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Ch – 33 Electromagnetic Induction Reading Quiz – Ch. 33 1. Currents circulate in a piece of metal that is pulled through a magnetic field. What is the correct name for these currents? a. Eddy currents c.Flux currents. b. Induction currents d.Faraday currents. 2. The magnetic flux is a measure of the magnetic field: a. while electromagnetic induction occurs b. parallel to a closed loop c. passing through a closed loop d. when it is changing Learning Objectives – Ch 33 • Understand how a changing magnetic field will induce a potential difference – this is the phenomenon of electromagnetic induction. • To observe the experimental evidence for electromagnetic induction. • To understand and use Lenz’s law for induced currents. • To learn of Faraday’s law as a new law of nature. • To understand basic applications of electromagnetic induction to technology. • To gain a qualitative understanding of electromagnetic waves. • To analyze circuits with inductors. Induced Current • A current creates a magnetic field • Is the opposite true? • Can a magnetic field create, or induce a current? Faraday’s Discovery • Faraday was hoping that the magnetic field generated by the current on the left circuit would induce a current in the wire on the right. • But no luck, as the meter shows. Faraday’s Discovery • So he shut down for lunch. • But when he opened the switch to cut off the current in the left circuit, the meter suddenly moved, showing a momentary current in the wire on the right. • It quickly went back to zero. Faraday’s Discovery • Baffled, he closed the switch again. • And noticed that the meter jumped again momentarily, and this time the meter needle went the other way. • I bet he actually did it a bunch of times before he noticed that it went the other way.. Faraday’s Discovery • Faraday found that he could induce a current in a closed wire, but only if the magnetic field through the coil is changing. • This is an informal statement of Faraday’s Law. Motional emf An induced current in a circuit can be Changing Stationary circuit created 2 ways: magnetic field (since current is with meter changing) 1. Change the strength of the magnetic field through a stationary circuit. That’s what Faraday did by opening and closing the switch. Motional emf Circuit changes Constant B size due to The other way to induce field into the page a current: 2. Change the size or orientation of the circuit in a stationary magnetic field. We shall look at this second method first. This is motional emf. Motional emf • The external B field causes a magnetic force on positive and negative charges moving to the right. • The electron holes move up and the electrons move down. Motional emf • Since the rest of the loop isn’t moving, there is no magnetic force on it, so the electron holes will flow along the wire to get back to the more negative side. • Wow, electric current with no power source except a moving wire! Why aren’t we all fat and happy? Motional emf • We’ll find out soon. • Another result of the charge separation is an E field. • This E field also causes a force, which is in the opposite direction as the force due to the B field. Motional emf • Electron holes continue to move up, but only until FE down equals FB up, at which charge separation ceases. • As long as the wire keeps moving, there will be a charge separation. • The magnetic force is doing work to maintain that charge separation. Workbook exercises • 1 b,d,e Workbook exercises - answers • b: ccw • d: ccw • e: 0 Motional emf Recall emf (motional or otherwise) is the work done per unit charge, i.e. a potential difference. ∆V = - Eds When FE equals FB (assume α = 900): qE = qvB or E = vB Motional emf For a wire of length L, moving in a direction perpendicular to an external B field: l ∆V = - vBds 0 emf = v LB Motional emf • The emf due to the charge separation exists whether or not the loop is closed (battery analogy). • If there is a closed loop: I = vLB/R • The induced current is due to magnetic forces on moving charges. Why aren’t we fat and happy? • Once there is a current flowing through the wire, charges move in 2 directions. – They all move right with the moving wire. – Current moves up. • Yet another magnetic force generated, this time to the left. Why aren’t we fat and happy? • This magnetic force opposes the original velocity of the moving wire. • The moving wire will slow down and stop. • Need a constant Fpull to the right to make the contraption work. • That’s why we’re not fat and happy. Why aren’t we fat and happy? • To keep the wire at constant speed, and continue the emf and current Fpull = Fmag F = ILB (Ch. 32) Fpull = Fmag = ILB I = vLB/R F = vL2B2/R Why aren’t we fat and happy? Fpull = Fmag Wpull = Wmag The rate at which work is done on the circuit equals the power dissipated by the circuit: P = I2R = v2L2B2/R Numerical Example • #3 = end of chapter Magnetic Flux m A B Φm = AB cos θ Units: 1 weber = 1Wb = 1Tm2 Magnetic Flux in a non-uniform field • Divide the loop into many small pieces d m B dA • The flux is the sum of all these: m B dA Lenz’s Law • There is an induced current in a closed conducting loop only if the magnetic flux is changing (either B, A or θ). The direction of the induced current is such that the induced magnetic field opposes the change. Using Lenz Law 1. Determine the direction of the external magnetic field. 2. Determine how the flux is changing. Is it increasing, decreasing, or staying the same? 3. Determine the direction of an induced magnetic field that will oppose the change in the flux. – Increasing: induced magnetic field points opposite the external magnetic field. – Decreasing: induced magnetic field points in the same direction as the external magnetic field. – Constant: no induced magnetic field. 