Physics 012 1st EditionExam # 2 Study Guide Lectures: 11 - 20Lecture 11 (February 9)Magnetic Forces and FieldsProblem: If a particle of charge -1.6 x 10-19 C is moving at 8 x 106 m/s at 30 degrees relative to a 0.5 T magnetic force field, what is the net force on the particle?Answer: FB = qvBsinθ = (-1.6 x 10-19 C)( 8 x 106 m/s)sin(30) = 3.2 x 10-3 NLecture 12 (February 11) Velocity SelectorProblem: A particle of charge q and velocity v passes through a parallel plate capacitor containing an electric field E pointed downward. A magnetic field B is present within the capacitor perpendicular to v and E going away from the observer. Find an expression for v so theparticle passes through the plates of the capacitor without curving upward or downward.Answer: ΣFy = qvBsinθ – qE = 0qvB = qEvB = Ev = E/BLecture 13 (February 13)Ampere’s LawProblem: A wire carrying a current I contains a circular loop of radius a. Find an expression for the net magnetic field at the center of the loop.Answer: Bnet = Bwire + Bloop = [(μ0I)/2a] + [(μ0I)/2πa]Lecture 14 (February 18)Magnetic FluxProblem: Find an expression for Fapp so the conducting bar moves at a constant velocity.Answer: ΣFy = Fapp – FB ; Fapp = FBIind = |εind|/RΦ = ΣBΔAcosθ = BΣΔA = BLyεind = -N(ΔΦ/Δt) = -(ΔBLy)/Δt = -BL(Δy/Δt) = -BLvIind = (-BLv)/RFB = (BLv/R)LBsinθ = (B2L2v)/R = FappLecture 15 (February 20)Induced CurrentsProblem: A rectangular circuit of dimensions L by w enters a magnetic field B (directed away from the observer), L end first at velocity v. The angle between v and B is 90 degrees. Find an expression for the force F exerted on the circuit by the magnetic field.Answer: By entering the magnetic field, the area is increasing, thus causing and increase in flux, an induced εmf, and an induced current.ΦB = ΣBΔAsinθ = BLyεind = -N(ΔΦ/Δt) = -NBL(Δy/Δt) = -NBLvIind = εind /R = (NBLv)/RF = NILBsinθ = N[(NBLv)/R]LB = (N2B2L2v)/RLecture 16 (February 23)Lenz’s LawProblem: A metal bar of length L is making contact with a rectangular circuit with resistance R. The circuit is oriented vertically so the bar falls along the circuit (which has infinite length). The circuit exists within a magnetic field B that points away from the observer. Find an expression for the terminal speed v of the metal bar.Answer: ΣFy = ILB – mg = maILB = mg + ma ; a = (ILB)/m - gεind = -N(ΔΦ/Δt) = -BLvIind = εind /R = (BLv)/Ra = ([BLv/R]LB)/m – g = (B2L2v)/(Rm) – gThe bar will reach terminal velocity when a = 0.(B2L2v)/(Rm) – g = 0v = (gRm)/(B2L2)Lecture 17 (February 25)TransformersProblem: In a transformer, the primary current is 20 A and the secondary current is 40 A. Find the ratio of the potential difference of the primary circuit to the secondary circuit.Answer: Ip/Is = Vs/Vp20/40 = ½ = Vs/VpLecture 18 (February 27)Energy Density of Electromagnetic WavesProblem: What is the total energy density of an electromagnetic field whose electric field waveshave an amplitude of 19.0 N/C?Answer: utotal = (ξ0E2max)/2 = ([8.85 x 10-12 C2/(N*m2)][19 N/C]2)/2 = 1.60 x 10-9 N/m2Lecture 19 (March 9)Polarized Light IntensityProblem: An beam of light is sent through a vertical polarizer, then a filter whose transmission axis is tilted 30 degrees from vertical, then through a third filter with a horizontal transmission axis. If the initial intensity of the light was 100 W/m2, find the final intensity.Answer: I = ([I0/2]cos2θ)cos2(90-θ) = ([100/2]cos2[30])cos2(90-30) = ([50][0.75]) cos2(60) = (37.5)(0.25) = 9.375
View Full Document
Unlocking...