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vikram_gupta11 · Sep 10, 2016, 01:57 AM

How much output I can get if a 3000 gauss neodymium magnet moves pass a 20000 turns coil after each 2 second and can I store this output in a super capacitor?

smoothy · Sep 10, 2016, 04:32 AM

What do you think, and why, how did you arrive at your answer? Site rules prohibit us doing you homework for you.

vikram_gupta11 · Sep 10, 2016, 07:23 PM

Originally Posted by smoothy

What do you think, and why, how did you arrive at your answer? Site rules prohibit us doing you homework for you.

Sir.
I asked you this question as I have a unique Idea regarding an Energy technology and want to develop this technology. If you are interested to know my Idea then I can post it for your review.
Please answer me.
I shall be very grateful to you.

Vikram

ebaines · Sep 12, 2016, 05:57 AM

The formula you need is:

$V(t) =n \frac {d \Phi}{dt}$

where V is the output voltage, n is the number of turns and $\Phi$ is the magnetic field in webers. One weber = 10^8 Gauss-cm^2. So, to convert 3000 gauss to webers we need to know the area enclosed by one turn of the coil.

vikram_gupta11 · Sep 12, 2016, 10:27 PM

Originally Posted by ebaines

The formula you need is:

$V(t) =n \frac {d \Phi}{dt}$

where V is the output voltage, n is the number of turns and $\Phi$ is the magnetic field in webers. One weber = 10^8 Gauss-cm^2. So, to convert 3000 gauss to webers we need to know the area enclosed by one turn of the coil.

AREA is 4 centimeter / square

ebaines · Sep 13, 2016, 05:48 AM

First convert gauss-cm^2 to webers:

3000 Gauss x 4 cm^2 x 10^-8 weber/gauss-cm^2 = 1.2 x 10 ^-5 weber

The magnetic flux is a sinusoidal function with period = 2 seconds. So:

$\Phi = 1.2 \times 10^{-5} \sin (\pi t)$ webers

and

$\frac {d \Phi} {dt} = 1.2 \times 10^{-5} \pi \sin (\pi t)$ weber/second

So the voltage produced would be:

$V(t) = 20000 \times 1.2 \times 10^{-5} \pi \sin (\pi t) = 2.4 \pi \sin(\pi t)$ volts.

Hence the peak voltage would be about 7.5 volts.

vikram_gupta11 · Sep 13, 2016, 06:02 AM

Originally Posted by ebaines

First convert gauss-cm^2 to webers:

3000 Gauss x 4 cm^2 x 10^-8 weber/gauss-cm^2 = 1.2 x 10 ^-5 weber

The magnetic flux is a sinusoidal function with period = 2 seconds. So:

$\Phi = 1.2 \times 10^{-5} \sin (\pi t)$ webers

and

$\frac {d \Phi} {dt} = 1.2 \times 10^{-5} \pi \sin (\pi t)$ weber/second

So the voltage produced would be:

$V(t) = 20000 \times 1.2 \times 10^{-5} \pi \sin (\pi t) = 2.4 \pi \sin(\pi t)$ volts.

Hence the peak voltage would be about 7.5 volts.

Dear sir ,
thank you very much for your answer.
Sir I' m presenting a unique concept before this community to know the response .Please reply me that will it work or not.
Dear Sir,
I 'm presenting a unique concept regarding Energy generation which is based on Magnetic induction law.
Sir In this device there will be a SEESAW system and two high tesla magnet will be attached with the both side of this SEESAW system as per sketch .
There will be 3 volt toy motor and this toy motor will be attached with the right or left arm of seesaw system with the help of a 'L ' form crank .This crank will work to convert rotatory motion of this toy motor into linear motion.

There will a coil having 10000 turns and each coil be in 4 centimeter/ square in Diameter.

when the 3 volt toy motor will press the seesaw system's ARM then it will come down and will work to move pass the coil and energy will be generated .In this way second arm of seesaw system will work to generate energy.

Now my question is that what will be the INITIAL OUTPUT if each side of seesaw system moves pass the coil after taking break of one second?

(2)Can we store this electricity in the super capacitor ?

(3)if the output is more than input then can we reuse this output again to increase the speed of 3 volt toy motor so that it could press the seesaw system with more force?

(4) I know that coil will work to oppose the movement of magnet of seesaw system due to reverse magnetism but if output is more than input then can we use this output to give more power to the motor so that it could work against the reverse magnetism?

Please answer me.

vikram_gupta11 · Sep 13, 2016, 06:14 AM

Attachment 48485

Dear Sir,
I 'm presenting a unique concept regarding Energy generation which is based on Magnetic induction law.
Sir In this device there will be a SEESAW system and two high tesla magnet will be attached with the both side of this SEESAW system as per sketch .
There will be 3 volt toy motor and this toy motor will be attached with the right or left arm of seesaw system with the help of a 'L ' form crank .This crank will work to convert rotatory motion of this toy motor into linear motion.

