F-1 Faraday's Law Faraday's Law is one of 4 basic equations of the theory of electromagnetism, called Maxwell's Equations. We have said before that charges makes electric fields. This is the truth, but not the whole truth. Michael Faraday (British physicist, c.1850) showed that there is a second way to make an electric field: a changing magnetic field makes an electric field. Faraday's Law (in words): An induced emf (E) is created by changing magnetic field. Definition: emf, E = a voltage difference ( = E d ) capable of doing useful work, generating power. Think of emf as a battery voltage. Batteries have an emf, but resistors do not, even though a resistor R can have a voltage difference across it ( = I R ) Definition: magnetic flux through a loop of area A = Φ B = BAcosθ = B A B is component of B perpendicular to the area A. If B-field is perpendicular to the area A, then Φ = B A. Units [Φ] = T m = weber (Wb) B θ B = B cos θ area A Faraday's Law (in symbols): loop of wire E (1 loop) Φ = t If B = constant voltage = E = 0 B(in) uniform voltmeter Φ If B is changing with time E = 0. t Φ If have several loops, E ( loop) = t = Dubson hys00 otes, University of Colorado
F- We can change the magnetic flux Φ in several ways: 1) change B (turn the magnetic field up or down) ) change A (by altering shape of the loop) 3) change the angle θ between the B-field and the plane of the loop (by rotating the loop, say) Example of Faraday's Law: We have a square wire loop of area A = 10 cm 10 cm, perpendicular to a magnetic field B which is 10 cm B increasing at a rate =+ 0.1 T / s. What is the magnitude of the t 10 cm emf E induced in the loop? Answer: E Φ (BA) B = = = = = = t t t What is the emf if = 1000 loops? (BA) E = = = t 3 1000 10 1 3 A (0.01m )(0.1T / s) 10 1m B Electrical Generators Convert mechanical energy (KE) into electrical energy (just the opposite of motors). A wire loop in a constant B-field (produced by a magnet) is turned by a crank. The changing magnetic flux in the loop produced an emf which drives a current. B crank I Lenz's Law The minus sign in Faraday's law is a reminder of Lenz's Law: the induced emf E induces a current that flows in the direction which creates an induced B-field that opposes the change in flux. light bulb Dubson hys00 otes, University of Colorado
F-3 Example: a loop of wire in an external B-field which is increasing like so B increasing induced B B ind induced I OR?? I ind Answer: B induced downward opposes the increase in original B. B decreasing B ind I ind Here, induced B is upward to oppose the decrease in the original B. Lenz's Law says "Change is bad! Fight the change! Maintain the status quo." Example of use of Lenz's Law A square loop of wire moving to the right enters a region where there is a uniform B-field (in). What is the direction of the current through the wire: CW or CCW? Answer: CCW The flux is increasing as the loop enters the field. In order to fight the increase, the induced B-field must be out-of-the-page. A induced CCW current will produce a B-field pointing out. Does the magnetic field exert a net force on the loop as it enters the field? Answer: Yes. The upward current on the right side of the loop will feel a force to the left (from F wire = ILB and R.H.R.). otice that the direction of the force on the wire loop will slow its motion. wire loop B = 0 here There is a subtlety in this problem that we have glossed over. To get the direction of the force on the right-hand side of the wire, we assumed that the direction of the (imaginary positive) moving charges in the wire is upward, along the direction of the current, and not to the right, along the I F v B (in) here Dubson hys00 otes, University of Colorado
F-4 direction of the motion of the entire loop. ow, it is really the negative conduction electrons that are moving within the wire, but we still have the problem of understanding which velocity v we should pick when we apply the force law F = q "v cross B". hould we pick the direction of the electron current (downward, parallel to the wire), the direction of the motion of the loop (to the right), or some combination of these directions? The conduction electrons in the right half of our wire are actually moving to both downward and to the right. But only the downward motion matters, because the motion to the right is effectively canceled by the motion of the positive charges within the wire. Remember that the wire is electrically neutral; there are as many fixed positive ions in the wire as there are mobile negative electrons. The force on the electrons due to their rightward conventional current I net force on wire electron current motion is exactly canceled by the force on the positive charges, which have exactly the same rightward motion. But the force on the conduction electrons due to their downward motion is not canceled out, and this is the cause of the net force on the wire. B(in) uniform motion of wire and of fixed positive ions in wire motion of conduction electron in wire Eddy Currents If a piece of metal and a B-field are in relative motion in such a way as to cause a changing Φ through some loop within the metal, then the changing Φ creates an emf E which drives a current I. This induced current is called an eddy current. The relative direction of this eddy current I and the B-field are always such as to cause a magnetic force (F = I L B sinθ ) which slows the motion of the metal (as in the example above). Again, if metal moving in B-field produces a changing Φ and the direction of the force always slows the motion. (Faraday) I F on metal ( = ILB) eddy current Dubson hys00 otes, University of Colorado
