Lecture 10 Induction and Inductance Ch. 30 Cartoon - Faraday Induction Opening Demo - Thrust bar magnet through coil and measure the current Topics Faraday s Law Lenz s Law Motional Emf Eddy Currents LR Circuits Self and mutual induction Demos Elmo Problems 15 and 34 Chapter 30 Polling
Demos 1. Galvanometer with bar magnet and 3 coils. 2. 2. Gray magnet, solenoid, two LED s, push and pull, shows different LED s light up. Lenz s Law. 3. Coil connected to AC source will induce current to light up bulb in second coil. 4. Jumping aluminum rings from core of solenoid powered by an AC source. Cool a ring. 5. Slowing down of swinging copper pendulum between poles faces of a magnet. Slitted copper pendulum. Eddy Currents 6. Inductive spark after turning off electromagnet. Inductance. 7. Neodymium magnet swinging over copper strip. Eddy currents 8. Neodymium magnet falling through copper pipe. Then cool with liquid nitrogen. Eddy currents 9. Hanging aluminum ring with gray magnet. Lenz s Law 10. Two large copper disks with one magnet levitating above another
Introduction Stationary charges cause electric fields (Coulombs Law, Gauss Law). Moving charges or currents cause magnetic fields (Biot-Savart Law). Therefore, electric fields produce magnetic fields. Question: Can changing magnetic fields cause electric fields?
Faraday's Law Emf =! d" m Discovered in 1830s by Michael Faraday and Joseph Henry. Faraday was a poor boy and worked as a lab assistant and eventually took over the laboratory from his boss. Faraday s Law says that when magnetic flux changes in time, an Emf is induced in the environment which is not localized and also is non-conservative. Lets look at various ways we can change the magnetic field with time and induce an Emf. If a conductor is present, a current can be induced. 1. What is magnetic flux! m? 2. What is an induced Emf?
Demo 1 Thrust a bar magnet through a loop of wire Magnetic flux Faraday s Law Emf =! d"! # =! B " nˆ da Field from Bar magnet A Units: B is in Tesla ˆn i A is in m 2! m is in (Webers) Wb Current flows in the ring in a direction that produces a magnetic field that oppose the bar magnet.
Lenz s Law B produced by bar magnet µ B i 0 2 R = at center Produced by current flowing in the wire. Lenz s Law The current flows in the wire to produce a magnetic field that opposes the bar magnet. Note North poles repel each other.
More on Lenz s Law: An induced current has a direction such that the magnetic field due to the current opposes the change in the magnetic flux that induces the current Question: What is the direction of the current induced in the ring given B increasing or decreasing? B due to induced current B due to induced current
Demo 2: Gray magnet, solenoid, Red and Green LEDs magnet B solenoid i Green N repel N Red Push gray magnet in, B field points to the right. Current starts to flow in coil to produce a field to oppose the initial field. Red LED lights up. Pull magnet out, the opposite happens and the Green LED lights up.
Demo 3: Coil connected to AC source Light bulb connected to second coil Coil 2 Iron core (soft) means lots of inductance in wire so AC Coil 1 doesn t heat up wire. Shows how flux changing through one coil due to alternating current induces current in second coil to light up bulb. Note no mechanical motion here.
Demo 4: Jumping aluminum ring from core of solenoid powered by an AC source. When I turn on the current, B is directed upward and momentarily the top of the iron is the North pole. If the ring surrounds the iron, then the flux in it increases in the upward direction. This change in flux increases a current in the ring so as to cause a downward B field opposing that due to the solenoid and iron. This means the ring acts like a magnet with a North pole downward and is repelled from the fixed coil. induced current induced B field Try a square-shaped conductor Try a ring with a gap in it Try a ring cooled down to 78 K N N AC source Iron core (soft) Coil Repulsive force because 2 North poles
Motional Emf- field fixed but conductor moves Pull a conducting bar in a magnetic field. What happens to the free charges in the material? Moving bar of length L and wih W entirely immersed in a magnetic field B. In this case an Emf is produced but no current flows!! W B L F + v Work = qvbl U = q! emf = qvbl! - v F! F m = qv! B Positive charges pile up at the top and negative charges at the bottom and no current flows, but an Emf is produced. Now let s complete the circuit.
Pull the rectangular loop out of the magnetic field. A current i will be induced to flow in the loop in the direction shown. It produces a magnetic field that tries to increase the flux through the loop. Motional Emf What force is required to keep current flowing in the circuit? Uniform magnetic field into the screen Wire F A A= area of magnetic field enclosed by the wire Close up of the wire F m = qv! F = il! B 1 B + v emf =! d" m =!B da
Motional Emf Continued F 1 = il! B Lorentz Law F F F 1 2 3 Faraday and Lenz's Law emf =! d" m dr =!B d(lx) = BiL = Bix =! F 2 =!B da F 1 = B2 L 2 v R cancel =!BL dx =!BLv emf = ir Ohms Law BLv = ir i = BLv R Uniform magnetic field into the screen R is the resistance F A This is the force you need to pull at to achieve constant speed v. F 1 =F A
Motional Emf : Work Done by me How much work am I doing in pulling the circuit? (Note that the magnetic field does not do any work,) W = F 1 d = B2 L 2 vd R What is the rate at which I am doing work? P=Fv P = F 1 v = B2 L 2 v 2 R What is the thermal energy dissipated in the loop? Circuit diagram for motional Emf. R is the resistance of the wire P = i 2 R i = BLv R P = B2 L 2 v 2 R Note that the rate at which I do work in pulling the loop appears totally as thermal energy.
