PHYS-2020: General Physics IICourse Lecture NotesSection VIDr. Donald G. LuttermoserEast Tennessee State UniversityEdition 3.3AbstractThese class notes are designed for use of the instructor and students of the course PHYS-2020:General Physics II taught by Dr. Donald Luttermoser at East Tennessee State University. Thesenotes make reference to the College Physics, 9th Edition (2012) textbook by Serway and Vuille.VI. Induced Voltages and InductanceA. Magnetic Flux.1. Michael Faraday demonstrated that electr ic currents can be pro-duced by a changing magnetic field =⇒ a chan ging B-field canproduce an induced emf .2. Consider a loop of wire in a uniform B-field. The magnetic fluxin the loop is the sum of the B-field strength across the entire(cross-sectional) surface area of the loop:ΦB=~B ·~A = B⊥A = BA cos θ . (VI-1)B>Front viewof loopUniformB-fieldLoop ofwireB>B>⊥B>||θSide viewof loopVI–1VI–2 PHYS-2020: General Physics I Ia) ΦBis the magnetic flux measur ed in T·m2= Wb (weber).b) The value of the magnetic flux is p roportional to the totalnumber of lines passing through the loop.B. Faraday’s Law of Induction.1. The emf induced in a circuit equals the rate of change of themagnetic flux through the circuit:E = −N∆ΦB∆t. (VI-2)a) E ≡emf, N ≡ number of loops in t he wire circuit, ∆ΦB/∆t ≡change in magnetic fl ux over time.b) Eq. (VI-2) is known as Faraday’s law of magnetic in-duction.c) The negative sign is included to indicate polarity of theinduced emf .d) Faraday’s law, as it is written in Eq. (VI-2), representsthe a verage induced emf. The instantaneous induced emf(using calculus), which describes the more precise versionof Faraday’s law, would be given byE = −N lim∆t→0∆ΦB∆t= −NdΦBdt.2. The polarity mentioned in Faraday’s law can be found fromLenz’s law:a) Any induced electromotive force will be in the directionsuch that the magnetic flux it creates will oppose thechange in the flux that produced it.b) More simp ly, the induced current tends to maintain t heoriginal flux throu gh the circuit.Donald G. Luttermoser, ETSU VI–3c) Lenz’s Law is one consequence of the principle of conser-vation of ener gy. We can see why through the followingexample:i) Move a permanent magnet towards the face of aclosed loop of wire (e.g ., a coil or solenoid).ii) An electric curr ent is induced in the wire, becausethe electrons within it are subjected to an increas-ing magnet ic field as the magnet approaches.iii) This produces an emf that acts upon them. Thedirection of the induced curr ent depen ds on whetherthe north or south pole of the magnet is approach-ing:• An approaching north pole will produce an anti-clockwise current (from the perspective of themagnet).• A south pole approaching th e coil will produce aclockwise current.Example VI–1. Problem 20.5 (Page 716) from the Ser-way & Vuille textbook: A long, straight wire lies in the plane of acircular coil with a radius of 0.010 m. T he wire carries a current of2.0 A and is pl aced along the diameter of the coil. (a) What is thenet flux through the coil? (b) If the wir e passes through the centerof the coil and is perpendicular to the plane of the center of the coil,find the net flux through the coil.VI–4 PHYS-2020: General Physics I II = 2.0 Ar = 0.010 m(a) Face-on ViewI = 2.0 Ar = 0.010 mcoil(b) Side ViewSolution (a):As shown in the diagram above, every field line that comes upthrough the area A on one side of the wire, goes back downthrough area A on the other side of the wire. Thus, the net fluxthrough the coil is zero.Solution (b):The magnetic field is p arallel to the plane of the coil as shownin the diagram above, so θ = 90.0◦. Therefore, ΦB= BA cos θ =BA cos 90.0◦= 0.Example VI–2. A 25-turn circular coil of wire has a diameterof 1.00 m. It i s placed with its axi s along the direction of Earth’smagnetic field of 50.0 µT, and then in 0.200 s it is flipped 180◦. Anaverage emf of w hat magnitude is generated in the coil?Solution:First, the change in flux results from the change of orientation ofthe loop with the magnetic field. Using Eq. (V I-1) we get:∆ΦB= BA cos θ2−BA cos θ1= BA[∆(cos θ)] ,where B = 50.0×10−6T, A = πD2/4 = π(1.00 m)2/4 = 0.785 m2,∆(cos θ) = cos 0◦−cos 180◦= 1−(−1) = 2. As such, the averageDonald G. Luttermoser, ETSU VI–5emf generated for N = 25 loops from this changing m agnetic fluxover a ∆t = 0.200 s time interval is given by E q . (VI-2):|E| = N∆ΦB∆t=NBA[∆(cos θ)]∆t=25(50.0 × 10−6T)(0.785 m2)(2)0.200 s= 9.82 × 10−3V = 9.82 mV .C. Motional emf.1. The potential differen ce is maintained across a conductor movingthrough a B-field. If the motion is rever sed, the polarity of thepotential difference is also reversed.XXXXXXXXXXXXXXBin>v>+-+-+-+-Positive chargeaccumulates hereNegative chargeaccumulates hereFe>FB+>FB->a) Note that the magnetic f orce, FB= qvB, points d own-ward for electron s an d upward for positive ions in thediagram above.b) The charge separates which set s up an electric field withforce, Fe= qE.VI–6 PHYS-2020: General Physics I Ic) The charges at either end build up until downward FBbalances upward Fefor the electrons (and op posite thisfor t he positive ions), thenFB= Fe=⇒ qvB = qE or E = vB .d) Since~E is constant, we can write ∆ V = E = E` (where∆V ≡ potential difference = emf), or∆V = E = E` = B`v . (VI-3)2. Let’s assume this bar moves a distance ∆ x in time ∆t, then across-sectional area A = ∆x ·` is cr eated as the bar moves in t heB-field (see diagram below):XXXXXXXXBin>v>to= 0 t∆xa) As the bar moves, the magnetic fl ux increases:∆ΦB= BA = B` ∆x .Donald G. Luttermoser, ETSU VI–7b) If we were to hook this bar into a circuit, we can derivean induced emf due to the moving bar.XXXXXXXXBin>v>R RIThese 2 circuitsare equivalentÖc) The magnitude of t he induced emf is then|E| =−N∆ΦB∆t=∆ΦB∆t(N = 1, one loop)or|E| = B`∆x∆t= B`v . (VI-4)d) Induced emf is sometimes called motional emf.e) If the resistance of the circuit is R, thenI =|E|R=B`vR. (VI-5)Example VI–3. Problem 20.30 (Page 718) from the Ser-way & Vuille textbook: Consider the arrangement shown in FigureP20.30 (and in the diagram below). Assume that R = 6.00 Ω and` = 1.20 m, and that a uniform 2.50 T magnetic field is directedinto the page. At what speed should the bar be moved to producea current of 0.500 A in the resistor?VI–8 PHYS-2020: