used in his thought experiment is wrong. Unless the voltage source is a 50-60Hz AC source and in the range of MVolts, a filament lamp or LED at 1m away from the voltage source will never light up! This path of electromagnetic energy Poynting vector is not possible or not sufficient to power up these lamps in this case described. The longer path, less resistive since he assumes in the beginning of the video that the cables have zero resistance (air is an insulator) will still need to be taken. Thus the a.c. electromagnetic field still needs 300,000 Km after the switch is closed, to travel and reach the lamp and start the oscillations of the electrons in the lamp that will start to emit visible light and heat.

So the correct ideal case answer (i.e. zero resistance wires but still a realistic small consumption incandescent lamp) for Ve's experiment setup and for an a.c. voltage source which can be turned ON/OFF for this type of lamps is 1s to actually light up and not (1/c)m thus ~3.3 ns at 1m as stated in the video. (Ve's video has too many unrealistic assumptions, zero resistance 300,000Km wires, incandescent lamp which lights up with infinitesimal small current! This circuit and specs he used is inappropriate to demonstrate the concept of wireless EM transmission of electric power).

On the other hand a fluorescent lamp with a few hundreds of KVolts a.c. voltage source will probably light up at 1 m distance and without being connected to any cables or switch.


As for the case actually Ve tried to demonstrate of a d.c. voltage source and a transient signal caused by closing the switch in the circuit shown, in an ideal case scenario conditions as also described in the previous paragraph for a.c., using a realistic small consumption incandescent lamb, again, unless the voltage source is in the range of MVolts, X(L)=ωL, X(C)=1/(ωC) inductive and capacitive reactance is so large in this circuit that only a very small fraction of voltage drop is generated on the lamp by wireless or wired power transmission, not sufficient realistically to power any incandescent lamp. The power on the lamp would be completely dumped by the enormous total 600,000Km circuit loop inductance and 1m distance air dielectric tiny capacitance even assuming zero ohmic resistance wires were used. Therefore, the last choice in Ve video, thus nothing of the above and specifically the lamp will not light up is the most plausible result in this case.

As an indication, the inductance L of a 300,000Km loop is enormous about 8E9 H for a 2cm wire thickens and single loop and the distributed segment capacitance for 2cm thick parallel wires at 1m distance and air as a dielectric, is less than 3nF which is a tiny small capacitance, not sufficient to transfer the needed amount of charge to the lamp in order to light it up during the switch d.c. transient. There might be some objections at this point regarding the capacitive coupling of the transmission line demonstrated in this circuit of Ve, that a capacitor at the start of its charging behaves like a short-circuit? This is an abstract generalization and leads to misconception. Although, for very large capacitors this might be true in some extend, all capacitors have a small internal ohmic resistance R in series. Further, even if we accept Ve explanation of the capacitive coupling of the power transfer at 1m distance sufficient to light up the lamp, the realistic time will not be 1m/c~3.3ns but relative much larger, constraint by the RC time constant of the capacitive coupling where assuming for example a R=100 Ohms, ohmic resistance of the lamp and C=3nF capacitance between the two parallel wires at 1m distance, RC time delay is 330nS.

Even further, taking the Antenna transmission-reception argument for the 3.3ns signal transient period thus effectively an a.c. signal with ~λ=1m wavelength, this would result in an 3E8λ transmission antenna dipole! Meaning this kind of antenna would not transmit EM waves even at 1cm away.

In short, Ve's circuit used in his thought experiment is invalid and unsuitable IMO to begin with. Although, he has good intentions to demonstrate the wireless transfer of EM power. However, there are no lies told as he says in his video. EM fields prefer to travel always in the more conductive path or medium.

As for the specific question asked here, regarding violation of the speed limit of information transfer due to the position of the switch in the circuit, the relative position of the wire switch does not matter , because the potential difference is already transferred from the source battery at the open end leads of the switch before any attempt is taken to close this switch.

Update 1st Dec 2021:

Despite all the above mentioned and willing to adopt this time, all the conditions and assumptions of Veritasium's video thought experiment and totally ignoring any capacitive and inductive reactance delays and signal losses, I cannot take it out of my mind that the given explanation result in the Ve video is still not correct from a pure electrical-circuit theory point of view.

