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We consider a flat rectangular plate moving horizontally in a vertical magnetic field,motion being in a direction perpendicular to the length of the plate. We have an emf=BLV between the tips,in the lenhgth wise direction[the axial emf]. During the formation of the axial emf a current flows along the length of the conductor. This current should get deflected in the lateral direction due to the existing magnetic field, producing a transverse emf between the lateral edges.If the tips are connected by a wire we have a closed circuit condition--we should simply have the Hall voltage between the lateral edges[in the direction of the motion].

Query: Would it be possible to use these voltages as supplementary power sourses in moving vehicles---in moving trains,cars ,aeroplanes etc? [We seem to have several ready-made parallel cells in these vehicles]

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Relevant Link: www.sugarsync.com/pf/D7596576_2251256_67367 –  Anamitra Palit Dec 30 '11 at 23:08
The presentation in the link[in the comment above] is based on the paper represented in the following link: eurojournals.com/ejsr_39_1_11.pdf –  Anamitra Palit Dec 31 '11 at 15:15

2 Answers 2

They wouldn't be supplementary sources They have to get energy from somewhere, right? And the magnetic field isn't being changed, is it? Eventually, you'd have to supply the energy.

Otherwise, stuff like hydroelectric plants would just need an initial push and they'd give free energy forever.


Lets consider a simple system: A stick in the y direction is moving along the x-axis. A uniform magnetic field exists in the -ve z direction (into the paper).

| x  x
| x  x
|----> v
| x  x
| x  x B
| x  x

^  (.)z

An emf $Blv$ will be induced in the upwards direction. So far, so good.

Now, lets put it in a simple circuit with a bulb or something. The total resistance of the circuit is $R$. So we get an upwards current (+ve y direction)=$Blv/R$. The total obtainable power from this circuit is $\frac{(Blv)^2}{R}$.

Ok. So we managed to get some energy out of it. But, there's something else: There's a current in a magnetic field. That means that there's a force. Let's assume that you do whatever's necessary to keep velocity constant (Otherwise doing the problem completely correctly would require an infinite series and a differential equation). So, we have a current $Blv/R$, and a perpendicular magnetic field $B$. The force on it will be $ilB=B^2l^2v/R$. It's opposite to velocity (verify this if you want). To keep velocity constant, the car has to apply an equal and opposite force. Power that we must apply will be $Fv=F^2l^2v^2/R$.

Surprise, surprise! We have to input the same power as we get out. Effectively, it's useless for getting power. Practically, there are other losses, so we basically ended up losing energy. If we don't attempt to keep the velocity constant, the car will slow down. Also bad.

Note that there will be no energy lost if the circuit is not connected and if we just have a moving rod, as there is no current. So, the minute we try to obtain energy from a moving rod, we lose the same amount of energy in keeping it moving.

A better use

Of course, this can be used as a way to slow down the car and obtain extra energy. But, as @Anna has calculated, it's not much energy, so it would be a pretty ineffective way.


Trying to obtain energy in this manner will slow down your car, and you will have to burn extra gas to keep it moving. In a perfect world with no other resistive forces, we would lose exactly the amount of energy we obtain. In our not-so-perfect world, this makes us lose more energy. We can use it to slow the car down, but it's not effective.

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We can always use what we are throwing away.If you consider a running train ,electrons should be flowing out of the wheels due to the flow of an induced current--this is slowing down the train. And we are tolerating this.There are two clear options:(1)Stop the flow of this induced current to increase the speed of the train slightly(2) Utilize the induced current. The train itself is a source of a multitude of millivolt cells. I have already suggested this in one of my comments –  Anamitra Palit Mar 1 '12 at 1:53
The induced voltage itself is very small. But the "cell" will run for an indefinite period of time.You may compare two 2-volt cells [each having the same type of electrolyte and electrodes of the same composition] --a large one and a small one. The larger one will run for a longer period of time –  Anamitra Palit Mar 1 '12 at 1:57
@AnamitraPalit Aah I see what you're saying about stuff being thrown away. Remember that the actual value is pretty small, and it's effect on the train is negligible. If you could figure out how to amplify the effect, that would be nice. Remember, adding a device that taps this will reduce the trains efficiency as well. –  Manishearth Mar 1 '12 at 2:03

I'm assuming that the magnetic field you're referring to over here is the Earth's field, then we have the field strength of $6.5 * 10^{-5}$. If we assume the vehicle to be a bullet train, then we get the velocity of $300 km / h$ which is $83.33 m/s$ Assuming a scenario where a rectangular plate of $1m$ is being used and the current is being generated along its diagonal of length $\sqrt2 m$ Then as you stated,

$ \varepsilon = BLV $

$ \varepsilon = 6.5 * 10^{-5} * 83.33 * \sqrt2 $

Then this setup will generate $.00766$ $V$ of electricity, which is too low to be applied for most practical purposes.

If you notice over here then you already have an engine powering the vehicle, and in effect you're trying to capture some of its output. Wouldn't it be much better to capture it at the source?

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Actually the cells are in parallel. But if we could separate the cells by insulator-strips and get them into a series combination by some suitable curcuit arrangement the emf could be increased.In the existing conditions current should be flowing out through the wheels to the ground------ a part of the engine power is already being consumed. –  Anamitra Palit Dec 31 '11 at 6:49
In the existing conditions it would be interesting to know how much current is flowing out of the wheels in moving trains. If the charge flowing out, could instead be stored ,say in capacitors and be put to future use... –  Anamitra Palit Dec 31 '11 at 7:07
Our interesting objective would be to utilize what we throw away----just like saving drops of petrol even if the drops are tiny and putting them to use.If we could use the tidal power or wind power to move conductors in the earths field it could be useful in obtaining energy –  Anamitra Palit Dec 31 '11 at 7:31
@AnamitraPalit : As far as using something like this for capturing tidal power goes, then I agree that it is an intriguing thought, but it's trivially easy to add a moving plate between a large array of magnets, which is essentially what most such efforts do. After all, what is a dynamo, but an abstracted form of this plate? However everything comes back to efficiency & I wager that experimentation will find the conventional method is far more efficient than using the earth's magnetic field. –  Anna Dec 31 '11 at 21:42
@AnamitraPalit : It would actually be far more efficient if you focused your efforts on increasing the efficiency of the engine. As at the end of the day what you're simply doing is extracting energy from the engine, you might as well directly work on increasing the efficiency with which the engine converts one form of energy to another to extract work. I really do admire your sense of perspective, and I would encourage you to take a look at improving our current energy conversion systems. –  Anna Dec 31 '11 at 21:48

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