Sayonara Posted April 2, 2007 Share Posted April 2, 2007 Partly a thought experiment, partly inspired by the "Line to the Moon" thread... Imagine if you will that YT is building a lamp in his work room. He has a battery terminal connected to a bulb with a short piece of wire. He connects his last piece of wire to the other terminal but is shocked to find that this piece is millions of miles long. YT decides to be fancy and launches the end of the wire into space with one of his home made rockets, then recovers it a year later (pretend this is at all possible). The larger piece of wire should now be assumed to occupy the same solar orbital in which Earth travels. The question is, when YT connects this end of the wire to the light bulb, will the circuit be completed (and hence the bulb illuminated) instantaneously? Give reasoning and neglect the heat effects of solar radiation. Link to comment Share on other sites More sharing options...
Klaynos Posted April 2, 2007 Share Posted April 2, 2007 Ignoring resistance. Well dissipation due to it. Nope, there is a fixed velocity (the speed of light in the material) for the propagation of the field within the wire. Link to comment Share on other sites More sharing options...
Sayonara Posted April 2, 2007 Author Share Posted April 2, 2007 I am going to add SUMS to this to make it more FUN for the bored ones. If you say "no - not instantaneous", then calculate time difference between connection of circuit and illumination of bulb. With and without resistance (assume copper wire of 1.4mm diameter). Link to comment Share on other sites More sharing options...
insane_alien Posted April 2, 2007 Share Posted April 2, 2007 i would have to disagree with you klaynos. the voltage provided by the battery would already be producing a local current. unless the current magically doesn't start till the voltage jump goes round the entire wire. that would mean that the battery could be disconnected and the bulb would stay lit wouldn't it? sayonara, if you include resistance then you must explain how he got a hold of a 3TV battery. oh wait, its YT isn't it. silly me, all is explained. Link to comment Share on other sites More sharing options...
Sayonara Posted April 2, 2007 Author Share Posted April 2, 2007 sayonara, if you include resistance then you must explain how he got a hold of a 3TV battery. oh wait, its YT isn't it. silly me, all is explained. lol You raise a good point though, indirectly. People trying to solve for this might want to declare some values (like, assume average radius of Earth's orbit, select arbitrary voltage, etc). Link to comment Share on other sites More sharing options...
Klaynos Posted April 2, 2007 Share Posted April 2, 2007 Well ignore IA, because I'm too tired to think it through properly. Useing the information that I and other contributed in: http://www.scienceforums.net/forum/showthread.php?t=25456&highlight=copper swansont's post #7, the given velocity of electric field in copper is: 225,000 km/s From wp (http://en.wikipedia.org/wiki/Earth), the orbital circumference is: 924,375,700 km which gives us: 1.1 ish hours. Link to comment Share on other sites More sharing options...
Jacques Posted April 3, 2007 Share Posted April 3, 2007 He connects his last piece of wire to the other terminal but is shocked to find that this piece is millions of miles long. If one end of the long wire is already connected, the potential will arrive at the other end after 1.1hr like Klaynos calculated. During all the years it took to streatch the wire on earth orbit the potential was there so when he connect the wire, the bulb will be illuminated instantaneously. Link to comment Share on other sites More sharing options...
Klaynos Posted April 3, 2007 Share Posted April 3, 2007 If one end of the long wire is already connected, the potential will arrive at the other end after 1.1hr like Klaynos calculated. During all the years it took to streatch the wire on earth orbit the potential was there so when he connect the wire, the bulb will be illuminated instantaneously. That's the bit I couldn't be bothered to think through properly Link to comment Share on other sites More sharing options...
YT2095 Posted April 3, 2007 Share Posted April 3, 2007 actually Propagation delays and standing waves are exploited all the time in RF circuitry. oddly enough, if I disconnected the other end once set up, it would also take that length of time for the bulb to extinguish, therein lies the basic principal behind fullwave, halfwave etc... antennas. the only difference is I would connect it in reverse polarity next time Link to comment Share on other sites More sharing options...
Sayonara Posted June 6, 2007 Author Share Posted June 6, 2007 That's the bit I couldn't be bothered to think through properly Did we ever decide on this? Link to comment Share on other sites More sharing options...
Klaynos Posted June 6, 2007 Share Posted June 6, 2007 From the point where the power supply is attached it will take 1.1h for the potential to propagate down the wire, after this point as soon as you connect the circuit into a loop current will flow. Link to comment Share on other sites More sharing options...
Sayonara Posted June 6, 2007 Author Share Posted June 6, 2007 So if YT connected both ends at the same time, it would take 1.1 hours, but as the question is phrased, it would be instantaneous. Link to comment Share on other sites More sharing options...
Klaynos Posted June 6, 2007 Share Posted June 6, 2007 So if YT connected both ends at the same time, it would take 1.1 hours, but as the question is phrased, it would be instantaneous. Yep Link to comment Share on other sites More sharing options...
YT2095 Posted June 6, 2007 Share Posted June 6, 2007 Totally Link to comment Share on other sites More sharing options...
swansont Posted June 6, 2007 Share Posted June 6, 2007 You probably would not even need the battery. If the loop was "cutting" magnetic flux lines, you would get a potential difference. At that size loop, probably a pretty damn big one, too. Link to comment Share on other sites More sharing options...
Sayonara Posted June 6, 2007 Author Share Posted June 6, 2007 Looks like you might need to rethink that design YT *bang* Link to comment Share on other sites More sharing options...
elas Posted June 11, 2007 Share Posted June 11, 2007 Current decays as it travels through a wire. The voltage needed at the start in order to get even a fraction of a volt at the end of a journey of this distance, would melt the equipment and probably a good bit of the surroundings. It just is not a practical thought experiment. elas Link to comment Share on other sites More sharing options...
Klaynos Posted June 11, 2007 Share Posted June 11, 2007 Current decays as it travels through a wire. The voltage needed at the start in order to get even a fraction of a volt at the end of a journey of this distance, would melt the equipment and probably a good bit of the surroundings. It just is not a practical thought experiment.elas I think we ignored resistive losses or put in some nice signal boosters along the transmission line. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now