# Applications of Magnetism

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Applications of Magnetism

PHY122 – Physics for the Life Sciences II Lecture 15 ERLC AC-circuits Review Chapters 24 - 26 Note: Clicker Channel 41 03/31/2015 Lecture 15 1 Chapter 26 – AC Electricity • AC Voltage Sources • AC Resistor circuits – AC currents in a resistor – AC power and RMS voltage and current • AC Capacitor circuits – AC voltage and current in a capacitor – Reactance of a capacitor • Inductors – AC voltage and current in an inductor – Self Inductance and Reactance • Transformers – Electrical Power in your home and Electrical safety • AC circuits with R, C, and L – ERLC circuits 03/31/2015 Lecture 15 2 Specific AC Circuits with EAC, R, C, and L • We can now construct circuits with all passive elements we learned about, • and calculate (using Kirchhoff’s Rules!) the currents and voltages … i(t) • We will only consider the simplest circuits, consisting of a SERIES connection of EAC, R, C, and L : EAC L R C • For this, we need to be able to add voltages TAKING CARE OF THE PHASES of the voltages vR(t), vC(t) and vL(t), with respect to the common current i(t) in the circuit … Phasors! 03/31/2015 Lecture 15 3 Phasors… Thus, the current i(t) (common to all elements in series) and voltage v(t) have two aspects: – an AMPLITUDE/MAGNITUDE I, V – a time-dependent PHASE angle θ=ωt I θ=ωt Phasor i(t)=I cosωt these two properties can be illustrated in a simple way by representing the current and voltage as PHASORS: – time-dependent vectors, rotating with constant angular velocity ω • Note: i(t), v(t) are NOT vectors; phasors are just a graphic and useful tool to work with time-dependent quantities of this sort! in L: in R: in C: V in sync Phasors I ωt v(t)= RIcosωt 03/31/2015 i(t)= Icosωt V phasor 90° ahead of I phasor I v(t)= –ωLI sinωt Lecture 15 ωt i(t)= Icosωt V phasor 90° behind I phasor v(t)= (1/ωC)I sinωt I V ωt i(t)= Icosωt 4 AC Circuits i(t) L EAC Series LRCEAC circuit, V vs. I: R C – Phasor diagram: VL V φ VR v(t) The total impedance Z: I ωt i(t) VL VC V V VC VR Z L I I XL I XC I RI 2 I VC 2 VR2 I 2 XL XC 2 R2 – the total voltage v(t) across L, C, and R (i.e. proper sum of VL, VC, and VR, equal to VE, can be found by phasor addition, analog to vector addition: 2 v(t ) L 1 C I sin t RI cos t L 1 C R 2 I cos t XL XC Z I cos t ; arctan L 1 C R If L is absent: XL=0; if there is no capacitor: XC=0 … 03/31/2015 Lecture 15 5 Resonance Let’s look at the current in a series ELRC circuit for a given AC voltage (like the 120 V household power outlet): v(t ) Z I cos t L 1 C 2 R 2 I cos t ; arctan L 1 C R V0 I 1 V0 Z 1/R1 1 L 1 C R 2 The I/V0 ratio is a function of ω, and in fact, has a maximum 1 1/R for XL=XC: X L L X C C 2 I/V0 0 0 1 LC 1/R2, R2>R1 0 ω0 ω the NATURAL frequency ! The height of the resonance peak depends inversely on R … At resonance: ω=ω0, XL=XC, and φ=0: voltage and current are in sync at the resonance frequency! Current: i(t)=v(t)/R; I=V/R, and Irms=Vrms/R ! 03/31/2015 Lecture 15 6 The natural frequency ω0 of a series ELRC series circuit with E=9.0 Vrms, L=2.0 H, and R= 4.0 Ω, and C= 8.0 μF equals … Responses 250 0.25 375 0.002 4 Other 0 1 LC 82% 10% 6% 03/31/2015 Lecture 15 1% 4 00 2 1% 0. 37 5 0. 