There will a coil having 10000 turns and each coil be in 4 centimer/ square in Diameter.

When the 3 volt toy motor will press the seesaw system's ARM then it will come down and will work to move pass the coil and energy will be generated .In this way second arm of seesaw system will work to generate energy.

Now my question is that what will be the INITIAL OUTPUT if each side of seesaw system moves pass the coil after taking break of one second?

(2)Can we store this electricity in the super capacitor ?

(3)if the output is more than input then can we reuse this output again to increase the speed of 3 volt toy motor so that it could press the seesaw system with more force?

(4) I know that coil will work to oppose the movement of magnet of seesaw system due to reverse magnetism but if output is more than input then can we use this output to give more power to the motor so that it could work against the reverse magnetism?

Please answer me.

ebaines · Sep 13, 2016, 06:21 AM

Sorry, but as with all perpetual motion schemes there is always a flaw. In this case the flaw in your system is that the power required to make the seesaw move at the speed you want is greater than the power generated by magnetic induction. In a perfectly designed system the input and output power would be equal at best, which would mean the mechanism could run (at least for a while) but 100% of the electrical energy produced by induction would be required to be converted into mechanical energy by the motor, and there would be no excess power available. In real life no process is 100% efficient - there are losses due to some amount of inefficiency of the motor, resistance in the wiring, and mechanical friction of the seesaw. So after being started up the mechanism will slow down and eventually come to a stop.

vikram_gupta11 · Sep 13, 2016, 11:20 AM

Originally Posted by ebaines

Sorry, but as with all perpetual motion schemes there is always a flaw. In this case the flaw in your system is that the power required to make the seesaw move at the speed you want is greater than the power generated by magnetic induction. In a perfectly designed system the input and output power would be equal at best, which would mean the mechanism could run (at least for a while) but 100% of the electrical energy produced by induction would be required to be converted into mechanical energy by the motor, and there would be no excess power available. In real life no process is 100% efficient - there are losses due to some amount of inefficiency of the motor, resistance in the wiring, and mechanical friction of the seesaw. So after being started up the mechanism will slow down and eventually come to a stop.

Sir ,Input energy is very very less to move the seesaw system as it is in balanced state or equilibrium state .we can move this seesaw system with just our finger tips or in other words input is almost ZERO.

I used 3 volt motor just as to explain its external source to move it and we can take even 100000 turns of coils and several hundreds neodymium magnets to move pass each other in this system as its input energy will same .

If we take 50000 turns coils and 2 tesla magnets and each turn has a diameter of 4 cm/square then output will be 160 volt as per magnetic induction formula so how can you say that input is more than output even output is 100 times more than input.

Please answer me.

ebaines · Sep 13, 2016, 12:23 PM

The seesaw mechanism may have very little friction in it. But you are forgetting that the more voltage produced the higher the force required to move the magnets through the coils. The current flow that is induced into the windings creates a magnetic field that opposes the movement of the sliding magnets. The motor has to overcome this opposing force to move the magnets, and the amount of power the motor must consume to do this is at least as great as the amount of power that the generator produces. The best case overall result is zero net power from the system, but that's only if everything works at 100% efficiency. In reality the net power produced will be negative.

vikram_gupta11 · Sep 13, 2016, 09:39 PM

Originally Posted by ebaines

The seesaw mechanism may have very little friction in it. But you are forgetting that the more voltage produced the higher the force required to move the magnets through the coils. The current flow that is induced into the windings creates a magnetic field that opposes the movement of the sliding magnets. The motor has to overcome this opposing force to move the magnets, and the amount of power the motor must consume to do this is at least as great as the amount of power that the generator produces. The best case overall result is zero net power from the system, but that's only if everything works at 100% efficiency. In reality the net power produced will be negative.

Sir, you are correct and I have mentioned about this in no. 4 point but my question is that can we reuse this high voltage output with the help of capacitor to increase the torque of motor so that it could overcome the opposing force?
suppose a magnet with 2 tesla and having 4 cm/square diameter is passing through a coil which has 10000 turns .Now according to formula this system will produce 32 volt as a output .we will store this 32 volt in a capacitor now my question is that can we use this 32 volt again in the motor to press the seesaw system so that it could overcome the opposite force again and again in every cycle or in other words the output that we gain 32 volt can we reuse as an input?

ebaines · Sep 14, 2016, 05:40 AM

You can certainly store a charge in a capacitor that you can later bleed off to power the motor. In a perfect system the energy stored in the capacitor from the first stroke of the generator would be just enough to provide energy to the motor to push the magnets through the coils on the second stroke. You could store the electrical energy generated on the 2nd stroke in a 2nd capacitor, so that you have energy available for the 3rd stroke. You could keep doing this, but note that there is no excess power being created - all the energy created on each stroke must go to moving the magnets on the subsequent stroke. Plus you had to provide the energy for that first stroke from some external source.

