Lecture 18: Faraday s Law & Motional EMF

Transcription

1 Outline: Lecture 18: Faraday s Law & Motional EMF More on Faraday s Law. Motional EMF. E = dφ B dt loop E dl = d dt surface B da

2 Iclicker Question A circular loop of wire is placed next to a long straight wire. The current I in the long straight wire is increasing. What current does this induce in the circular loop? A. a clockwise current B. a counterclockwise current C. zero current D. not enough information given to decide

3 Iclicker Question The drawing shows the uniform magnetic field inside a long, straight solenoid. The field is directed into the plane of the drawing, and is increasing. What is the direction of the electric force on a positive point charge placed at point a? A. to the left B. to the right C. straight up D. straight down E. misleading question the electric force at this point is zero

4 Loop Rotating in Magnetic Field φ = ωt ω = angular frequency = angular velocity Φ m = B da = BA cos ωt ℇ = dφ m dt = ωba sin ωt

5 Alternator / DC Generator Alternator: generator of AC current. ℇ = dφ B dt t Φ B DC generator: generator of DC current (DC motor in reverse). ℇ t Φ B

6 Iclicker Question A flexible loop of wire lies in a uniform magnetic field B directed into the plane of the picture. The loop is pulled as shown, reducing its area. The induced current A. is zero because the magnetic field is time-independent. B. flows upward through resistor R. C. flows downward through resistor R. D. does not flow through resistor R.

7 Iclicker Question The flux of the magnetic field in the wire loop varies with time as shown in the Figure. Which dependence(s) Φ B t correspond(s) to the largest magnitude of the induced current? Φ B A. 1 B. 2 & 3 C. 5 D. 4 & 5 Φ B t

8 iclicker Question loop 1 loop 2 Two circular loops are stacked on the same axis as shown in the figure. One loop has a steady current flowing through it (provided by a power supply). When the power is turned off, will the two loops briefly attract or repel each other? A. They will attract each other. B. They will repel each other. C. There is no induced current. D. They will neither repel nor attract each other. E. The answer depends on the direction of current in loop 1.

9 F B I v F us Motional EMF Primitive model of a DC generator. Part of the loop is in a uniform B field. The loop is pulled to the right. Find the induced current I. 1. Faraday s Law approach: ℇ = dφ dt = d dt BLx = BLv Current: I = ℇ R = BLv R Thermal power (Joule heat ) released in the wire: P = ℇ2 R = BLv 2 R Who does the corresponding work? Of course, not the magnetic field! We do, by pulling the loop with constant force: F us = IL B = B2 L 2 v R P = Fv = BLv 2 R

10 Motional EMF I 2. Lorentz force approach: F B v F us There are two components to the velocity: The pulling velocity v and the drift velocity v d. The Lorentz force due to the pulling velocity gives rise to an EMF: ℇ = E NC dl loop = loop F L q dl = vert.segment in B field qv B q dl = vbl The Lorentz force from v d gives rise to a resistive force (braking force).

11 Motional EMF General expression for the motional EMF: ℇ = v B dl loop Sometimes (e.g. Fig. (b)) the motional EMF can be successfully described using Faraday s Law. Sometimes (e.g. Fig. (a)) we have to go back to the Lorentz force description given above.

12 Example A metal disc rotates about a horizontal axle. A uniform magnetic field B is directed parallel to the axle. A circuit is made by connecting one wire end of a resistor to the axle and the other end to a sliding contact which touches the outer edge of the disc. Find the current through the resistor. F L = qvb directed along the radial direction v = ωr ℇ = E NC dl loop = loop F L q dl a = vbdr 0 a = ωb rdr 0 = ωb a2 2 I = ℇ R = ωba2 2R

13 Eddy Current (Foucault) Brake Linear eddy current brake (useless at low speed, efficient at high speed) The current direction in the coils positioned along the rail is alternating: Demonstrations: brakes for linear and rotational motion

14 Iclicker Question A. 1, 2, 3, 4 B. 4, 3, 2, 1 C. 1, 4, 3, 2 D. 2, 3, 4, 1

15 Faraday Law Example Consider a wire loop (radius a) and a long solenoid (radius b) with a uniform B(t) = B 0 (t/ ) inside and B = 0 outside of the solenoid. Find current I in the wire if its resistance is R. E dl loop = d dt B da surf 2πa E = πb 2 db t dt = πb 2 B 0 τ Note that B=0 at the wire s location! Still, there is an induced E 0. E = b2 B 0 2 aτ I = b2 B 0 2aτR A wire loop has a total resistance R. If the total flux through it changes from i to f, show that the magnitude of the total charge that will flow through the loop is given by Q = i f I t dt Q = Φ i Φ f R = 1 R i f dφ t dt dt = Φ i Φ f R