25 25 0 1% Ot he r Rank 1 2 3 4 5 6 7 Series LCR Circuit Quantative discussion: a – Apply Kirchhoff #2 Note that i= –dq/dt: if q(t) decreases, i(t)>0, as sketched 2 di q d q dq VC L Ri 0 L 2 R 0 dt C dt dt 2 d q dq 1 damping term L 2 R q0 dt dt C t – Trial solution: q (t ) q0 e d cos t – Then: E oscillating term S b C i(t) L R ONLY correct for “underdamped” case dq q q0 e t d sint dt q 2 t d d 2q q t d t d 2 q e cos t q0 e t d sin t 2 q q e sin t q e sin t 0 0 0 2 2 2 d d d d d dt q (t ) – Note the appearance of sinωt and cosωt terms; must cancel independently 2 eq’ns: sin t : L 03/31/2015 2 d R 0 2L d R d 2L R R2 1 R2 R 1 R2 R2 1 1 2 2 2 cos t : 2 02 2 02 d2 L 2 R 0 L LC 4 L 2L C 4L 2L d C 4L Lecture 15 8 Discussion: LCR Circuits Solution: This is an exact analog to damped Simple Harmonic Motion! Damping: ωd=b/(2m), natural frequency: ω0=√(k/m) q(t ) q0 e t d cos t q0 e d t cos t R 1 2 2 , 0 with 0 d , 0 d 2L LC extreme case: R=0 ωd=0 NO damping , and ω2=ω02=1/LC … – the quantity ω0 is the “natural frequency” of the system – charge, current and (capacitor) voltage vs t: qR 0 (t ) q0 cos 0t q0 cos i (t ) 1 t LC dq 0 q0 sin 0t i0 sin 0t dt vC (t ) Numerical example: R=1.0 Ω, L=20μH, C=500pF: q q 0 cos 0t C C 2 L 40 μ s – The damping is small: d 40 μs d 25 103 s 1 R 1.0 1 1 1 0 0 1.0 107 s 1 LC 20 106 500 1012 1.0 1014 s 2 – The circuit, once started, oscillates with frequency: 1 1 d2 25 103 s 1 5 1013 s 1 7.1106 s 1 LC 40 μ s 500 pF f 1.1MHz 2 03/31/2015 Lecture 15 9 A capacitor C, initially charged to q0, is connected in series an inductor L. What would happen to the period of oscillations T and the maximum current Imax if the same charged capacitor were connected to a larger inductance? A. B. C. D. E. F. G. H. i (t ) 0 q0 sin 0t i0 sin 0t T and Imax both increase T and Imax both decrease T increases and Imax decreases T decreases and Imax increases T remains the same and Imax increases T remains the same and Imax decreases T increases and Imax remains the same T decreases and Imax remains the same L 0 2 T T ; i0 0 q0 03/31/2015 44% Lecture 15 16% 10% 4% 12% 6% 6% 2% T a nd Im T ax a b nd Im oth T in in ax cr cr b ea ea ot s se T es h d d a ec nd ecr re Im ea as se ax T es re a d nd ec m ai ... Im ns T a x th re in e m c. sa ai .. m ns T e th in a nd e cr sa ea ... m T se e d s a ec nd and re ... Im as es ax a re nd ... Im ax re ... 0 1 LC 10 Example Consider an inductor L=0.400 H and a capacitor of C=2.0 μC: – Calculate XL and XC for f=60 Hz and f=600 Hz: X L (60 Hz) L 2 fL 120 0.400 151 X L (600 Hz) 1200 0.400 1.51k 1 1 1 X C (60 Hz) 1.33k 6 C 2 fL 120 2.010 1 X C (600 Hz) 133 6 1200 2.010 – Calculate the frequency for which XL=XC : X L XC L f XC=1/(ωC) 1 1.12 krad/s LC 1 C 1 2 X XL=ωL 1 178 Hz LC 0 03/31/2015 Lecture 15 1/√(LC) ω 11 Problem A C=21 nF capacitor is connected across an AC generator that produces a peak voltage of Vpeak=5.3 V. – At what frequency is the peak current Ipeak=42 mA? – What is the instantaneous value of the emf at the instant when iC=IC ? 03/31/2015 Lecture 15 12 Problem In a nuclear magnetic resonance spectrometer, a L=15 mH coil is designed to detect an emf oscillating at f=400 MHz. The coil is part of an RLC circuit. – What value for the capacitance C should be used so that the coil circuit resonates at 400 MHz? 03/31/2015 Lecture 15 13 Note of Caution… Note: any REAL circuit element (R, C, L) will be a mixture of the ideal elements; – a resistor (in fact any wire) has capacitance (to ground and other parts of the circuit) and inductance; coaxial cable – a solenoid has resistance and capacitance (e.g. between windings); a capacitor has inductance; etc. Example: the self inductance of a straight wire (radius a) with a return (taken as a hollow cylinder of inner radius b) with uniform current i: 0i 0 r 2 0i – The magnetic field inB(r a) i 2 r , B (a rb) 2 /outside the wire: 2 r 2 r a 2 a 0i a 0i b ldr 1 0 b – magnetic flux: B BdA r ldr 0li li ln 2 2 a 0 2 a r 2 2 a – self-inductance: 10/22/2013 L B 1 1 b 0 1 ln Ο( H) l 2 a li Lecture 15 14 Example A series ELRC circuit: An AC generator voltage, VAC = 30 V and ω=250 rad/s, is applied to the following elements in series: R=200 Ω, L=0.400 H, and C=6.00 μF. – Calculate the current amplitude I: X L L 250 0.400 100 ; X C 1 C 1 250 6106 667 Z X L XC 2 R 2 567 2 2002 601 I V Z 30.0 601 49.9 mA – Calculate the phase angle φ between voltage and current: arctan X L XC 567 arctan 70.6 200 R i.e. the voltage LAGS the current by 71° – Calculate the voltage amplitudes across L, C, and R: VR RI 200 49.9 mA 9.98 V EAC VL X L I 100 49.9 mA 4.99 V (leads the current) VC X C I 667 49.9 mA 33.3V (follows the current) – Note that the Phasor sum of voltages gives: 2 VAC VL VC VR 2 28.02 9.982 30.0 V 10/22/2013 Lecture 15 i(t) L R C 15 Mechanical Analog! system of mass m on a spring k with v-dependent damping, see PHY121: In the special case of small “damping” proportional to velocity v, Fd ≈ –bv (a good approximation for drag at small velocity), we can again solve the motion: dx d 2x m d 2 x b dx FNet kx bv kx b ma m 2 x 0 2 dt dt k dt k dt q LC RC 1 L m, R b, C k – this particular differential equation can be solved, with solution: 2 b t 2m 2 k b x(t ) Ae cos( ' t ) , with ' m 2m – this is similar to SHM, now with a “damping factor” exp(–b/2m) multiplying the amplitude A, and a slightly different frequency ω’ – Amplitude and phase factor again depend on initial conditions… – “Critical Damping” occurs when ω’= 0, i.e. for b = 2√(km); i.e. for LCR circuit: R = 2√(L/C) 10/22/2013 Lecture 15 16 Example: Filter Circuit Filter circuits are used to pass signals of different frequency with different efficiency; e.g. a low-pass filter passes low frequencies, blocking high frequencies … i(t) VS Vout 0 V S Thus: lim 1 C L 1 C 1 0 1 0 2 2 R2 Vout 1 LC 1 R C 2 2 2 2 1 2 1 Vout/VS 2 Vout 1 1 1 0 and: 2lim 2 4 2 2 2 2 2 4 2 2 1 LC V LC 2 L C R C S LC 10/22/2013 L C – A simple low-pass filter can be constructed by R (from the leads and windings), L, and C : – Analyze: IX Vout VC C VS IZ VS R Lecture 15 0 ω0 =1/√(LC) ω 17 Problems 10/22/2013 Lecture 15 18 Superconductivity and Critical Field The element niobium, which is a metal, is a superconductor (i.e. has zero resistance) at temperatures below 9 K. However, the superconductivity is destroyed if the magnetic field at the surface of the metal reaches or exceeds B=0.10 T. – What is the maximum current in a straight, 4.0 mm diameter superconducting niobium wire? 10/27/2011 Review Midterm 2 19 Superconductivity and Critical Field The element niobium, which is a metal, is a superconductor (i.e. has zero resistance) at temperatures below 9 K. However, the superconductivity is destroyed if the magnetic field at the surface of the metal reaches or exceeds B=0.10 T. – What is the maximum current in a straight, 4.0 mm diameter superconducting niobium wire? 0 I B 2 r 10/27/2011 I crit 2 RB Review Midterm 2 0 1000 A 20 A Proton in a Uniform Magnetic Field At t=0 a proton is moving with a speed v=5.0×105 m/s at an angle of θ=30° from the x-axis. A uniform magnetic field B=1.5 T is pointing in the positive y-direction. – What will be the y-coordinate of the