1 Part 4: Electromagnetism 4.1: Induction A. Faraday's Law The magnetic flux through a loop of wire is Φ = BA cos θ B A B = magnetic field penetrating loop [T] A = area of loop [m 2 ] = angle between field vector and a vector perpendicular to the plane of the loop mks units of flux [T-m 2 = Weber = Wb] If we define the area vector A as a vector perpendicular to the plane of the loop and with a size equal to the area of the loop, then the flux can be written as Φ = B A Faraday s Law of Induction states that a voltage difference is induced across the ends of a coil of N turns if the magnetic flux through the coil changes. V ind = N Φ Δt If the time interval approaches zero, then you can write that V ind = N dφ dt V = absolute value of the voltage difference [V] N = number of turns in the coil = magnetic flux through one coil [Wb] If the ends of the coil are connected such that a current can flow through the coil and the coil has resistance, then there is an induced current given by Ohm s Law. I ind = V ind
2 Example: A coil has 100 turns and each turn has an area of 5 cm 2. The coil is sitting in a uniform magnetic field of 0.3 Teslas where the field direction is perpendicular to the plane of each loop. The field is turned off in half of a second. Find the average induced voltage across the coil and the average induced current that flows through it. Ans. 30 mv and 2 ma The direction of the induced current is given by Lenz s Law. The induced current will flow in a direction that produces a magnetic field that opposes the change in the magnetic flux. Example: We did many examples of Lenz s law in class. You can also practice applying it in the computer problem entitled Induction (Lenz s Law).
3 B. AC Generators A generator consists of a coil that is rotated in a magnetic field. A time-varying voltage difference that regularly changes polarity is produced of the form V(t) = V M sin(ωt) V M = amplitude of the voltage [V] = angular frequency of the rotation [rad/s] Note that = 2 f where f is the frequency in units of cycles per time [cycle/s = Hertz =Hz]. The amplitude of the voltage depends on the number of turns in the coil, the area of each turn, the size of the magnetic field, and the frequency of the rotation. V M = NBAω The voltage is sinusoidal in time as the plot shows. V T 0 V M t A positive voltage difference is for one polarity and a negative voltage difference for the other polarity. T = period = time for one cycle [s] T = 1/f The root mean square or rms voltage represents an average that is useful in analysis. For a sinusoidal voltage signal V rms = V M 2
4 4.2 Maxwell s Equations The integral forms of Maxwell s Equations are written below. The emphasis in this class is not on calculating the integrals but rather on understanding the physics expressed in the equations. Gauss s Law for Electricity Electric field lines E da = q in ε The Physics: The electric flux through a closed surface is proportional to the net charge inside the surface. The source of electric field is charge. The charge of a single object can be positive, negative, or zero. Gauss s Law for Magnetism B da = 0 The Physics: The magnetic flux through a closed surface is zero. Magnetic field lines form closed loops. There are no magnetic monopoles. Faraday s Law E ds = d dt ( B da ) Magnetic field lines Magnetic field lines The Physics: An electric field is induced in a region where there is a changing magnetic field. Ampere-Maxwell Law I B ds = μi + με d dt ( E da ) Electric field lines The Physics: A current (moving charge) creates a magnetic field. A magnetic field can also be induced in a region where there is a changing electric field.
