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Number of wire turns to make a Magnet Generator

  1. Jan 28, 2016 #1
    I'm trying to make my own simple magnet generator project

    Like these:

    http://amasci.com/amateur/coilgen.html
    http://www.wikihow.com/Make-a-Simple-Electric-Generator

    with the following materials:

    2 1.5X1.5X1/8" square magnets
    https://www.amazon.com/gp/product/B011FNB7WK?psc=1&redirect=true&ref_=oh_aui_detailpage_o01_s00

    Copper wire (gauge shouldn't matter right?)
    https://www.amazon.com/gp/product/B007TUQW1K?psc=1&redirect=true&ref_=oh_aui_detailpage_o01_s00

    Blue LED (that requires 3-4.5 volts)
    https://www.amazon.com/gp/product/B004UZCADG?psc=1&redirect=true&ref_=oh_aui_detailpage_o01_s00

    I have a small cardboard box with no bottom or top. Inside is a magnet connected to a rod that sticks out of the box. I can turn the rod to spin the magnet. I wrapped the copper wire around the box and will attach the ends to a light bulb. Assuming this works (and if you believe I am missing something please do tell), I can spin the magnet to induce a current in the coil and to the light bulb, thus powering a light bulb without a battery.

    My question is, how do I know how many turns/wrapping of copper wire around the box I need?

    I don't have a voltmeter and the copper wire is thin like human hair, so I might accidentally rip the wire if I take off the insulating material for testing. In other words, I only have one try to figure out how many times I need to wrap the wire and make it so it can supply 3-4.5 volts.

    I attempted using Faraday's law:

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

    Voltage = - (Number of turns)(Change in Magnetic Flux over time)
    or
    Voltage = -(Number of turns)(Magnetic Field Strength)(Surface Area of Magnet)/ (time)

    I used an app on my phone to get a general idea of the magnetic field (it read 200 micro teslas).
    B = 200 * 10^-6 Teslas

    Magnet surface area is 1.5X1.5" so in meters is: 0.0381 ^2 meters or 0.00145161 square meters

    A = 0.00145161 m^2

    Voltage needed = 4.5 Volts

    Time = I felt this would be an arbitrary number, but would be in the range of 2 - 10 seconds. I wouldn't want to spin the magnet forever.

    So N (number of turns) should equal to = Volts * time / B*A = 4.5 Volts * 10s / 200*10^-6 Teslas * 0.00145161 m^2 = 155000310.001

    I'm really hoping I made a mistake. Please help me figure this out.

    Thank you!
     
  2. jcsd
  3. Jan 28, 2016 #2

    Hesch

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    Gold Member

    200 μT doesn't seem much. At least 0.1 T should be expected. Also the time ( 10s ), seems very slow indeed ( 0.1 rad/sec ). You'll have to train a lot!

    Anyway I'd suggest that you redesign the generator, adding an iron core to increase the magnetic flux through the coil. I think the winding process will become easier, and the number of turns needed, will be decreased.

    Make the width of the airgaps at the ends of the magnet small.
     
  4. Jan 28, 2016 #3

    BvU

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    Science Advisor
    Homework Helper
    Gold Member

    Hi there ! :welcome:

    I notice the magnet product description (link questionable; I clicked see more product details) mentions a much higher magnetic field strength: 13500 Gauss (1.35 Tesla). Do you trust the phone app ? Even the earth magnetic field should already be around 0.5 Gauss !

    Then: time is not the time you crank the thing, but the time in which the magnetic flux changes. So if you turn it 5 times/second you want to calculate
    Change in flux / time = 2 x 1.35 x area / 0.2 seconds

    And check if the LED doesn't already light up a bit earlier (2 V ? - don't forget the resistor!)

    Did you notice the magnet poles are the flat surfaces ? So you want the rotation axis parallel with the surface.

    Let us know how it goes ! Good luck!
     
  5. Jan 28, 2016 #4

    Hesch

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    #2: Forgot the sketch ( attached )
     

    Attached Files:

  6. Jan 28, 2016 #5

    Hesch

    User Avatar
    Gold Member

    200 μT doesn't seem much. At least 0.1 T should be expected. Also the time ( 10s ), seems very slow indeed ( 0.1 rad/sec ). You'll have to train a lot!

    Anyway I'd suggest that you redesign the generator, adding an iron core to increase the magnetic flux through the coil. I think the winding process will become easier, and the number of turns needed, will be decreased.

    Make the width of the airgaps at the ends of the magnet small.
     
  7. Jan 28, 2016 #6
    The only materials I have currently are what I linked to but next time I'll look into getting an iron core. For now I'll definitely reduce the distance between the magnet and coil. Completely forgot about that, thanks!
     
  8. Jan 28, 2016 #7

    Thank you for pointing out the Gauss description in the link (and the fact that I'll be using 2 magnets) and explaining the time. That makes things a lot easier and in fact with new calculations reveal more reasonable numbers:

    2 magnets with 1.35 Teslas, area of 0.00145161 m^2, and 4.5 volts needed

    N (wire wrapping) = Volts / change in flux = Volts / (B*A / time) = 4.5 volts / [(2*1.35*0.00145161) / time for one revolution]

    1 turn per second = about 1200 wire wrappings
    2 turns per second = about 600 wire wrappings
    3 turns per second = about 400 wire wrappings
    4 turns per second = about 300 wire wrappings
    5 turns per second = about 250 wire wrappings

    Of course these numbers sound a lot nicer for me but does it sound reasonable to anyone else?

    Could you explain how using resistors would help reduce the voltage needed to power the LED?

    Would it be simply because P = V^2/R ? -> V = sqrt (PR) Increase resistance to reduce voltage needed for the LED? I thought something that absorbs energy like an LED would need a specific amount/range of volts. Does adding a resistor lower that range?

    I found a toy that used the same LED and it required 3 LRA1 batteries - thus my assumption for the need of 3 to 4.5 volts. It didn't have a resistor or anything else attached.

    And also how I can physically apply the resistor? Would the resistor be attached to the positive end of the wire? And would it be attached before the LED?
     
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