I think it's important to discuss how voltage produced by the generator related to power. They are not the same thing. The power produced by the generator is equal to the voltage produced times the current flow, I. The current flow in turn is dependent upon the load of the circuit it is attached to. If there is no load (i.e. an open circuit), then I = 0 and the power produced by the generator is zero. In this case it is very easy to slide the magnets through the coils, because no energy is being created and hence no work is required to move the magnets. But suppose you attach a load - let's say a 1 ohm resistor to keep the math easy. Given a 32 volt output of the generator (per your example) the current flow will be I=32 amps, and the power produced is 32 volts x 32 amps = 1024 watts. If it takes 1 second for one stroke of the seesaw, the total energy produced is 1024 watt-seconds = 1024 joules. This means 1024 joules of work must be expended by the motor to push the magnets through the coils. The force required to do this would be 1024 J divided by the length of the coils. For example if the coils are 0.1 meters long, the force required to push the magnets through in 1 second is 1024/0.1 = 10,240 newtons. That's a pretty big force, one that your toy motor certainly can't provide. A quick calculation shows that if we increase the load of the circuit to 1000 ohms, then the current is reduced to 32 milliamps, the power produced becomes approximately 1 watt, and the force that the motor must overcome to move the magnets in the coils is about 10 N. This would be a more practical system for a small motor to handle.

vikram_gupta11 · Sep 14, 2016, 09:37 AM

Originally Posted by ebaines

You can certainly store a charge in a capacitor that you can later bleed off to power the motor. In a perfect system the energy stored in the capacitor from the first stroke of the generator would be just enough to provide energy to the motor to push the magnets through the coils on the second stroke. You could store the electrical energy generated on the 2nd stroke in a 2nd capacitor, so that you have energy available for the 3rd stroke. You could keep doing this, but note that there is no excess power being created - all the energy created on each stroke must go to moving the magnets on the subsequent stroke. Plus you had to provide the energy for that first stroke from some external source.

I think it's important to discuss how voltage produced by the generator related to power. They are not the same thing. The power produced by the generator is equal to the voltage produced times the current flow, I. The current flow in turn is dependent upon the load of the circuit it is attached to. If there is no load (i.e. an open circuit), then I = 0 and the power produced by the generator is zero. In this case it is very easy to slide the magnets through the coils, because no energy is being created and hence no work is required to move the magnets. But suppose you attach a load - let's say a 1 ohm resistor to keep the math easy. Then given a 32 volt output of the generator the current flow will be I=32 amps, and the power produced is 32 volts x 32 amps = 1024 watts. If it takes 1 second for one stroke of the seesaw, the total energy produced is then 1024 watt-seconds = 1024 joules. This means 1024 joules of work must be expended by the motor to push the magnets through the coils. The force required to do this would be 1024 J divided by the length of the coils. For example if the coils are 0.1 meters long, the force required to push the magnets through in 1 second is 1024/0.1 = 10,240 newtons. That's a pretty big force, one that your toy motor certainly can't provide.

So what kind of motor and circuit we can use for this device .
Sir,please guide me.
I shall be very grateful to you.

vikram.

ebaines · Sep 14, 2016, 10:24 AM

I edited my previous response to show how a circuit with higher resistance (i.e less current flow) would lighten the force required to move the magnets. But again I must emphasize that the system will not run all by itself, no matter how you design the coils, magnets, capacitors and load.

vikram_gupta11 · Sep 19, 2016, 10:53 AM

Originally Posted by ebaines

I edited my previous response to show how a circuit with higher resistance (i.e less current flow) would lighten the force required to move the magnets. But again I must emphasize that the system will not run all by itself, no matter how you design the coils, magnets, capacitors and load.

Dear sir,

In this sketch you will find that I have attached 10 fingers on the both side of seesaw system and this seesaw again moving with 3 volt toy motor but this time I have used 10 no. coil+ magnet and each coil+ magnet system will produce only 2 volt as a output which will be less than input and on the first hand each magnet+ coil system is producing only 2 volt and on the second hand the total 10 no. magnet+ coil systems are producing 20 volt . If we try this concept than input is more than output but if we add total output then it is greater than input.

Further more if I connect a motor with exhaust fan with each coil+magnet then these total 10 no. fan will work to pass the air through a blow pipe and this exhaust air will work to run a 12 volt motor again .In this way can we get more output than input.

In this case as far as I think I'm not violating any thermodynamics laws as we are getting less output than input with each coil+magnet system. But total 10 no. magnet +coil systems are producing 12 volt as a output and no reverse magnetism work in this system also due to more input than output.

Please guide me about your views.

I shall be very grateful to you.

