A Solution to the RWP for Exam 1 - Stealing Power

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.

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