PHYS 202 1nd EditionExam # 3 Study Guide Lectures: 12-19Lecture 12 (February 26)Answer true or false for the following statements regarding Earth’s magnetic field “B”;A) The magnitude of earth's B is about 1E-4 Tesla. B) B does work on a charge only when the charge has a velocity component parallel to B. C) For a bar magnet, the B - lines exit through the N pole and return through the S pole . D) The vector B acts only on charges at rest. E) B does work on a charge only when the charge has a velocity component perpendicular to B. F) The earth's B lines are parallel to the earth's surface at all points. G) Unlike magnetic poles attract.Answers:A) TrueB) FalseC) TrueD) FalseE) FalseF) FalseG) TrueLecture 13 (March 3)The four wires that lie at the corners of a square of side a= 5.50 cm are carrying equal currents i= 1.00 A into (+) or out of (-) the page, as shown in the picture:- X +a Y - +A) calculate the y component of the magnetic field at the center of the square. B) Calculate the x-component of the force on a 1.0-cm long piece of the lower right-hand wire, due to the other three wires.Answer:A) 1.45x10-5TB) 5.45x10-8NExplanation:A)Because the magnetic field due to all four wires is the same, By=4(B)*Cos(45)Because the current carrying wires are in the corners of the square, there is a 45 angle between any B vector and the x or y axis of the square.B=µi/[2πr]r=a/(√2) = .055/√2=.03889B=(1.26x10-6)(1)/(2π*3.89)=5.155x10-6By=4(5.155x10-6)*cos(45)=1.45x10-5B)The force of wire 2 on wire 1 isF2<-1=µi2L/2πaThe force of wire 2 on wire 4F4<-2=µi2L/2√(2πa)Therefore, the x component would be;Fx=F2<-1+F4<-2*Cos(45)Lecture 14 (March 17)A proton with a speed of 2.20×106 m/s is shot into a region between two plates that are separated by a distance of 0.195 m. A magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. What must be the magnitude of the magnetic field (in T), so the proton just misses colliding with the opposite plate?Answer: .118TExplanation:F=q(vxB)=ma =m(v2/d)The minimum field required to avoid hitting the plate is;qvB=m(v2/d) => B=mv/qdB=(1.67x10-27)(2.20x106)/(1.6x10-19)(.195)Lecture 15 (March 19)Two parallel wires spaced a distance d apart experience an attractive force per unit length of 8.60×10-4 N/m. If the currents have a magnitude of 8.0 A each, what is the distance d?Answer: .0149mExplanation:F/l=(µ0i1i2)/(2πd).00086=[(4πx10-7)(8)(8)]/2πdd=[(4πx10-7)(8)(8)]/2π(.00086)Lecture 16(March 24)The rectangular loop shown below, carrying the current I, is located in a uniform magnetic field of 0.35 T pointing in the positive x direction. A) What is the magnitude of the magnetic dipole moment of the loop? B) Calculate the torque about the z axis on the loop. Use the following data: H=4cm, W=7cm, I=6A,φ=37⁰ Z Y φ XAnswer:A) .0168A*m2B) .00354N*mExplanation:A)Magnetic moment(µ)=NIA=(1)(6)(.04*.07) =.0168B)T=µB=µBSin(θ) where θ is the angle by which the loop turns:How we find the value of θ depends on the axis that B runs along:For B along a –x or +x, θ=φFor B along a +Y, θ=90-φFor B along a –Y, θ=90+φT=µBSin(φ)=(.0168)(.35)Sin(37)=.003538Lecture 17 (March 26)A very long solenoid of circular cross section with radius a= 3.30 cm has n= 86.0 turns/cm of wire. An electron is sitting outside the solenoid, at a distance r= 3.80 cm from the solenoid axis. What is the magnitude of the force on the electron while the current in the solenoid is ramped up at a rate of 28.0 Amps/sAnswer:6.94x10-22NExplanation:B=µ0ni (only inside the solenoid)Outside, B=0E=ΔφB/Δt = Δi/Δt(Bπa2)E=πa2 (ΔB/Δt)B/t=µn(i/t)E=[πa2 (µn(i/t))]/2πrE=[π(.0332)(µ(8600)(28)]/(2π(.038))E=.0043F=EqF=(.0043)(1.6x10-19)Lecture 18 (March 31)In the diagram is shown an RL circuit with a switch. ε = 105.0 V, R1 = 45.0 Ω, R2 = 90.0 Ω and L = 40.0 H.A)What is i1 just after the switch is closed? B) What is i2 just after the switch is closed? C) What is the value of the current in the switch just after the switch is closed? D) What is the the potential difference across R2 just after the switch is closed? E) What is the the potential difference across L just after the switch is closed? F) What is the rate of change of the current di2/dt in the time just after the closing of the switch? G) What is i1 a long time after the switch is closed? H) What is i2 a long time after the switch is closed? I) What is the value of the current in the switch a long time after the switch is closed? J) What is the the potential difference across R2 a long time after the switch is closed? K) What is the the potential difference across L a long time after the switch is closed?L) What is the rate of change of the current di2/dt a long time after the closing of the switch?Answers:A) 2.33AB) 0AC) 2.33AD) 0VE) 105VF) 2.62A/sG) 2.33AH) 1.17AI) 3.5AJ) 105VK) 0VL) 0A/sExplanation:For an RL circuit, immediately after the switch is closed(t=0), the current through the inductor will be equal to 0 and the potential difference (V) through the inductor will be equal to –E of the circuit. After the switch has been closed for a long time (t=∞), the current through the inductor will equal to E/R and the potential difference through the inductor will be equal to zero. Lecture 19 (April 2)We have a large coil with a small coil concentric to it. The large coil has a radius of 1m. and the small coilhas an area(πr2) of 7cm. There is a current of 200A running counterclockwise through the large coil, which is brought down to 0A and then brought down again to -200A(meaning it is now running clockwise). All of this happens in a time(Δt) of 1 second. Find the magnetic field(B) at the center when A) t=0. B) t=.5 and C) t=1. D) What is the EMF of the small coil?Answers:A) 1.257x10-4TB) 0TC) -1.257x10-4TD) 1.75x10-5VExplanation:A) B=µi/2rL = µ(200)/2(1)B) B=µi/2rL = µ(0)/2(1)C) B=µi/2rL = µ(-200)/2(1)D) E=φ/tΦ=aΔB/tΦ=a(Bt=0 -