F-5 If the eddy current force did not slow the motion, but instead aided the motion, then we would have runaway motion free energy violation of energy conservation. Transformers The entire electrical power distribution system in the civilized world depends on a simple device called a transformer. A transformer is a device for transforming AC voltage from one value (say 10 AC) to another value (like 10 AC or 000 AC). A transformer is made of coils of wire, usually wrapped around an iron core. It is a simple device with no moving parts. rimary coil = input coil, with turns in = econdary coil = output coil, with turns out = We will show below that = = or = out Fe core This "Transformer Equation" says that, for AC voltage, the voltage ratio is equal to the turns ratio. OTE: the transformer only works for AC voltage. If in is DC, then out = 0. = > 1 "step-up transformer" = < 1 "step-down transformer" (A step-down transformer gives a smaller, but a larger current I.) Transformers work because of Faraday's Law: (AC) I (AC) B (AC) B (AC) + Faraday E = Dubson hys00 otes, University of Colorado
F-6 roof of the Transformer Equation: If we apply Faraday's Law to the primary and secondary coils, we get: (1) () Φ t = Φ t = same Φ = B A in each turn of primary and secondary because the iron core "guides flux" from to. (1) () = (End of proof.) If a transformer is well-designed, only 1 to 5% power in is lost to heating of coils and eddy currents in the iron core. I out in I = I = = I A step-down transformer produces a smaller voltage, but a bigger current (same = I ). Light bulbs and appliances with motors (vacuum cleaners, blenders) use AC voltage to operate. But devices with electronic circuits (T's, computers, phones, etc) need DC voltage to function. The "power supply" in computers and T's converts the AC voltage from the wall socket into DC voltage (usually 10-15 ) that the electronic circuitry needs. Example of use of transformers: uppose you want to melt a nail by putting a big current through it. What happens if you try to melt the nail by putting 10 AC (from your wall socket) across the nail? Answer: you will blow a fuse or trip a breaker. The resistance of a nail is quite small: R nail 10 3 Ω. The current produced by a 10 voltage difference across the nail is I = = 10 = 10000 A. This will never happen since your breaker will Ω huge: nail 3 Rnail 10 trip when the current exceeds 15 A. (Here's an experiment you should never try at home: Bend a nail into a U shape and plug it into your wall socket. Watch the lights go out.) o how do we melt that nail? olution: Use a 100-to-1 step-down transformer. Dubson hys00 otes, University of Colorado
F-7 1 100 = =, = in = 10 AC out 1 = = = 10 = 1. AC 100 1. I = I = = = 100 A Ω out 3 Rnail 10 (enough to melt the nail) How much current will this draw from the wall socket? Recall that in = out or I = I. 1 = (ot enough to blow the fuse.) I = Iin = I = I = (100 A) 1 A 100 Circuit diagram: = 10 I = 1 A 100:1 symbol for transformer R nail 0.001 Ω I = 100 A ower dissipated in nail = I R = (100) (10 3 ) = 1440 W will melt nail. Transformers and ower Distribution Economical power distribution is only possible because of transformers. Electrical power is transmitted from the power plant to the city by big aluminum cables (power lines). ome energy is inevitably wasted because the power lines have a resistance, and so they get hot: lost = I R cable. In order to minimize this waste, the power must be transmitted from the plant to the city at very high voltage (typically 300 k). A high voltage allows a small current, at a given power (since = I). And a small current means small I R losses in the cable. When the high-voltage power is delivered to the city, step-down transformers are used to transform the very dangerous high voltage down to the not-so-dangerous 10 before it enters your home. The voltage is stepped down in stages as it is distributed throughout the city. Dubson hys00 otes, University of Colorado
F-8 0 k step-up step-down step-down 300 k 8 k 10 Home power plant transmission line substation neighborhood A very simplified model of power distribution (transformers not shown): R cable ( made as small as possible) ower plant (It's really AC) I R city (adjusted to keep total power fixed) ome typical numbers: ower output of plant = out = 100 MW to 1 GW = 10 7 to 10 8 W (fixed by demands of the city) out = I, lost = I R cable Using I = out /, we get lost Fraction of power wasted = If R cable 10Ω and out = 10 8 W, then out out = R cable lost out = R cable If = 50,000 : If = 00,000 : 10 = 10 = 0.4 8 lost out 5 10 4 ( ) 10 = 10 = 0.05 8 lost out 10 5 ( ) (40% lost!) (.5% lost) Boosting the voltage at which the power is transmitted makes the losses acceptably small. Dubson hys00 otes, University of Colorado