proton at t=1.0 μs? 10/27/2011 Review Midterm 2 21 A Proton in a Uniform Magnetic Field At t=0 a proton is moving with a speed v=5.0×105 m/s at an angle of θ=30° from the x-axis. A uniform magnetic field B=1.5 T is pointing in the positive y-direction. – What will be the y-coordinate of the proton at t=1.0 μs? The proton will spiral around the B-field direction, and describe a helical path! The velocity component in the y -direction remains constant! v y v sin 2.5 105 m/s Thus y (t1.0 μs) v y t 25 cm 10/27/2011 Review Midterm 2 22 A Particle in E and B Fields An antiproton (which has the same mass as a proton but a charge –e) is moving in the combined electric and magnetic fields of the figure. – Calculate the acceleration of the antiproton at the instant shown … 10/27/2011 Review Midterm 2 23 A Particle in E and B Fields An antiproton (which has the same mass as a proton but a charge –e) is moving in the combined electric and magnetic fields of the figure. – Calculate the acceleration of the antiproton at the instant shown … Two forces: 1. Electric: FE qE eE 1000e N/C(up!) 2. Magnetic: FB qvB evB 1250e N/C(down!) Net Force: Fnet 250e N/C(down!) F 250e N/C(down!) 10 2 acceleration: a net 2.40 10 m/s mp 1.67 1027 kg 10/27/2011 Review Midterm 2 24 Torque on a Current Loop A current loop in a motor has an area A=0.80 cm2 and N=100 windings. It carries an I=240 mA current in a uniform field B=0.60 T. – What is the magnitude of the maximum torque τ on the current loop? 10/27/2011 Review Midterm 2 25 Torque on a Current Loop A current loop in a motor has an area A=0.80 cm2 and N=100 windings. It carries an I=240 mA current in a uniform field B=0.60 T. – What is the magnitude of the maximum torque τ on the current loop? NAI B cos A ,B 10/27/2011 max NAIB Review Midterm 2 26 Induced emf A potential difference ΔV=4.0×10−2 V is developed across an L=10 cm long wire as it moves through a magnetic field B at v=5.0 m/s. – The magnetic field B is perpendicular to the axis of the wire and to the direction of motion … – Calculate the magnitude and direction of B 10/27/2011 Review Midterm 2 27 Induced emf A potential difference ΔV=4.0×10−2 V is developed across an L=10 cm long wire as it moves through a magnetic field B at v=5.0 m/s. – The magnetic field B is perpendicular to the axis of the wire and to the direction of motion … – Calculate the magnitude and direction of B Force on a charge carrier in the rod: FL qvB; F The emf that results from this: E EL L L vLB q B E vL The direction of B must be OUT OF THE PAGE for this E dB dA B Also note that E vLB dt dt 10/27/2011 Review Midterm 2 28 A m=50 g horizontal metal bar, L=10 cm long, is free to slide up and down between two very tall, parallel, vertical metal rods 10 cm apart. A uniform B=0.60 T magnetic field is directed perpendicular to the plane of the rods. The bar is raised to near the top of the rods, and a R=0.50 Ω resistor is connected across the two rods at the top. Then the bar is dropped. – Calculate the terminal velocity v of the metal bar. 10/27/2011 Review Midterm 2 29 A m=50 g horizontal metal bar, L=10 cm long, is free to slide up and down between two very tall, parallel, vertical metal rods 10 cm apart. A uniform B=0.60 T magnetic field is directed perpendicular to the plane of the rods. The bar is raised to near the top of the rods, and a R=0.50 Ω resistor is connected across the two rods at the top. Then the bar is dropped. – Calculate the terminal velocity