Magnetic field dependent on current only....

Jun 19, 2005 4 Replies

right?



This means the 12volts/18ampere setup produce larger magnetic field magnitude (from the current flow) than the 110volts/2Ampere setup (noting that both produce the same 220 Watts power), right??



Or does the larger 110 volts in the latter give more push to the 2 ampere resulting in similar magnetic field magnitude for both setups??



If the answer is that the 12volts/18 ampere indeed produce larger magnetic field, do designers find it necessary to use larger voltage instead of larger current to prevent magnetic field interference?



If the answer is that both produce the same magnetic field magnitude, then it means in larger voltage such as 110 volts, it pushes the 2 ampere more so it moves faster compared to the 18Ampere with 12 volts that don't push it faster. If not, how can the magnetic field magnitude in both be the same (assuming they are the same).



Thanks.



qude



The 18 amp wire will have a larger magnetic field, at least in the vicinity of the wire. It is further complicated because the wires are usually coiled or double-coiled, and the geometry of the lead wires could have an effect as well.

HOWEVER: magnetic fields are of no consideration whatsoever in light bulb design. Designers use the voltage source that is available: if the bulb is to be used in a household in the USA, this is 110-120 V. If for a household in most other countries, it is 220 V (or somewhere thereabouts). If it's for a car, then 12 V.

Correct!

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Indeed.

No. Fewer charges need to move past a point in a second to transfer the same energy because they are moving faster. Tho' charges have energy, (angular momentum) it is not created nor destroyed. The energy that does the work is the mass(equivalent energy) of the charge in motion.

Hmmm... Frequently higher voltages are used to avoid thick wire. I can't think of an example. There are other techniques that are more effective, but in principle you are correct.

But it isn't. You got it right above.

Sue...

that both produce the same magnetic

--------- Who were you answering? There is no indication of what you are saying "right" to. Please include at least some of the material you are responding to.

The magnetic flux density due to a wire carrying current I is proportional to the current and inversely proportional to distance. The presence of other current carrying wires (such as the necessary return path(s) affect the flux density so consideration of a single wire in space is not a realistic option(effectively you have a single turn coil with a return conductor infinitely far away). The field due to a current of 2A is 1/9 that due to a current of 18A at a point external to the wire.

Is this of importance to decisions as to what voltage is to be used? Not really.

There is an optimum voltage level depending on the power to be transferred and the distance it is to be transferred (at a given frequency). Rule of thumb- more power and or longer distance- go to higher voltage. This is not a physical rule- but an economic one. That is why AC transmission put Edison's DC systems out of business. You can transmit 100MW at 200V for a distance of 200 miles but the cost would be horrendous compared to doing it at 140,000V. However, to use 140,000 V to supply a home would be rather dangerous, expensive, and ridiculous so 120/240 is a good practical and economic balance. (Sue, you should know this- examples are all around :))

-- Don Kelly snipped-for-privacy@peeshaw.ca remove the urine to answer

He was saying "right?" to the header of this thread, "Magnetic field dependent on current only...."

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