A farmer
reportedly stole electrical power by
strategically placing a large coil of wire beneath the high-voltage
transmission lines that crossed his field. For several years, the
farmer obtained free electricity to operate equipment for his farm,
until the power company finally discovered the theft. Eventually
the farmer was convicted of stealing power even though no physical
connections were made to the transmission lines. Use this
information and your knowledge of physics to answer the following
questions: (Each question is worth 2 points, except for #4, which
is worth 5 points, for a total of 25 points.)

1) Without examining the farmer's property, how could the power company recognize that energy was being stolen?

In principle, the power company could calculate the difference between the power that was being delivered and the power that was being purchased. An unexplained difference (beyond the usual power loses via resistive heating along transmission lines) might be a result of someone stealing power. In practice, the power stolen by the single user in this scenario would most likely be too small to detect.

2) What
principle did the farmer utilize to steal electrical
power from the transmission lines?

The farmer used Faraday's
Law of Induction to induce a voltage and current in the coil of
wire placed near the high-voltage transmission line. This effect
is possible because the alternating current through the transmission
line changes the magnetic field around it, and if the inductive coil is
positioned so that the magnetic field lines penetrate its area, then an
emf will be induced, and an alternating current will flow through the
coil.

3) How
must the farmer have positioned and oriented the coil in
order to most effectively achieve his goal?

The coil should be positioned as close as possible to the transmission
line since the magnetic field is inversely proportional to the
distance from the line, and the coil should be oriented upright
in a vertical plane aligned with the transmission line to
maximize the magnetic flux from the
magnetic field lines that encircle the transmission line.

4) Assume
that the lowest transmission line was 10 m above the
ground and carried current alternating at 60 Hz with a maximum of 150 A
at 230 kV. If the coil was in the shape of a square 5 m on a side
and touching the ground, approximately how many turns (loops) of wire
were needed for the coil to produce a standard voltage of 120
V?

The maximum emf induced in the coil depends on the
change in magnetic flux: emf = Nd(phi)/dt = NAdB/dt.

For this problem, the magnetic flux changes because the strength of the
magnetic field varies due to the alternating current, which peaks in
each direction twice per cycle or once every 1/120 of a second.
This means that the magnetic field changes from its maximum value to
zero four times per cycle, so dt = 1/240 of a second = 4.2
ms. The maximum magnetic field around the transmission line can
be estimated from the maximum current through the wire, using the
equation for a long, straight wire: B(wire) = uo*I/(2pi*r).
Since the square induction coil extends a distance of 5 to 10 m away
from the
transmission line, the magnetic field will be weaker at the far end of
the coil and stronger at the nearest end, but using the average radial
distance of 7.5 m gives the average magnetic flux for a specific
current. Therefore, the maximum magnetic field at a distance of
7.5 m from the transmission line is B = (4pi*10^-7 T*m/A)(150
A)/(2pi)/(7.5 m) = 4.0 uT, which is consistent with the mean magnetic
field of 6 uT near a 230-kV line indicated in Reference 2.

The number of turns required to produce an rms voltage of 120 V (or a
peak voltage of 170 V) is then:

N = (170 V)(4.2 ms)/(25 m^2)/(4.0 uT) = 7140 turns

5) Why would
the voltage in the coil vary depending on the time of
day? When would you expect it be largest and smallest?

The voltage induced in the coil depends on the
strength of the magnetic field produced by the transmission line, which
depends on the amount of current passing through the line, and this
depends on the demand for power (load) by users of electricity from
this line. We should expect this demand for power to be greatest
during the afternoon and evening hours, and smallest during the early
morning hours of each day. This daily variation in peak power
demand can be seen in the graph in Reference 2.

6) What is the
frequency of the current in the coil? What is the phase
angle between the voltage in the transmission line and the coil?

Since the current induced in the coil results from
changes in the current in the transmission line, they should both have
the same frequency of 60 Hz.
However, the phase angle between these currents (or the corresponding
voltages) will be 90 degrees
since the induced current in the coil reaches a maximum value when the change in
current in the transmission line is greatest, and this occurs 1/4 of a
cycle, or 360/4 = 90 degrees before and after the current peaks in the
transmission line.

7) If the coil
and all of the equipment connected to it had a total
impedance of 200 ohms, what is the maximum rate at which energy was
consumed?

The maximum rate at which energy is consumed
(power) is: Pmax = Irms*Vrms = (Vrms)^2/Z = (120 V)^2/(200
ohms) = 72 W

A more accurate estimate of the average
power could be found by including an appropriate power factor = R/Z, and this could
be determined by calculating the inductance of the coil and determining
the inductive reactance. Using Reference 3, the inductance for
this coil is approximately 1600 H (which is a very large
inductance). The inductive reactance is then X(L) = 2(pi)(60
Hz)(1600 H) = 603,000 ohms! This is much greater than the
specified impedance of 200 ohms, so it seems that the power factor
would be nearly zero!

8) Assuming a
cost of $0.10/kW-h, what is the approximate value of the
energy stolen by the farmer over the period of one year?

If we assume that the farmer used the equipment
half of the time (an average power consumption of 36 W over the course
of a day), then the total cost for one year would be: cost = (36
W)(24 h/day)(365 days/year)($0.10/kW-h) = $31.54/year. Even at
the maximum power rate of 72 W, the value of the energy stolen would be
only $63/year. This hardly seems worth the effort, risk, and cost
(see below).

9) Estimate the
cost to make the coil, assuming that the farmer used
12-Gauge copper wire at a cost of $0.15/ft. Evaluate the
cost-effectiveness of the farmer's design and calculate the approximate
pay-back period for his investment.

With 7140 loops of wire required and a length per
loop of 20 m, the farmer would need 143 km of wire! At $0.15/ft =
$0.50/m, this wire would cost $71,400!
This means that even at the maximum energy consumption rate, the payback period would be at least 1000
years!

10) What advice
would you give to the farmer to
maximize the efficiency
of his design?

To maximize the emf induced in the coil while
using the least amount of wire to save on cost, the farmer should place
the coil as close as possible to the transmission line and use a coil
with the largest possible area (since doubling the area of a square
coil quadruples the induced emf while only doubling the cost).
Placing iron inside the coil would increase the efficiency of this
transformer, but the effect would be rather small and probably not
worth the effort.

11) What other insights can be learned from this
problem?

Based on the calculations above, it seems that
this method of stealing power is highly inefficient and
impractical. I seriously doubt that any farmer has actually done
this (perhaps this is a myth suitable for the Mythbusters to
investigate!). This application of Faraday's law is essentially a
power transformer with very low efficiency. There are many other
applications of Faraday's law that pervade our modern lives; this is
just one that is not advised.

1) http://sci-phys-plasma.caeds.eng.uml.edu/1999/01-99-011.htm

2) http://infoventures.com/private/federal/q&a/qaenvn2a.html

3) http://www.technick.net/public/code/cp_dpage.php?aiocp_dp=util_inductance_rectangle