Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84; site watdcsu.UUCP Path: utzoo!watmath!watnot!watdcsu!sgcpal From: sgcpal@watdcsu.UUCP (P.A. Layman [EE-SiDIC]) Newsgroups: net.sci Subject: Re: Question about Electricity Message-ID: <1821@watdcsu.UUCP> Date: Fri, 1-Nov-85 10:51:32 EST Article-I.D.: watdcsu.1821 Posted: Fri Nov 1 10:51:32 1985 Date-Received: Sat, 2-Nov-85 05:58:54 EST References: <621@hlwpc.UUCP> <662@petrus.UUCP> Reply-To: sgcpal@watdcsu.UUCP (P.A. Layman [EE-SiDIC]) Distribution: net Organization: U of Waterloo, Ontario Lines: 60 Summary: In article <662@petrus.UUCP> mwg@petrus.UUCP (Mark Garrett) writes: >++ >> If you put a live electrical wire into a large >> swimming pool, what happens to the current? > >This is just an educated guess but, I would think that the current >would flow from the point of the end of the wire to the ground in >a quickly widening cone. Since the electons repel each other, the >main current would take up as much of the conductor as it can while still >having some component in the direction toward the ground. In addition, >there might be eddy currents all over the place, especially if it is >AC or (worse) lightning, which is why you would probably get electrocuted >no matter where you were in the water. >-Mark Guess is right. Althougth their is no simple answer to the question, we can quickly approximate 2 cases, using the point form of ohm's law, which is: _ _ J = 1/p E. _ _ In this equation J and E are current density(A/(cm**2)) and electric field (volts/cm) and are Vectors. p is the resistivity(ohm-cm) and is a scalar. If we assume the walls of the pool to be an equipotential surface of 0 volts, and the conductor to be insulated except for the end which is situated the middle of the pool at some potential V, we find that near the end of the conductor the equipontential surfaces are roughly spherical, and thus E is roughly uniform, and J will be constant in all directions. Since the total current flowing through an equipotential surface will be constant we quickly see that J will decrease as 1/(x**2) as x, the distance from the conductor increases. Near the pool walls things are quite different beacause the equipotential surface now follows the pool walls rather than being spherical. The current density will be much lower in the corners, than along flat walls. If we now consider the walls of the pool to be insulating, and the drain pipe as the zero potential we get the second simple case. Remember shaking iron powder over a bar magnetic and observing the magnetic field pattern. The lines of electric field observed between the conductor and the drain will be very similar to the magnetic field lines observed between the north and south poles, except they will be 3-dimensional. Highest magnitude of electric field will be in the straight line joining the wire and the drain, and thus so too will the current density. As we move out from this line, the lines of the electric field become longer, and thus the magnitude of the field is reduced, and so too the current density. As far as being electrocuted is concerned for 120 volts in a reasonably sized pool you probably wouldn't be anywhere *inside* the pool. However in a plastic lined pool, the danger is stepping out and providing a lower resistance than the drain, in which case you might. The only eddy currents that will be observed are those near the filter outlet, and those will be water, not electron current. There is no magnetic field to induce currents in a conductor in the pool. You might want to have a look at "Engineering Electro-Magnetics" by Hayt, a basic EE text, which describes some simple graphical methods of solving this type of problem Paul L. (EE at waterloo) sgcpal@watdcsu.UUCP