5 4.3 Inductance & Inductors A inductor is a device that develops a voltage difference between the ends of the device when the current (I) through it changes. The inductance (L) of the inductor determines the size of the induced voltage (V). V = L I Δt If the time interval approaches zero, then you can write that V = L di dt In this equation, I is the current and V is the size of the induced voltage. mks units of inductance [Henry = H = V s/a = s] The inductance of an inductor depends on the shape of the inductor and the material at its core. For a long solenoid, L = μn2 A l A N turns where N is the number of turns, A is the area of one of the turns, is the length of the coil, and is the magnetic permeability of the core. If the core is free space, = o = 4 x10-7 T m/a. An inductor stores energy in its magnetic field as long as a current flows through the inductor. U = 1 2 LI2
6 Transformer A transformer consists of two separate coils wrapped a common ferromagnetic core. An ac voltage difference supplied to the primary coil results in an induced ac voltage signal across the secondary coil. V 2 = N 2 N 1 V 1 V 1 = average voltage supplied to primary coil N 1 = number of turns in primary coil V 2 = average voltage across secondary coil N 2 = number of turns in secondary coil The transformer must obey the Law of Conservation of Energy. Some energy is always lost in the core due to the heating of the core from induced eddy currents. Thus, the average output power of the secondary coil is less that the average input power used by the primary coil. P 2 = ηp 1 = power efficiency of transformer ( < 1) The average power can be written as P = IV so the transformer power equation can be written as I 2 V 2 = ηi 1 V 1 Example: A step-up transformer with a turn ratio of 4 is used to power a motor. The primary side of the transformer uses 10 V. When connected to the transformer, the motor draws 237.5 ma from the secondary coil. The power efficiency of the transformer is 95%. Find (a) the voltage difference supplied to the motor and (b) the current that flows into the primary side. Ans. (a) 40 V (b) 1 A
7 4.4 DC Circuits with Inductors L Circuits Close switch at t = 0 Turning on a current: Current increases in size to its final value of V o /. V o L I(t) = V o (1 e t τ) Turning off a current: Current through resistor decreases with time to zero. Simultaneously open switch 1 and close switch 2 at t = 0 1 I(t) = V o e t τ V o 2 L In both cases, is the L time constant: = L / [s] The current reaches its final value after about 5 time constants. LC Circuit A fully charged capacitor with an initial charge of Q on its plates is discharge through a pure inductance. The charge on a plate and the current flowing through the circuit oscillate at the same angular frequency of according to q(t) = Q cos(ωt) I(t) = dq dt = ωq sin(ωt) where the natural oscillation frequency is ω = 1 LC C Close switch at t = 0 L
8 4.5 AC Circuits Consider a sinusoidal voltage signal from an ac supply of the form V(t) = V M sin(ωt) When this supply is connected to a single device (a resistor, capacitor, or inductor), the ac current that flows will be of the form and where I(t) = I M sin(ωt + ) I M = V M X = phase angle between current and voltage [rads] X = reactance of device [ ] Device X ( ) (rads) esistor 0 current in phase with voltage Capacitor 1/( C) /2 current leads voltage Inductor L - /2 current lags voltage If the supply is connected to more than one device, then the current will be of the same form and the reactance of the circuit is called the impedance. and where I(t) = I M sin(ωt + ) I M = V M Z = phase angle between current and voltage [rads] Z = impedance of circuit [ ] The phase angle and impedance depend on the type of devices in the circuit and how they are connected. Since the rms value is proportional to the amplitude, we can also write I rms = V rms Z
9 Average power delivered to circuit: P ave = I rms V rms cos φ Example: LC Series Z = 2 + (ωl 1/ωC) 2 tan φ = ωl 1/ωC P ave = V rms 2 2 Z 2 = I rms At resonance, the impedance is minimized and the rms current and average power are at their maximum values. The current and voltage signals are in phase ( = 0). esonant frequency: ω o = 1 Minimum impedance: LC Z min = Maximum rms current: V rms L I rms max = V rms C Maximum average power: P ave max = V 2 rms
10 Example: An LC series circuit uses a 250-Ohm resistor, a 0.6-Henry inductor, and a 3.5-microFarad capacitor. It is connected to an ac voltage supply operating at 60 Hz with an amplitude of 150 V. Find the following: (a) rms voltage of the supply (b) reactances of the individual devices. (c) impedance of the circuit. (d) rms current flowing through the circuit (e) average power delivered to circuit (f) resonant frequency of the circuit in Hz (g) rms current at resonance (h) average power at resonance Ans. (a) 106 V (b) : 250 L: 226 C: 758 (c) 588 (d) 180 ma (e) 81 W (f) 110 Hz (g) 424 ma (h) 81 W