**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=I**

(a) Find the maximum power that can be extracted.

(b) Will I extract more power when I form multiple loop instead of single?

_{0}cos(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

|----------R

_{wire}-----------|

| |

| | a

| |

|------------R

_{load}---------|

**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 P**

(b) The problem here is that I found that for N loop wire, I would have V

and the resistant of the wire should be

R

Hence, I concluded that

P= [tex]\frac{V_{total}^{2}*R_{L}}{4R_{total L}^{2}}[/tex]

= [tex]\frac{(NV)^{2}}{4NR_{L}}[/tex]

= [tex]\frac{NV^{2}}{4R_{L}}[/tex]

_{rms}=190W(b) The problem here is that I found that for N loop wire, I would have V

_{total induced}=N*V_{1 loop induced}and the resistant of the wire should be

R

_{total L}=N*R_{load}Hence, I concluded that

P= [tex]\frac{V_{total}^{2}*R_{L}}{4R_{total L}^{2}}[/tex]

= [tex]\frac{(NV)^{2}}{4NR_{L}}[/tex]

= [tex]\frac{NV^{2}}{4R_{L}}[/tex]

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