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Author Topic: Why is the voltage high and current low in a rapidly changing magnetic field?  (Read 1700 times)

Offline Ashish Bhatia

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Hi Experts,

   I'm doing an experiment of Electromagnetic Induction (producing current by changing magnetic field i.e. by circulating magnetic wheel of N52 Neodyium Magnets around copper coil),

but what i found

when the speed of wheel is HIGH, the Output Current is LOW (in range of .02 microAmp) &

when wheel is about to stop i.e. when its speed is LOW, the Output Current is HIGH (in range of .11 microAmp)

check the video once

Kindly suggest where i'm doing wrong or what i need to change in my experiment, because as per Faraday's law, more frequent change in magnetic field produces more current, but its not seen in my experiment


Kindly help

Thanks in Advance
Ashish
"Solar Roof on Roads with Rain Water Harvesting & Advertisements"
lightenmyways.blogspot
« Last Edit: 21/03/2016 20:00:45 by chris »


 

Offline Colin2B

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Re: Electromagnetic Induction Help Required
« Reply #1 on: 21/03/2016 17:37:06 »
From the video it is difficult to see how you have connected your meter, but here are a few thoughts:
It looks as though the meter is connected across the coil terminals, ammeters have very low resistance and will effectively short out the output. You would do better with a load resistor and measure voltage drop across it.
The meter shows -ve occasionally, is it set to DC? Remember this type of generator produces AC so the average DC value will be zero, this may be why you are seeing low average current. The reason you see a higher value when the rotation is slow could be because the meter is just showing a flip one way (or part of). Try putting it on a 'scope to see the detail.
As I say, hard to see exactly what you are doing, but hope this helps.
 

Offline evan_au

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Re: Electromagnetic Induction Help Required
« Reply #2 on: 21/03/2016 20:38:40 »
I agree; I also assume that the meter is on DC; it should be on AC.

Electronic meters operate in an environment which is full of AC picked up from power cords, power cables embedded in the walls and AC room lighting; this occurs at 50Hz or 60Hz, depending which country you are in.

To measure small DC signals accurately, meters are designed to cancel out AC signals that are close to mains frequency (or multiples of it).

Last time I looked at these, handheld multimeters used some form of dual-slope integrator; the initial fixed integration time is made a multiple of the AC cycle, so it has no overall impact on small DC measurements.

So the worst thing you could do in DC mode is to spin the wheel so the magnets are producing an AC signal at a multiple of the integration time - the DC output will register as exactly zero.

You should also try setting the meter to AC Volts rather than DC Amps.
 

Offline Ashish Bhatia

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Hi Colin2B & evan_au,

    Thanks for such quick reply,

yes, both copper coil terminals (which is connected to voltmeter) are for measuring DC current & meter is on DC, & as you suggested, i'll try to measure the output but putting meter on AC Volts & let you know the findings,

& I like to know, if suppose i get good AC output, than how can i convert it into DC,
1 way I know is full-wave rectifier but will it work ?? or any other process, you like to suggest, kindly reply.

Thanks in Advance
Ashish
lightenmyways.blogspot
 

Offline Colin2B

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Full wave rectifier is cheap and easy. Depending on the application you might want to consider smoothing the DC.
 

Offline Ashish Bhatia

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Hi,

   I tried putting multimeter at AC volts, but its showing zero reading, maybe because output voltage is too low, as output DC current was in microampere.

I tried with putting load (resistor) at the copper wire ends, to measure output voltage or current, but condition remains the same.

I also tried to put magnets close to each other with opposite polarity, but still condition is same.

Kindly suggest, if anything else i can do, to make experiment working properly.

Thanks in Advance
Ashish
lightenmyways.blogspot
 

Offline evan_au

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Quote from: Colin2B
Full wave rectifier is cheap and easy.
A bridge rectifier with 4 silicon diodes has a forward voltage drop of about 1.4 V (0.7V per diode).
I would be surprised if this generator could produce a peak of over 1.4V, in which case the output of the bridge rectifier will be zero.

You may be better trying a germanium diode, or even a Schottky diode, as these have a lower voltage drop.


Quote from: Ashish Bhatia
I tried with putting load (resistor) at the copper wire ends
If you cannot measure a voltage on the AC volts range, a load resistor will just further reduce the voltage.
Some meters have a separate AC millivolts range, which will be much more sensitive.

Quote
if suppose i get good AC output, than how can i convert it into DC
It depends what you mean by "good". What do you want to use it for?

If you cannot measure any AC voltage on the meter, then you have big problems - you may not be generating a voltage that you can rectify.
  • try putting the coil on a "U" shaped soft iron core, so it will direct more of the magnetic field through the coil
  • try mounting the permanent magnets at 90 to their present mounting, so they pass through the open ends of the "U"; this directs the magnetic field straight into the iron core. 

I am guessing that in its present form, this generator can only produce milliwatts.

If you just want to see a rectified waveform, you can use an operational amplifier with a rectifier in the feedback loop. This overcomes the forward voltage drop of a diode, but this circuit probably consumes more energy than the generator produces.

Old-style DC generators used mechanical contacts to reverse the voltage every time the coil passed a magnet. This removes the need for diodes, but introduces the mechanical problems of moving electrical contacts.
« Last Edit: 23/03/2016 08:47:42 by evan_au »
 

Offline highvoltpower

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I can't understand what u said, so please define more . . .
 

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