Eddy Currents A solid piece of copper is moving out of a magnetic field. While it is moving out, an emf is generated forming millions of current loops as shown. Eddy currents are also formed in a copper pendulum allowed to swing across a magnet gap cutting magnetic lines of flux. Note that when the copper plate is immersed entirely in the magnet no eddy currents form.
. Demo 5: Show a copper plate swinging through a magnet. For example, in pulling it out, that part of the plate that was in the B field experiences a decrease in B and hence a change in magnetic flux in any loop drawn in that part of the copper. An emf is developed around such loops by Faraday's Law and in such a direction so as to oppose the change. Horseshoe magnet induced current S B Copper pendulum N B induced Pull back pendulum and release. Pendulum dampens quickly. Force acts to slow down the pendulum.
Demo 5 Cont. Also try copper plate with slits. What happens now? Demo 5 Cont. Note inductive spark when turning power off Demo 6: Show neodymium magnet swinging over copper strip.
Demo 7: Copper pipe and neodymium-ironboron magnet with magnetic dipole moment S N Copper pipe Two Norths repel so the magnet drops more slowly. N N d i a B F D N i B W.M. Saslow Am. J. Phys. 60(8)1977
Cool down the copper pipe with liquid nitrogen 28 K. What do you expect to happen? F D =Magnetic Drag Force ~!µ 2 d a 4! is the conductivity of copper! = 1 "! =! 0 (1+ "(T # T 0 )) = 1.69 $ 10 #8 (1 + 0.0043(68 # 293)) = 1.69 $ 10 #8 (1 # 0.96) = 0.067 $ 10 #8 % # m Decreases by a factor of 25 not quite right. Why!!
Demo 8: Hanging aluminum ring with gray magnet i induced current N N induced B field repel B Note 2 N poles. Move magnet toward ring they repel Current induced in ring so that the B field produced by the current in the ring opposes original B field. This means the ring current produces a N pole to push away the N pole of the permanent magnet. When magnet is pulled back, it attracts the ring. i N S B N Current in ring is opposite to that above
Changing magnetic field generates electric field The orange represents a magnetic field pointing into the screen and let say it is increasing at a steady rate like 100 gauss per sec. Then we put a copper ring In the field as shown below. What does Faradays Law say will happen? Current will flow in the ring. What will happen If there is no ring present? Now consider a hypothetical path Without any copper ring.there will be an induced Emf with electric field lines as shown above. In fact there will be many concentric circles everywhere in space. The red circuits have equal areas. Emf is the same in 1 and 2, less in 3 and 0 in 4. Note no current flows. Therefore, no thermal energy is dissipated
We can now say that a changing magnetic field produces an electric field not just an Emf. For example: Work done in moving a test charge around the loop in one revolution of induced Emf is Work = emf! q 0 " " = Work done is also F # ds = q E # ds q E(2! ) 0 0 r Hence, Emf = 2πrE or more generally for any path Emf =! E " ds d! B E # ds = " Faraday s Law rewritten $ But we can not say f V # V = #! E " ds because it would be 0. f i i Electric potential has no meaning for induced electric fields
Summary Characteristics of the induced emf The induced emf is not localized such as at the terminals of a battery. It is distributed throughout the circuit. It can be thought of as an electric field circulating around a circuit such that the line integral of the electric field taken around a closed loop is the emf. Since the line integral is not 0, the field is non-conservative. There are no equipotential surfaces. If there is a conductor present, a current will flow in the conductor. If no conductor is present, there is no current flow, only emf. Energy is dissipated only if charges are present.
