Physics 2113 Jonathan Dowling Lecture 30: WED 04 NOV Induction and Inductance II Fender Stratocaster Solenoid Pickup
F a r a d a y ' s E x p e r i m e n t s I n a s e r i e s o f e x p e r i m e n t s, M i c h a e l F a r a d a y i n E n g l a n d a n d J o s e p h H e n r y i n t h e U. S. w e r e a b l e t o g e n e r a t e e l e c t r i c c u r r e n t s w i t h o u t t h e u s e o f b a t t e r i e s. T h e c i r c u i t s h o w n i n t h e f i g u r e c o n s i s t s o f a w i r e l o o p c o n n e c t e d t o a s e n s i t i v e a m m e t e r ( k n o w n a s a " g a l v a n o m e t e r " ). I f w e a p p r o a c h t h e l o o p w i t h a p e r m a n e n t m a g n e t w e s e e a c u r r e n t b e i n g r e g i s t e r e d b y t h e g a l v a n o m e t e r. 1. A c u r r e n t a p p e a r s o n l y i f t h e r e i s r e l a t i v e m o t i o n b e t w e e n t h e m a g n e t a n d t h e l o o p. 2. F a s t e r m o t i o n r e s u l t s i n a l a r g e r c u r r e n t. 3. I f w e r e v e r s e t h e d i r e c t i o n o f m o t i o n o r t h e p o l a r i t y o f t h e m a g n e t, t h e c u r r e n t r e v e r s e s s i g n a n d f l o w s i n t h e o p p o s i t e d i r e c t i o n. T h e c u r r e n t g e n e r a t e d i s k n o w n a s " i n d u c e d c u r r e n t "; t h e e m f t h a t a p p e a r s i s k n o w n a s " i n d u c e d e m f "; t h e w h o l e e f f e c t i s c a l l e d " i n d u c t i o n."
loop 2 loop 1 In the figure we show a second type of experiment in which current is induced in loop 2 when the switch S in loop 1 is either closed or opened. When the current in loop 1 is constant no induced current is observed in loop 2. The conclusion is that the magnetic field in an induction experiment can be generated either by a permanent magnet or by an electric current in a coil. Faraday summarized the results of his experiments in what is known as " Faraday's law of induction. " An emf is induced in a loop when the number of magnetic field lines that pass through the loop is changing.
Lenz s Law The Loop Current Produces a B Field that Opposes the CHANGE in the bar magnet field. Upper Drawing: B Field from Magnet is INCREASING so Loop Current is Clockwise and Produces an Opposing B Field that Tries to CANCEL the INCREASING Magnet Field Lower Drawing: B Field from Magnet is DECREASING so Loop Current is Counterclockwise and Tries to BOOST the Decreasing Magnet Field.
30.5: Induction and Energy Transfers: If the loop is pulled at a constant velocity v, one must apply a constant force F to the loop since an equal and opposite magnetic force acts on the loop to oppose it. The power is P=Fv. As the loop is pulled, the portion of its area within the magnetic field, and therefore the magnetic flux, decrease. According to Faraday s law, a current is produced in the loop. The magnitude of the flux through the loop is Φ B =BA =BLx. Therefore, The induced current is therefore The net deflecting force is: The power is therefore
d = c > b = a
30.5: Induction and Energy Transfers: Eddy Currents
Example: Eddy Currents A non-magnetic (e.g. copper, aluminum) ring is placed near a solenoid. What happens if: There is a steady current in the solenoid? The current in the solenoid is suddenly changed? The ring has a cut in it? The ring is extremely cold?
Another Experimental Observation Drop a non-magnetic pendulum (copper or aluminum) through an inhomogeneous magnetic field What do you observe? Why? (Think about energy conservation!) Pendulum had kinetic energy What happened to it? Isn t energy conserved?? N S Energy is Dissipated by Resistance: P=i 2 R. This acts like friction!!
