L12 - FARADAY’S LAW • Michael Faraday and Joseph Henry...

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Unformatted text preview: FARADAY’S LAW • Michael Faraday and Joseph Henry discovered it in 1831 • Changing magnetic flux produces an emf • Or Changing B-Field produces E-Field • The rate of change of magnetic flux is required Induced EMF produced by a changing Magnetic Flux! dΦ ε =− dt Φ= area ∫ r r B ⋅ dA Changing Flux due to moving permanent magnet dΦ ε = −N dt Nature of a changing flux r r Φ = B ⋅ dA area • Since flux is defined as a dot product – B can change – A can change – θ can change ∫ d ε = −N dt ∫ A r r B ⋅ dA Sample prob. 30-1: Calculate induced emf in coil C LENZ’S Law (“no such thing as a perpetual motion machine” The direction of the emf induced by changing flux will produce a current that generates a magnetic field opposing the flux change that produced it LENZ’S LAW The polarity of the emf induced by a changing flux will produce a current that generates a magnetic field opposing the flux change that produced it Sample prob. 30-2: Example what is the emf in loop and what is the net current ? Φ= area ∫ r r B ⋅ dA dΦ ε = −N dt Sample 30-3 Find emf when B is not uniform r 2 2 ˆ B = 4t x j Induced electric fields • The electric field due to an emf is NOT conservative – Net work must be done over a closed path (circuit) • Therefore, the closed path integral of E is non-zero – Charges will accelerate parallel to E. Induced electric fields • The electric field due to an emf is NOT conservative – Net work must be done over a closed path (circuit) • Therefore, the closed path integral of E is non-zero – Charges will accelerate parallel to E. dΦ ε =− dt r r ε = E ⋅ ds ∫ Induced electric fields We will assume here that B is increasing into the page r r W = F ⋅ d s = q0 E 2π r around loop Φ = area ∫ r r B • dA ∫ W ε= = 2π rE q0 More generally, r r W = F ⋅ d s = q0 ∫ ∫ s r r E ⋅ d s , or ε= ∫ s r r E ⋅ ds so, ∫ s r r dΦ E ⋅ ds = − dt Sample problem If R=8.5 cm and dB = 0.13 T/sec dt a) Find E when r=5.2cm b) Find E when r=12.5 cm E= E= r dB 2 dt R 2 dB 2 r dt LR circuit - initial current is Apply Kirchhoff”s Loop Rule ε L − IR = 0 dI − L − IR = 0 dt L τ= R I= ε R − t e τ Time dependence of an LR circuit At t = 0, i = 0, and switch is just closed ε − iR + ε L = 0 di R ε + i = dt L L Apply Kirchhoff’s Loop Rule t ⎛ − ε ⎜ i = ⎜1 − e τL R⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ L τL = R INDUCTANCE An Application of Faraday’s Law • Definition • Units • Self inductance – Examples • Mutual inductance – Example • Energy in Inductor • LR circuit – exponential growth and decay dΦ ε = −N dt Definition and Units L= NΦ i di ε = −L dt Bsolenoid = μ 0 n i – based on Faraday’s Law – assume A and θ are constant when determining flux – B is proportional to i ε = − NA μ o n = − n 2 lA μ = − L where, per L = n l 2 o di dt di dt • Unit is henry (H) equals volt-second/meter di dt the inductance unit Aμ o length Ideal Solenoid r v ∫ B ⋅ ds = μoienc r r r r ∫ B ⋅ ds + ∫ B ⋅ ds a r r r r + ∫ B ⋅ ds + ∫ B ⋅ ds = μ o ienc c d b Bh + 0 + 0 + 0 = μ oienc Bh = μ o N turns i where N = nh B = μ o ni Constant field in x-direction r B = μ o ni ˆ x Inductors and self inductance L and Back EMF-voltage L = μo n Al 2 Changing flux induces emf in same element that carries current A “back” emf is generated by a changing current emf opposes the change causing it (Lenz’s Law) emf thus opposes the changing current (i.e., “back” emf) Motional emf The current follows the prediction from Lenz’s Law The experiment that tells you there is no perpetual motion machine. What is the power consumed by the resistor in terms of B and v? B 2 L2v 2 P= R The wire has resistance so we can think of an effective circuit Self inductance of a flat coil Determine the flux through the center of the coil Φ = BA ⎛ μ o Ni ⎞ 2 Φ=⎜ ⎟ πR ⎝ 2R ⎠ Φ L=N i N ⎛ μ o Ni 2 ⎞ πR ⎟ L= ⎜ i ⎝ 2R ⎠ πμ o N R L= 2 2 ( ) Self inductance of a coaxial cable r v Φ = ∫ B ⋅ dA A The flux is due to the field of the central wire passing through the “blue” area b μ oi Φ= l ln 2π a μoi Φ= ldr 2πr ∫ a b μ oi l Φ= 2π ∫ a b Φ L= i dr r μ ol b ln L= 2π a Mutual inductance • A changing flux in one element induces an emf in another • Multiple inductors can exhibit combined self and mutual inductance N 2Φ 21 M 21 = i1 Example Problem 30-9 Energy in an inductor Start with the power delivered to the inductor di PL = ε L I = − L i dt d W = ∫ Pdt = − L ∫ 0 0 dt t i ⎛ i2 ⎞ ⎜ ⎟ dt ⎜ 2 ⎟ ⎝ ⎠ di ε − L − iR = 0 − Kirchoff dt di εi − L i − i 2 R = 0 − power ! dt 1 2 U = −W = Li 2 1 2 U = Li 2 Energy density in an inductor Determine the energy in a solenoid 1 U = Li 2 2 B = μ0 n i L = μ o n 2 Al 2 ⎛ B ⎞ 1 2 ⎟ U = μ on A l⎜ ⎜μ n⎟ 2 ⎝ o ⎠ U U u= = V Al u= 1 2μo B2 This is a general result ...
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