[Faraday's Law] Power Extraction from Line in a country

1. Mar 11, 2009

sgod88

1. I have encountered a problem on power extraction in Faraday's Law application. Consider I live out in the country and I suddenly realized that a power line of current I=I0cos(2pi*60t) [A] is situated 20m from my farm. A copper wire of diameter 4.1mm, 200 m in length was found in the garage and it was used to form a rectangular loop with width a and length b with single loop.
(a) Find the maximum power that can be extracted.
(b) Will I extract more power when I form multiple loop instead of single?

I
--------------------------<--------------------------------

20m
b
|----------Rwire-----------|
| |
| | a
| |

2. (a) I have done this part and I found that to achieve maximum power, a= 38.08m and an assumption based on impedance matching was made that Rwire =Rload. I obtained the Prms=190W

(b) The problem here is that I found that for N loop wire, I would have Vtotal induced=N*V1 loop induced
and the resistant of the wire should be
Hence, I concluded that
P= $$\frac{V_{total}^{2}*R_{L}}{4R_{total L}^{2}}$$
= $$\frac{(NV)^{2}}{4NR_{L}}$$
= $$\frac{NV^{2}}{4R_{L}}$$

Then it would increase the power N times compared to one loop.. Is that correct?

2. Mar 11, 2009

dslowik

Are you assuming that the copper loop is 20 m from the power line and that B is constant there?
Why does the EMF induced in the loop depend on its shape?
For part b, the induced EMF depends on both N, and the area of the loop which decreases as N increases..

3. Mar 11, 2009

sgod88

Yes the copper loop is 20 m from the power line and B is assumed to be constant.
EMF induced in the loop should be depending on the area of the loop, where
$$V_{ind}=-\frac{d\Psi}{dt}$$
and
$$\Psi=\int{BdA}$$
$$=\frac{\mu_{0} I}{2\pi}\int\int{\frac{1}{y} dy}{ dx}$$

For part b,
you mean that the area should decrease by N times? So will more loop extract more power then?

4. Mar 11, 2009

dslowik

If B is constant, you would choose a shape that just maximizes area given the fixed wire length.
But that integral for magnetic flux has y dependence?

for b) as N increases, area decreases, but N times flux enclosed should be maximized.

5. Mar 12, 2009

sgod88

Sorry I that I mentioned the B is constant, it varies with time. I think so it has y dependence because B field changes as y changes (B field itself is varying with time and space cos of the current I and the distance between the wire and the copper wires) and the B field for infinitely long straight line is
$$B(t)=\frac{\mu_{0} I(t)}{2\pi y}$$

So if I substitue in the magenetic flux equation, it will be y-dependent.

So maximizing the area itself does not imply the maximum $$\psi$$ is obtained.

(b) For the N loops, how can I explain this quantitatively? Is there anyway that I can do that?

Thank you
Sgod88

6. Mar 12, 2009

dslowik

a) right, since B goes like 1/y, where y is the distance to the power line, it seems the loop would be near the power line to maximize the flux. Can you do that integral for flux you wrote down before to get it in terms of a, b, and the distance from the powerline? would give you a log, and you need 2a + 2b = 200m...

b) For one loop the perimeter is 200m. If N is 2, 2a+2b=100m, etc. This implies smaller loop areas.
On the other hand, the net induced EMF is proportional to N.
So it is a trade off between those two effects to find the best N.

7. Mar 12, 2009

sgod88

a) Yes I could do that. It is

$$\psi=\int_{20}^{20+a}{\frac{\mu_{0} I(t)(100-a)}{2\pi y}}dy}$$
$$= \frac{\mu_{0} I(t)(100-a) }{2\pi}}ln(\frac{20+a}{20})$$

Then I got the answer I mentioned by making assumption for impedance matching.

Thank you! =)

8. Mar 12, 2009

dslowik

So that's the magnetic flux through the rectangular single loop with the close edge 20m from the powerline. Can you put the close edge against the powerline? that would increase the flux.
In any case, next would be to maximize flux wrt shape of rectangle, i.e. wrt a, using derivative dpsi/da=0.

9. Apr 27, 2009

JohnBS

I'm afraid that in practice, you can't have a single wire carrying significant LF current without looking for the return path...... in this case it's probably a second power-line (single-phase, or two lines/3-phase) not more than a few feet from the first. When you do the flux-linking sums you'll find that the net magnetic field at 20m is much less than you thought.

John