The question you have first to answer is, is EM field propagation in a conductive loop circuit the same as EM wave propagation from a transmitting antenna for example?

In the presented circuit thought experiment of Ve the battery is already connected to the conductive circuit prior the switch is closed. Therefore, the positive lead battery potential as shown, is already present along the whole circuit up to the switch right side connecting lead as well as the negative potential of the battery at the left side connecting lead of the open switch prior to the switch being closed.

Ve's thought experiment, correct explanation

Note: Electron flow of current is used in the above schematic

Assuming our conductive circuit has ideally, inductance $L=0$ and capacitance $C=0$ and therefore $t=L/C=0$, as soon as the switch is brought in the closed position (assuming this is done instantaneous at time t=0 as shown in the figure), the potential difference is present at any segment of the circuit meaning an electric field vector E is generated simultaneously at any point of the circuit including the incandescent lamp which is actually a resistor connected in series in the circuit.

There is no time delay in this case described for the electric field to extend out from the positive lead of the battery and circulate along our circuit, as some would believe will hold also in this case described, confused with the EM waves radiation from an antenna and propagation of electromagnetic waves in space in general. The electric potential is applied as soon as the switch is closed and the d.c. transient signal due the switching is generated simultaneously at any point of the circuit shown.

There is no violation of any speed of information transfer limit c, since a potential is not energy although a potential difference between two connected points in a circuit can create an energy flow thus an electric current.

Therefore, the correct answer according to all technical circuit conditions and ideal assumptions described in Ve's video and herein is as following:

  1. t=0, when assuming that the lamp lights up as soon as the phase current enters the lamp Iin. There is no time delay, the lamp will light up instantaneously.

  2. t=d/c, time delay when assuming that the lamp lights up as soon as the phase current exits the lamp Iout. Where d is the total length of the filament inside the lamp and c the speed of light.

Notice here, that in case 2 above assumption that the lamp would light up only if the phase current would pass through the total resistive path of the lamp, although the electric field is instantaneously applied to all points in the circuit as described, the force $E=qE$ on each charge still propagates form charge to charge at the speed of light c ideally assumed in our system or realistically speaking, at the speed of light inside the wire.

Conclusion: Non-intuitively, the speed of the conductive circuit path in Ve thought experiment with all the above assumptions explained herein, will be faster than the 1m air distance cap, path! Thus the lamp will start to light up continuously (i.e. steady state) at time less than 1m/c, thus t<3.3 ns. Meaning, in this scenario explained, the lamp would reach steady-state operation before the transient wireless EM pulse generated covers the 1m distance at ~1m/c time and reaches the lamp.

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    $\begingroup$ Your assumption that a particular frequency is required is equivalent to an assumption about the inductance of the long cable-to-nowhere. I think your assumptions are reasonable for cables that could be produced by ordinary industrial processes. A cable of length 300,000 km is a different animal. $\endgroup$
    – rob
    Commented Nov 21, 2021 at 0:01
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    $\begingroup$ @rob The whole thought experiment with the long wires is wrong and a distraction. At sufficient radiated electromagnetic a.c. power, the lamp at 1m distance would light up anyway without any cables. You don't need to get in such extend to demonstrate electromagnetic induction. A simple transformer example would suffice. The lamp just needs to be close to the a.c. power source. $\endgroup$
    – Markoul11
    Commented Nov 21, 2021 at 1:10
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    $\begingroup$ The circuit diagram clearly shows a battery, so I'm not sure why you're talking about AC sources. $\endgroup$
    – Sandejo
    Commented Nov 21, 2021 at 1:38
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    $\begingroup$ That's true, which raises the question of why you're talking about that, rather than what the question actually asked about. $\endgroup$
    – Sandejo
    Commented Nov 21, 2021 at 3:39
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    $\begingroup$ @Markoul11 Your answer states correctly that, in this circuit, the duration of the transient behavior caused by the closing of the switch lasts for a timescale set by the propagation time down the long cable. Now, in the comments, you seem to be claiming that the battery makes it a direct-current circuit, in which inductance is irrelevant. You can’t have it both ways. The transient behavior is governed by the capacitance and inductance of the circuit, and current flows through the lamp during the transient, before the eventual steady-state behavior is reached. $\endgroup$
    – rob
    Commented Nov 21, 2021 at 12:14

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