Start Your Engines: The Ignition Coil The gap between the spark plug in a combustion engine needs an electric field of ~10 7 V/m in order to ignite the air-fuel mixture. For a typical spark plug gap, one needs to generate a potential difference > 10 4 V! But, the typical EMF of a car battery is 12 V. So, how does a spark plug even work at all!? 12V spark Breaking the circuit changes the current through primary coil Result: LARGE change in flux thru secondary -- large induced EMF! The ignition coil is a double layer solenoid transformer: Primary: small number of turns -- 12 V Secondary: MANY turns -- spark plug http://www.familycar.com/classroom/ignition.htm
Start Your Engines: The Ignition Coil Transformer: P=iV
30.6: Induced Electric Field:
30.6: Induced Electric Fields, Reformulation of Faraday s Law: Consider a particle of charge q 0 moving around the circular path. The work W done on it in one revolution by the induced electric field is W =Eq 0, where E is the induced emf. From another point of view, the work is Here where q 0 E is the magnitude of the force acting on the test charge and 2 r is the distance over which that force acts. π In general,
b = out c = out e = into d = into First loop 3 Next Loop 4 Last Loop 2
Changing B-Field Produces E- We saw that a time varying magnetic FLUX creates an induced EMF in a wire, exhibited as a current. Field! Recall that a current flows in a conductor because of electric field. Hence, a time varying magnetic flux must induce an ELECTRIC FIELD! A Changing B-Field Produces an E-Field in Empty Space! C! E d! A da! ds B = E i dφ To decide direction of E-field use Lenz s law as in current loop. B
Example The figure shows two circular regions R 1 & R 2 with radii r 1 = 1m & r 2 = 2m. In R 1, the magnetic field B 1 points out of the page. In R 2, the magnetic field B 2 points into the page. Both fields are uniform and are DECREASING at the SAME steady rate = 1 T/s. Calculate the Faraday integral for the two paths shown. Path II: C Path I:!! E ds = C!! E ds = dφ B! E ds! C 2 R 1 I II R 2 = ( πr1 )( 1T / s) = + 3.14V [ ] 2 2 ( πr )( 1T / s) + ( πr )( 1T / s) = 9.42V B 1 1 2 + B 2
Solenoid Example A long solenoid has a circular cross-section of radius R. The magnetic field B through the solenoid is increasing at a steady rate db/. Compute the variation of the electric field as a function of the distance r from the axis of the solenoid. B First, let s look at r < R: Next, let s look at r > R: 2 E (2π r) = ( πr ) db 2 E (2π r) = ( πr ) db magnetic field lines r E = 2 db 2 R E = 2 r db electric field lines
E r < R = r 2 r E(r) Solenoid Example Cont. db E r < R = µ 0 nr 2 r = R di E r > R = R2 2r 1 / r r db E r > R = µ 0nR 2 i Added Complication: Changing B Field Is Produced by Changing Current i in the Loops of Solenoid! 2r di B i B = µ 0 ni E db = µ n di 0
Summary Two versions of Faradays law: A time varying magnetic flux produces an EMF: C! E E = dφ B A time varying magnetic flux produces an electric field:! ds = dφ B
30.7: Inductors and Inductance: An inductor (symbol ) can be used to produce a desired magnetic field. If we establish a current i in the windings (turns) of the solenoid which can be treated as our inductor, the current produces a magnetic flux F B through the central region of the inductor. The inductance of the inductor is then The SI unit of inductance is the tesla square meter per ampere (T m 2 /A). We call this the henry (H), after American physicist Joseph
Inductors: Solenoids Inductors are with respect to the magnetic field what capacitors are with respect to the electric field. They pack a lot of field in a small region. Also, the higher the current, the higher the magnetic field they produce. Capacitance C how much potential for a given charge: Q=CV Inductance L how much magnetic flux for a given current: Φ=Li Using Faraday s law: E = dφ = L di Units : [ L] = Tesla m Ampere 2 H (Henry) Joseph Henry (1799-1878)
Inductance Consider a solenoid of length l that has N loops of N B! area A each, and n = windings per unit length. A current l i flows through the solenoid and generates a uniform l 2 L = µ 0n la magnetic field B = µ ni 0 inside the solenoid. The solenoid magnetic flux is Φ = NBA. 2 ( µ 0 l ) The total number of turns N = nl Φ = n A i. The result we got for the special case of the solenoid is true for any inductor: Φ = Li. Here L is a constant known as the inductance of the solenoid. The inductance depends on the geometry of the particular inductor. B B B Inductance of the Solenoid 2 Φ B µ 0n lai 2 For the solenoid, L = = = µ 0n la. i i
E di = L Self - Induction In the picture to the right we already have seen how a change in the current of loop 1 results in a change in the flux through loop 2, and thus creates an induced emf in loop 2. loop 2 loop 1 If we change the current through an inductor this causes a change in the magnetic flux Φ B = Li through the inductor dφ B di according to the equation = L. Using Faraday's law we can determine the resulting emf known as self - induced SI unit for L : dφ B di emf: E = = L. the henry (symbol: H) An inductor has inductance L = 1 H if a current change of 1 A/s results in a self-induced emf of 1 V. (30 17)
E = dφ = L di If i is constant (in time) then E = 0 a? b? c? e? d? f?
E = L di Example The current in a L=10H inductor is decreasing at a steady rate of i=5a/s. If the current is as shown at some instant in time, what is the magnitude and direction of the induced EMF? i (a) 50 V (b) 50 V Magnitude = (10 H)(5 A/s) = 50 V Current is decreasing Induced EMF must be in a direction that OPPOSES this change. So, induced EMF must be in same direction as current