Vikram

ebaines · Sep 19, 2016, 12:04 PM

I think you are again confusing voltage with power. The power required to turn the seesaw is still greater than the power output of the coils. Playing around with the number of coils doesn't help - if you have five times the number of coils as in your previous examples, each producing 1/5 the voltage as before, each coil requires 1/5 the force to move but the total force required is still the same. Hence the power that the motor need to provide must be at least as great as the power developed by the coils.

As for the exhaust blower being able to develop power - same problem. The power developed by the exhaust generator (note that this actually a generator, not a motor) will be less than the power consumed by the ten fans to produce the wind to turn the generator. It doesn't matter that the fans are operating at 2 volts while the generator produces 10 volts - again, voltage is not the same as power. Stated another way - the generator can't produce enough current to turn the fans quickly enough to create enough wind to keep the generator turning.

vikram_gupta11 · Sep 20, 2016, 12:49 AM

Originally Posted by ebaines

I think you are again confusing voltage with power. The power required to turn the seesaw is still greater than the power output of the coils. Playing around with the number of coils doesn't help - if you have five times the number of coils as in your previous examples, each producing 1/5 the voltage as before, each coil requires 1/5 the force to move but the total force required is still the same. Hence the power that the motor need to provide must be at least as great as the power developed by the coils.

As for the exhaust blower being able to develop power - same problem. The power developed by the exhaust generator (note that this actually a generator, not a motor) will be less than the power consumed by the ten fans to produce the wind to turn the generator. It doesn't matter that the fans are operating at 2 volts while the generator produces 10 volts - again, voltage is not the same as power. Stated another way - the generator can't produce enough current to turn the fans quickly enough to create enough wind to keep the generator turning.

Dear Sir,
In this design each circuit is separate with each other and each circuit is producing 2 or 1.5 volt output separately and you are correct that input power is still greater than output as input power is 3 volt but in this design each coil+ magnet system is generating 2 volt as a output and I have used 10 no. coil+ magnet system and attached 10 no. fans .Now these total 10 fans will work to exhaust air through a pipe and this pipe is widen from front side and narrowed to back side .In this way Can we move a 10 volt turbine to produce at least 8 volt or not. I did an experiment with this system and it is working.

I attach a fan with a 3 volt motor and used a 1.5 volt AA battery and tried to move a 3 kilogram weight (this 3 kilogram weight has wheels) with the help of air of this fan but this weight didn't move but when I used 10 no motors and fans and gathered total exhaust air with the help of a blower then this weight moved fastly or in other words I 'm running 10 no. 2 volt motors only with one 3 volt motor as far as I think this 3 kilogram weight is enough to produce more energy than 3 volt . So now I'm very much confused that can we use this Idea to produce more energy as I mentioned in my Idea as we are using only one 3 volt motor to move 10 no.coil+ magnet system and each coil+ magnet system is producing just 2 or 1.5 volt output .
The mass is not an issue in this seesaw design even we can take 100 or more coil+ magnet system . my question is only that will each coil+magnet system generate 2 or 1.5 volt energy and in this way way 10 no.coil+magnet system generate 2 or 1.5 volt energy separately or not and why we cannot get more output by gathering all 2 volt energy as whole system is operating on 3 volt toy motor?

Please read carefully about my experiment .

Thank you Sir.

Vikram

ebaines · Sep 20, 2016, 09:10 AM

I'm sorry that I am not able to communicate to you that voltage is not the same thing as power. Let's try a different tact - please see the attached circuit and tell me if you think the fan will run, and why or why not. Note that the wind turbine generates a greater voltage than the fan requires.

InfoJunkie4Life · Sep 20, 2016, 10:00 AM

Friend,

The concept of any perpetual motion machine is against all of the laws of physics. Primarily the Law of Conservation of Mass-Energy. You may also consider Newton's 3 Laws of Motion to observe how force in transferred between objects in the form of inertia and using acceleration. Similarly this can be observed in pressure systems where Bernoulli's Principle can be observed in transference of pressures and moment forces. Likewise in electrical systems there is a relation between voltage and amperage called Power. The Magnetic Force on wires can be defined in similar principles.

All types of energy transference share 2 things in common.

1.) The total power inputted into a system is equal to the total power expelled from the system.

2.) No system operates without resistance which dissipates these forces into heat or other ambient radiation, reducing the power output as much as the heat/light/magnetic loss that is emitted from the system.

Storing power in a state of Potential Energy is calculable in the same metrics. Energy input is less than output. We can never seem to break this rule.

Given this we have developed use of fairly dense potential energy sources, i.e.. Wood, coal, oil, radioactive metals, light, geothermic energy etc. But still some force had to exercise on that energy to produce the potential.

No matter how you look at it, energy has to come from somewhere. The concept of free-energy in physics is considered an absurdity, however, the collection of ambient energy can be free.