v of the metal bar. Terminal velocity: a 0 W FL mg ILB E d dt vLB I I ind R R R vLB mgR LB v 2 2 68 m/s mg R LB 10/27/2011 Review Midterm 2 30 Induction by Changing B Field The loop in the figure has an induced current as shown. The loop has a resistance of R=0.16 Ω. Assume that the direction of B remains the same. – Calculate the rate of change ΔB/Δt, with its sign … 10/27/2011 Review Midterm 2 31 Induction by Changing B Field The loop in the figure has an induced current as shown. The loop has a resistance of R=0.16 Ω. Assume that the direction of B remains the same. – Calculate the rate of change ΔB/Δt, with its sign … E I ind R B A t t B I ind R 3.75 T/s A t – the induced current itself produces an induced magnetic field that points out of the page … – this induced field opposes the change in the original field B, thus B must be INCREASING … 10/27/2011 Review Midterm 2 32 A Series LRC Circuit In a series LRC circuit, the resistance is R=500 Ω, the inductance is L=0.500 H, and the capacitance is C=20.0 nF. – What is the resonance angular frequency ω0 of the circuit? – The capacitor can withstand a peak voltage of VC,max=500 V. If the voltage source operates at the resonance frequency, what maximum voltage amplitude can the source have if the maximum capacitor voltage is not to be exceeded? 10/27/2011 Review Midterm 2 33 A Series LRC Circuit In a series LRC circuit, the resistance is R=500 Ω, the inductance is L=0.500 H, and the capacitance is C=20.0 nF. – What is the resonance angular frequency ω0 of the circuit? 0 1 LC 1.00 104 rad/s – The capacitor can withstand a peak voltage of VC,max=500 V. If the voltage source operates at the resonance frequency, what maximum voltage amplitude can the source have if the maximum capacitor voltage is not to be exceeded? – at resonance: Z=R VC ,max 10/27/2011 Emax 1 I max X C R 0 C Emax 0 RCVC ,max 50 V Review Midterm 2 34 Power, and Peak and rms Values … A toaster oven consumes P=1440 W of power when operating at 120 V/60 Hz. – What is the resistance of the oven’s heater element? – What is the peak current through it? – What is the peak power dissipated by the oven? 10/27/2011 Review Midterm 2 35 Power, and Peak and rms Values … A toaster oven consumes P=1440 W of power when operating at 120 V/60 Hz. – What is the resistance of the oven’s heater element? 2 Vrms P R 2 Vrms R 10.0 P – What is the peak current through it? I rms Vrms R I max Vrms R 2 17.0 A – What is the peak power dissipated by the oven? Pmax Vmax I max Vrms 2 I rms 2 2 P 2880 W 10/27/2011 Review Midterm 2 36 Reactance The peak current through an inductor L is Imax=40.0 mA. – What is the current if the emf peak voltage Emax is doubled? – What is the current if the emf frequency f is doubled? – What is the current if the frequency f is halved and, at the same time, the emf is doubled? 10/27/2011 Review Midterm 2 37 Reactance The peak current through an inductor L is Imax=40.0 mA. – What is the current if the emf peak voltage Emax is doubled? If frequency remains constant, X L L is unchanged; Emax I max ' 80.0 mA If E doubles, I doubles I max XL – What is the current if the emf frequency f is doubled? If frequency doubles, X L L doubles; E I max max If X L doubles, I halves I max ' 20.0 mA XL – What is the current if the frequency f is halved and, at the same time, the emf is doubled? If frequency halves, X L L halves; E I max max If X L halves and Emax doubles I max ' 160 mA XL 10/27/2011 Review Midterm 2 38