4. Determine the direction of the induced current. Use the right-hand rule. Faraday’s Law Recall that a current in a circuit can be created 2 ways: – Change the size or orientation of the circuit in a stationary magnetic field. – Change the strength of the magnetic field through a stationary circuit. Both of these create a changing magnetic flux. Faraday’s Law The current exists because the changing magnetic flux has induced an emf. In a closed circuit with a resistance, R: I = ε/R The current is a consequence of the emf. The emf is a consequence of Φm Faraday’s Law An emf is induced in a conducting coil of N turns if the magnetic flux through the coil changes. The magnitude of the emf is equal to the rate of change of the magnetic flux: d m N dt The direction is given by Lenz Law. Faraday’s Law Recall the expression for emf of a wire of length L, moving in an external B field: l ∆V = - vBds 0 emf = v LB Faraday’s Law Does Faraday’s Law give us the same? ε = dΦ/dt = d(xLB)/dt L, B are constant so ε = (LB)dx/dt ε = v LB Faraday’s Law For the case shown, the induced B field will be in which direction? Workbook exercise #14 (p.33-8) Draw a graph of the current, given the graph of the magnetic field Answer I0-1s: negative constant I1-2s: no current I2-3s: positive constant, smaller magnitude than I0-1s Numerical Problem The resistance of the loop is 0.10 Ω. A. Is the magnetic field strength increasing or decreasing? B. What is the rate of change of the the magnetic field? Numerical Problem A. Induced current is shown moving ccw. RH rule indicates a magnetic field out of the page, opposing external field. Therefore, external magnetic field must have been increasing. B. Rate of change is 2.34 T/s Inductors • An inductor is a device that produces a uniform magnetic field when a current passes through it. A solenoid is an inductor. • The magnetic flux of an inductor is proportional to the current. • For each coil (turn) of the solenoid: Φper coil = A•B Φsol = N(A•B) = NAB = NA(u0NI/ℓ) = (Au0N2/ℓ)Isol • This is actually a self-inductance Inductors • The proportionality constant is defined as L, the inductance: Lsol = Φsol /I = Au0N2/ℓ • Note that the inductance, L depends only on the geometry of the inductor, not on the current. • The unit of inductance is the henry 1 H = 1 Wb/Ampere The circuit symbol for an inductor: Potential difference across an inductor • For the ideal inductor, R = 0, therefore potential difference across the inductor also equals zero, as long as the current is constant. • What happens if we increase the current? Potential difference across an inductor • Increasing the current increases the flux. • An induced magnetic field will oppose the increase by pointing to the right. Induced Induced current • The induced current is field opposite the solenoid current. • The induced current carries positive charge to the left and establishes a potential difference across the inductor. Potential difference Potential difference across an inductor The potential difference across the inductor can be found using Faraday’s Law: Induced d m Induced current N field dt Where Φm = Φper coil Φsol = N Φper coil We defined Φ = LI dΦsol/dt = L |dI/dt| Potential difference Potential difference across an inductor • If the inductor current is decreased, the induced magnetic field, the induced current and the potential difference all change direction. • Note that whether you increase or decrease the current, the inductor always “resists” the change with an induced current. The sign of potential difference across an inductor ∆VL = -L dI/dt • ∆VL decreases in the direction of current flow if current is increasing. • ∆VL increases in the direction of current flow if current is decreasing. • ∆VL is measured in the direction of current in the circuit Conceptual Question - Inductors Which of the following statements could be true? a. I is from a to b and constant. b. I is from a to b and increasing. c. I is from a to b and decreasing. d. I is from b to a and steady. e. I is from b to a and increasing. f. I is from b to a and decreasing. Faraday’s Law Problem (#34, p.1078) The top figure shows a 5- turn 1.0-cm diameter coil with R=0.10 Ω inside a COIL 2.0-cm solenoid. The solenoid is 8.0 cm long, Solenoid cross section at t = 0.02s has 120 turns and carries the current shown in the graph. A positive current is cw when seen from the left. Determine the direction and magnitude of current in the coil. Setting up the problem The mathematical representation is Faraday’s Law: COIL d m N Solenoid cross section at t = 0.02s dt I = ε/R Icoil = 1/R (N dΦ/dt) For each of the following variables, decide whether the quantity refers to the solenoid or the coil: R, N, Φ Setting up the problem Icoil = 1/R (N dΦ/dt) For each of the following COIL variables, decide whether the quantity refers to the solenoid or the coil: Solenoid cross section at t = 0.02s R, N, Φ Answers: R – coil, N – coil Φ – coil Icoil = 1/Rcoil (Ncoil dΦcoil/dt) Setting up the problem Icoil = 1/Rcoil (Ncoil dΦcoil/dt) For the above expression: COIL Write an expression for Φcoil in terms of the magnetic field and the Solenoid cross section at t = 0.02s area. Once again, the pertinent question is solenoid or coil: Magnetic field (B) Area (A) And what about cos θ? Setting up the problem Area (A) – the current is induced in the coil, due to COIL the flux through the coil; therefore it’s the area of Solenoid cross section at t = 0.02s the coil. Magnetic field (B) – It’s the external magnetic field of the solenoid that induces a current in the coil Φcoil = Acoil Bsolenoid And what is the magnetic field of a solenoid? Setting up the problem Φcoil = Acoil Bsolenoid COIL Write an expression for the magnetic field of a solenoid. For any Solenoid cross section at t = 0.02s variable that has a value for both coil and solenoid (e.g I,R,N,r) specify to which you are referring. Setting up the problem Φcoil = Acoil Bsolenoid COIL Bsol = (u0NsolIsol)/ℓ (we weren’t given a length Solenoid cross section at t = 0.02s for the coil). Now we need an expression for dΦcoil /dt, so write one. Setting up the problem Φcoil = Acoil Bsolenoid COIL Bsol = (u0NsolIsol)/ℓ Solenoid cross section at t = 0.02s dΦcoil = (Acoil u0Nsol/ℓ)dI/dt dt We may actually be ready to solve this puppy. Icoil = 1/Rcoil (Ncoil dΦcoil/dt) Answer COIL Icoil = 1/Rcoil (Ncoil dΦcoil/dt) Solenoid cross section at t = 0.02s Icoil = 3.7 x 10-4A or .37 mA Direction: clockwise when seen from the left (induces a field to the right).
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