Example: x x B x R x x x x r A magnetic field is to the board (screen) and uniform inside a radius R. What is the magnitude of the induced field at a distance r from the center? x x x Field circulates around B field d l! E is parallel to dl $ Edl = E2" r = # " m = BA = B! r " 2 d! d m 2 = E2! r = "! r m d ( B! r 2 ) =! r 2 db r db E =! r < 2 R db Notice that there is no wire or loop of wire. To find E use Faraday s Law. 2! re =! R 2 db 2 R db E =! r > 2 r R
Example with numbers Suppose db/ = - 1300 Gauss per sec and R= 8.5 cm Find E at r = 5.2 cm r db E =! r < 2 R (0.052m) E = 0.13T = 0.0034 V = 3. 4 m 2 mv m Find E at 12.5 cm 2 R db E =! r > 2 r R 2 (0.085m) V E = 0.13T = 0.0038 = 3. 8 m 2(0.125m) mv m
What is an inductor? An inductor is a piece of wire twisted into a coil. It is also called a solenoid. If the current is constant in time, the inductor behaves like a wire with resistance. The current has to vary with time to make it behave as an inductor. When the current varies the magnetic field or flux varies with time inducing an Emf in the coil in a direction that opposes the original change. Suppose I move the switch to position a, then current starts to increase through the coil. An Emf is induced to make current flow in the opposite direction. Now suppose I move the switch to position b
Start with Faraday s Law What is inductance? L Emf =! d" =! NAdB =!µ 0 nna di Emf =!µ 0 n 2 la di Emf =!L di " = NBA B = µ 0 ni for a solenoid N=nl nˆ A i l B N turns L = µ 0 n 2 la (H=Henry) 1 H=1 T.m 2 /A L l = µ 0n 2 A (Henry/m) µ 0 = 4! " 10 #7 T$ m/a
Numerical example how many turns do you need to make a L = 4.25 mh solenoid with l = 15 cm and radius r = 2.25 cm? L = µ 0 n 2 la n = L µ 0 la N = nl N = Ll (.00425)(.15) nl = = 7 µ A 4#! 10 " # (.0225) 0 2 = 565 Turns Area nˆ i l B N turns A Show demo: Inductive spark after turning off electromagnet
Numerical Example You have a 100 turn coil with radius 5 cm with a resistance of 10 Ω. At what rate must a perpendicular B field change to produce a current of 4 A in the coil? Emf = IR = (4A)(10Ω) = 40 Volts Emf =!N d" m db 40 = N " r =!NA db =!N#r2 db = 40 40V 100! 3.15!(.05m) = = 2 2 51T s N B(t) N = 100 turns R = 5 cm Coil resistance = 10 Ω N coils so Multiply by N emf = " N d! m
RL Circuits V R = Ri V L Close the switch to a. What happens? Write down the loop rule. Loop Rule: Sum of potentials =0! + V R + VL = 0! " ir " L di = 0 The potential can be defined across the inductor outside the region where the magnetic flux is changing. Solve this equation for the current i.
i =! R (1" e" Rt L ) V R = 0.63! Rt! L V R = " ( 1! e ) V R = Ri R! = L LR RRiR Ri 10 V 2000 Ω 4.0H V L = " e Rt! L V L =!(e "1 ) = 0.37! Note τ = L/R = 4/2000 = 0.002 s, i =! R (1" e"1 ) = 0.63! R! L di and V R =!(1" e "1 ) = 0.63!
How is the magnetic energy stored in a solenoid or coil in our circuit?! " ir " L di = 0 Start with Loop rule or Kirchhoff's Law I! = ir + L di!i = i 2 R + Li di Solve it for ε Multiply by i Rate at which energy is delivered to circuit from the battery Rate at which energy is lost in resistor Rate at which energy is stored in the magnetic field of the coil du B = Li di
What is the magnetic energy stored in a solenoid or coil du B = Li di du B = Lidi U! B du B =! Lidi U B =! Lidi = 1 2 Li2 0 i 0 i 0 U B = 1 2 Li2 For an inductor L Now define the energy per unit volume u B = U B Al u B = 1 2 Li2 Al = L l i 2 2A L l = µ 0 n 2 A l Area A u B = 1 2 Li2 Al = 1 2 µ 0 n 2 i 2 B = µ 0 ni The energy density formula is valid in general u B = B2 2µ 0 u E = E 2 2! 0
What is Mutual Inductance? M When two circuits are near one another and both have currents changing, they can induce emfs in each other. I 1 2 2 I 1!! m 1 = L1 I1 + M 21I 2 m 2 = L2I 2 + M12I1 M M = 12 = 21 M On circuit boards you have to be careful you do not put circuits near each other that have large mutual inductance. They have to be oriented carefully and even shielded.
Chapter 30 Problem 15 In Figure 30-44, a stiff wire bent into a semicircle of radius a is rotated at constant frequency f in a uniform magnetic field B. (Use a, f and B as necessary, with all quantities in SI units.) Fig. 30-44 (a) What is the frequency of the emf induced in the loop? (b) What is its amplitude?
Chapter 30 Problem 34 Figure 30-56 shows two circular regions R1 and R2 with radii r1 = 12.0 cm and r2 = 34.0 cm. In R1 there is a uniform magnetic field of magnitude B1 = 50.0 mt into the page and in R2 there is a uniform magnetic field B2 = 75.0 mt out of the page (ignore fringing). Both fields are decreasing at the rate of 7.80 mt/s. Calculate the integral line int E ds for each of the three dashed paths. (a) path 1 (b) path 2 (c) path 3 Fig. 30-56
Chapter 30 Problem 35 A long solenoid has a diameter of 12.0 cm. When a current i exists in its windings, a uniform magnetic field B = 27.0 mt is produced in its interior. By decreasing i, the field is caused to decrease at the rate of 7.90 mt/s. Calculate the magnitude of the induced electric field at the following distances from the axis of the solenoid. (a) 2.20 cm (b) 8.20 cm