Find Max Power from 60Hz 4000A Power Line w/ 200m Wire

[seang]

Homework Statement

A 60Hz, 4000A power line passes near your house. You have 200m of wire. The closest you can get to the wire is 20m. Find the maximum power you can extract from the power line. The graphic shows that the loop has to be rectangular with side lengths A and B.

Homework Equations

The Attempt at a Solution

I took the wire to be in the Z direction, and so A is in the r direction. The B field from the wire is.

$$B = \frac{\mu_0 I}{2 \pi r}$$

Flux Linkage through the loop:

$$\psi = \frac{\mu_0 I}{2 \pi} \int_{0}^{B} \int_{20}^{20 + A} 1/r drdz$$

...

$$= \frac{\mu_0*4000sin(377t)*b}{2 \pi} * ln(\frac{(20 + A)}{20})$$

then,

$$200 = 2a + 2b , b = 100-a$$

$$V = \frac {d\psi}{dt} = .3016cos(377t)*(100-A) * ln(\frac{20+A}{20}$$

I used MatLab to find that the maximum voltage occurs when A = 31.8 meters. The corresponding voltage is 19.909 Volts.

So this is as far as I've gotten. I need either the Z of the load or the current in the loop to find the power, right? Is the induced current related to the source current?

 

[seang]

Maybe I can make this easier, I think I just need to know how current on one loop induced a current on another, given that I don't know the resistance of the second loop.

I mean, In this example, I know that the current on the second loop will be 60Hz, but how can I find the magnitude?

 

[m2003]

To find the maximum power that can be extracted from the power line, we need to consider the power equation P = VI, where V is the voltage and I is the current. From the solution provided, we can see that the maximum voltage that can be induced in the loop is 19.909 Volts. However, to determine the maximum current, we need to know the impedance of the load connected to the loop. If we assume a resistive load, then the current can be calculated as I = V/R, where R is the resistance of the load. Once we have the maximum current, we can calculate the maximum power as P = 19.909 * I.

Alternatively, if we know the current in the power line (4000A), we can use the power formula P = I^2R, where R is the total resistance of the loop. From the solution provided, we can see that the total resistance of the loop is 100-A. Therefore, the maximum power that can be extracted from the power line is P = (4000)^2 * (100-A).

It is important to note that in this solution, we have assumed a resistive load and have not taken into account any losses in the loop or the power line. In a real-world scenario, these factors would need to be considered to accurately determine the maximum power that can be extracted from the power line.