# Converting milliamp hours into work

1. Dec 9, 2007

### kmarinas86

Let's say we have a battery of with a given electrical capacity and we are trying to maximize the work being done with it. Then let's say we use the battery and some length of wire to generate a magnetic field which we will use to rotate a magnet. The total magnetic field generated for a given amount of current is determined by how long the wire is. However, the longer our wire, the greater the resistance is, so the more voltage we would need to maintain the same current. Is it true that if we increase the length of the wire in proportion to the voltage, that we can keep the same current but create more magnetic field as a result? Is our ability to convert milliamp hours into work limited only by voltage and the length of wire we choose? Is there a theoretical limit to milliamp hour to work conversion that can't be breached regardless of the technology?

2. Dec 9, 2007

### Staff: Mentor

A battery is based on a electro chemical reaction which has a specific voltage, so you cannot vary the voltage. If you take a voltage times a current, what kind of unit do you have?

3. Dec 9, 2007

### kmarinas86

You can choose the voltage by choosing which battery.

Power.

What if you connect the positive and negative ends of two batteries by direct contact. What kind of magnetic field does that produce versus having the given current remain as long as possible within a wounded coil of some unspecified length? Is the integral of the magnetic field along infinitesimal lengths of the wire proportional only to the current? No it is not. If you want the charges to go farther, and maintain the same current, you have to choose a different battery, one that has a higher voltage. Doing so generates a greater magnetic field.

$d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{r^2}$

(in SI units), where
$I$ is the current,
$d\mathbf{l}$ is a vector, whose magnitude is the length of the infinitesimal|differential element of the wire, and whose direction is the direction of conventional current,
$d\mathbf{B}$ is the differential contribution to the magnetic field resulting from this differential element of wire,
$\mu_0$ is the magnetic constant,
$\mathbf{\hat r}$ is the unit displacement vector from the wire element to the point at which the field is being computed, and
$r$ is the distance from the wire element to the point at which the field is being computed.

As you can see, the magnetic field contribution is determined by the product of the current AND the distance traveled by that current. If we choose the voltage by choosing the battery, we can maximize the power for the same amount of current, because we can make that current travel farther. Therefore, my point here is to show that the amount of magnetic field we can get from a battery's charges is limited by its voltage. There is no fundamental and natural limit that I am aware of that prevents high voltages from getting more magnetic field out of a given charge.

Take for example the product of volts and amps. Assuming that we have the same amount of charges (for whatever battery we choose here) and we're trying to produce the same amount power, we could choose a battery that will last twice as long by choosing a battery with twice the voltage and the same capacity and reduce the current in half (because half the current means half the drain). Reducing the current could be done by increasing the length of the coil wire by four times. But doing this also increases the magnetic field because while the current is only halved, the length of travel is quadrupled! How you can you increase the magnetic field yet have the same power? Since this is basically an electromagnet, one can switch the current to get a magnet rotating. Yet with a bigger magnetic field, we can take advantage of this and use a bigger magnet, for the same power! Does this make sense?

I believe its reasonable to assume that the magnetic field is produced by current momentum, the product of current and distance, not just current. Now my question is, in what ways is the total magnetic field that can be produced limited by current alone?

Last edited: Dec 9, 2007
4. Dec 9, 2007

### chroot

Staff Emeritus
You're trying to minimize the effect of losses in the wires, and trying to deliver the most power possible to the load.

This is always done by making the output resistance equal to the load resistance. In this case, the output resistance is the ESR of the battery, plus the resistance of the wires.

- Warren

5. Dec 9, 2007

### kmarinas86

I have a question. As long as you can increase the voltage and the output resistance in proportion to the length of the wire, is there any limit to how much magnetic field you can get from 1 amp as long as you are not limited by the number of turns of wire?

Last edited: Dec 9, 2007
6. Dec 9, 2007

### Staff: Mentor

I thought you had said "for a given battery". If the goal is to maximize the work done and you are free to choose your battery then simply choose the biggest battery. All of the rest is rather premature.

Once you have done that your voltage is fixed and you can continue with the problem from there. You really don't need a complicated Biot-Savart analysis. Just use conservation of energy. Either the energy goes to doing useful work or it goes to waste heat. So just minimize your Ohmic losses by impedance matching, as mentioned by chroot.

7. Dec 9, 2007

### alvaros

There is a misunderstanding in your post. You think:
more magnetic field -> more work
this is not true, yes, you need some energy to establish the magnetic field but, once the magnetic field is constant, you dont need any more energy, the energy you lost is for heating the wires, and you can use wires with a big diameter ( or you can use superconductor wires, or you can use permanent magnets )

The work done in a electric motor comes from pushing the current against the the fem that is generated by the motor in its movement.

And you cant get more power from a battery than V.I

8. Dec 9, 2007

### Staff: Mentor

From his description the magnet is part of a motor. But you are correct, if you don't have the magnet pushing on something then there is only energy stored in the magnetic field, not used for work.

9. Dec 9, 2007

### kmarinas86

** yes

This is an image of the first motor I built of this kind:
http://x55.xanga.com/951c3555d6732158770148/b119244434.jpg

The second motor I built was a longer telephone wire with four colored strands I connected together to create a circuit that was four times longer. It drew less current than my first one but was able to spin the magnet just as well as the first one did with the same 9V battery, and I know this because the first motor would heat up the 9V battery more quickly than the second.

The motors I built are such that the current is off during 80% of each rotation. The current is closed when two wires, one from the battery and the other from the coil, touch the middle bar simulatenously.

Last edited: Dec 9, 2007
10. Dec 10, 2007

### Staff: Mentor

Looks like a pretty fun project!

Seriously, you don't have to calculate Biot-Savart or anything complicated. All you have to do is match impedance ( http://en.wikipedia.org/wiki/Impedance_matching ). This should be pretty easy to do if you have a good multi-meter.

11. Dec 10, 2007

### kmarinas86

Will doing so maximize the magnetic field produced for a given current?

edit: I found what the term for this was. It's called inductance.

Now I realize that for a given current, a higher inductance means a stronger magnetic field.

But switching the magnetic field on and off can do work....

If you kept the voltage the same, increasing the turns and length of wire would increase the inductance but would also decrease the current. Doing so would decrease the power (volts*current). But for some reason, this doesn't keep the magnet from responding from a switchable magnetic field. If there was no coil, there wouldn't be inductance from it, and the permanent magnet would not respond. If limit of power transfered to the magnet really was voltage times current, then the ideal coil size would be medium sized and would produce the most power. ?

Last edited: Dec 10, 2007