# A loop of wire with circumference of a lightyear is connecting a bulb and a battery. How long will the bulb take to light up?

Imagine a circular loop of conducting non wire with a perimeter of a light year. There is a powerful battery (able to produce a voltage that is higher than the drop in wire resistance) and a bulb connected in series using this wire. The bulb and battery are placed diametrically opposite. Let's introduce a switch in between. If I turn the switch ON, how long will take the bulb to light up?

Now another scenario. Here also the bulb and battery are connected using a light year long wire, but the bulb and battery are only 10 meters apart. Instead the wire is irregularly coiled, so that it takes one light year of wire to connect them. In this case, how long will it take for the bulb to light up?

Since the wire has inductance and capacitance, does the geometry of the loop affect the time? Does the physical separation between bulb and source affect the time?

• The second scenario is more interesting, besides, the first one has mostly been covered already: physics.stackexchange.com/q/6674/75633 . Commented Nov 19, 2021 at 15:20

When you abruptly turn on a voltage source $$V$$, the leading edge of a voltage step function propagates out into the wires either direction ($$V/2$$ from one side, $$-V/2$$ from the other) at the effective speed of light. This effective speed is determined by things like the geometry of the wire/waveguide and the permittivity and permeability of the surrounding media, but it’s usually on the order of the speed of light on vacuum.

When the edge of either voltage step function reaches your bulb, then current flows through it, and it lights. Once both voltage steps reach it, then it lights to full brightness.

It may be interesting to consider what happens if the wire is disconnected from one battery contact. That is, your circuit is broken. Then, the bulb will turn on when the voltage edge first passes. But the leading edge will reflect at the end of the wire/waveguide, propagating back toward the source, all the while cancelling itself as it propagates back. So when the edge reaches the bulb for a second time (a year later, perhaps?) then it will turn off.

• That was informative especially the last paragraph. What about the second case I presented, is the observation same there too? Also, what if the wires were aligned like a rectangle, with bulb a battery on the opposite longer sides, in this case will the transformer action between the wires affect the result, you said there will be a voltage wave, which causes a current wave which in turn create a magnetic field which induce a voltage on the other side of the rectangle. Here is the picture, ibb.co/JrSJFGk
– User
Commented Nov 18, 2021 at 18:01
• The voltage is defined between the two wires; how can one arrive independently of the other? Also, one wire does not need to be broken for the voltage step to be reflected. If the characteristic impedance of the wire pair does not match the bulb there will be a reflection. Commented Nov 18, 2021 at 23:03
• @ASWINVENU You could include that picture in your question. Commented Nov 19, 2021 at 15:18
• @DavidG25 (A) in this case, since the wire spacing is much larger than the wavelength (with frequency approximately the inverse turn-on time of the battery), the ground reference is essentially at infinity. Perhaps it would be helpful to imagine a grounded pin in the center of the loop which serves as a reference for both transients. (B) I agree that there will in general be a reflection at the bulb itself. I considered it an unnecessary complication for my answer. Feel free to revise. Commented Nov 19, 2021 at 15:39

One part of an electrical circuit can influence another mainly in two ways:

1. The first part influences the local current or voltage, and this influence propagates along the wires to the second part.
2. The first part influences the local electric and magnetic field around the wires, and this influences propagates outwards through the space around the circuit, eventually meeting each other part of the circuit which it can influence through things like electromagnetic induction.

At the time of writing of this answer, existing answers have commented on process 1 but not process 2.

In process 1 the influence propagates along the wires at close to the speed of light. So if the wire is a lightyear long, then it will take a year for this influence to go from one end of the wire to the other. However, if the wire is coiled up in a coil then process 2 will offer a way to take a short-cut, so for example a change in the current at one end of an inductor will propagate along the inductor faster than you would think if you only took account of process 1.

In the case of a circuit with very long wires, but where two parts are physically close to one another (say just a metre or so apart) then process 2 will allow some sort of influence to propagate between those places in a very short time (just a few nanoseconds). One has to calculate each case to see how significant this is. With a 1 volt voltage difference across a switch, and an ordinary light bulb, then when the switch is closed, would the electromagnetic induction effect produce enough current to see a glimmer of light from the light bulb, a year before the main change in the current arrives after propagating down the wire? I don't know. I guess that was calculated in the video that was the source of the interest in this issue.

• I was curious about the second one, sadly the video does not comment on it, the presenter is solely relying on Poynting theorem and Poynting vector, I have created another question which is particularly focused on that question from the video with more details, would you mind taking a look? physics.stackexchange.com/q/678359
– User
Commented Nov 21, 2021 at 9:31

I think it's far simpler than it sounds. The bulb will light up when current flows through its filament. And current is just motion of electron. The motion of electron will only occur when the electric field from the battery reaches the electron. This electric field propagates at nearly the speed of light (depends of medium). So, when the battery's electric field reaches the bulb, it will light up.

For your first scenario, it will take a little over half a year for the bulb to light up.

For the second scenario, since the bulb is only 10 meters away, the electric field induced by the battery will reach it immediately. So, it will light up immediately.

The simple formula is to consider how long it will take the electric field to propagate through the conductor to reach the bulb from the battery. That is how long it will take the electrons in the bulb to be energized and move. It's not entirely wrong to think of that as the speed of current. But it's not entirely right either. The concept is a bit more complex. I would suggest looking into electromagnetic field propagation and current induction for complete understanding.

PS: Sorry, I was half asleep and messed up a lot.