Physics 2 study guide
Physics 2 study guide Physics 1260
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This 2 page Study Guide was uploaded by Briana Notetaker on Wednesday February 24, 2016. The Study Guide belongs to Physics 1260 at East Carolina University taught by Dr. Yong - qing Li in Spring 2016. Since its upload, it has received 154 views. For similar materials see General Physics 2 in Physics 2 at East Carolina University.
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Date Created: 02/24/16
Chapter 20 Direct Current: The current which flows through a battery in one direction. DC circuits: consists of circuit elements including resistors and batteries or even other sources of a constant emf which is connected in closed loops Branch Points: where three or more conductors join at points; so points C and D would be branch points. Open circuit: This would be an incomplete loop also known as a path that is not closed. Steadystate current: zero Short circuit: the ideal conducting path connecting across circuit element or element combination. Current is zero. These shorts can possibly cause a very dangerous large increase through the circuit in the current. (a) A one-loop circuit (b) A two-loop circuit The red wire make a short Equivalent Resistance the single resistor that carries the same circuit between a and b. current as the network as for the potential difference across it is the Kirchhoff’s first rule: sum of currents going in any equivalent as that across the network. branch point would be zero: Series Resistors: they Charge conservation connecting end to end, no Kirchhoff’s second rule: sum of voltage drops near branch points are in any closed path would be zero Parallel Resistors: connecting so potential difference will be the between, that way the same across them all. Find the sum of the inverses of the resistances current is the same in each. Energy Conservation The sum of the resistance is to find the inverse of the equivalent resistance for parallel resistors. the equivalent resistance of resistors. Ex.1 Find current I. Ex.2 Verify the sum of the voltage drop is zero or a counterclockwise loop ∑ I=0 beginning at the point a. I1+I2+I 30 I3= 3 2 5A10 A=15 A Ex. 5 Equivalent Resistance of a Network. a. Find equivalent resistance of network b. Find current through 4.0 Ω resistor when V = 3ab a Multiloop circuits more than one loop. Apply Ex. 7 Currents and Voltages in a Multiloop Ex. 8 Finding an Unknown Resistance by current CQ. 208 Kirchhoff’s rules in order to calculate currents and a.The circuit that will the measure the current in the resistor aonf d voltage measurements. potential differences in circuits. ammetea.Find currents Ammeter has a resistance of 1Ω and reads current in 3 b. Theb.Find voltage of the 15 V battery, Ve baher circuit? milliamperes also known as 10 A>mA. Voltmeter has 1.0ΩResistor is the battery internal resistance. resistance of 10 Ω.show meters being use to measure a)1. current through R, while voltage dropping across R, and the value R. Voltmeter is 11 V & ammeter is 50 b) ammeter in the second mA. diagram is short passing the resistor CQ.209 a)which circuit will V measure the potential difference between points a and b? Ammeter: measure electric current Ex.9 Charging a Capacitor in an RC Circuit b)what is the effect of the voltmeter in the other circuit? Connected in series with adjacent circuits R=50Ω elements. Voltmeter measure potential difference. a)diagram 2 Connected in parallel with adjacent circuit C=100 μF b)V in 1 diagram Ammeter should have zero resistance so current Ε=10 V lowers the current through circuit that doesn’t have the ammeter elements. st unchanged. Often smaller than 1Ω Should have an infinite resistance so the Find current when switch Is 1 closed, and the final equivalent resistance between points of charge is stored on the capacitor, the circuits time constant connection will be unchanged by the voltmeter. Resistance most times greater than P.2037 . 3 1 MΩ 1.00*10 μF capacitor has an initial charge of .100 C. When the resistor is connecting cross the capacitor plates, there would be an initial current going through the resistor of 1.00 A. So what would the current 1.00 s later? Ohmmetermeasure resistance Q 0.1 C Resistance can be indirectly measured through 3 3 6 3 C=1.0 ×10 μF=10 ×10 F= 10 F measurements of voltage and current just as it I0=1.0 A I (t=15) =? could very well be directly measure through the V =Q /C=.1 C/10 100 V I =V /R ohmmeter Discharging a Capacitor 0 0 0 0 Electric shock physiological reaction or injury that is >K=V /I0=000V/1.0 A=100Ω Multimeter can be an ammeter, voltmeter, caused by electric current that passes through the human T=RC=(100Ω)(10 )=.153 ohmmeter, or even capacitance meter just depends If at the beginning the capacitor is charged, we t/T 1.05/15 5 can discharge a charged capacitor by having it body. I(t)=I0e =1.0Ae =4.5×10 A on dial setting. Body is sensitive to external imposed electric voltages connected to a resistor directly across its plates. and currents. Effects on body would depend on what the Ground connecting a point to earth b Average electrical RC circuit have a battery E, resistor R, capacitor using a conductor of negligible power where 1200 C, and switch. current’s magnitude, direction and the path is. Switch closed Begin to be dangerous in body by about 1mA. resistance> potential of 0 V. grounding in kWh of ee used in @ t=0 an electric circuit stand for by the symbol a 30day period? Electric Shock and Electricity safety I=V/R ΔE=1200 kWh e The body’s electric resistance usually of the range of Δt=30d=30d*24h= 3 4 The Earth good conductor, carries a net 720 10 to 10 Ω, depends on whether the skin is wet or dry. negative charge on its surface of about Exponential decay But perhaps a fatal current of 10mA can sometimes end 9 2 Power If originally the capacitor, C is uncharged & up form a potential difference as small as 10 V. 10 C/m . p=ΔE/Δt=1200 switch is open, to close the switch would produce Power Lines conducting wires maintained at some potential relative to kWh/720 k a current time t: P=1.67kw ground =1.67×10 W Charging a Capacitor the initial current the time constant 21-53 53 Chapter 21 Magnetism Ex.1 find instantaneous Chapter 22 222 Inductance Magnet exerts force on an object with touching not involved. Electrostatics: stationary charge When a circuit carries a time varying current, this creates a time acceleration of an Magnetic Fieldphenomen,magnetic field lines electron moving at distribution produces electric fields varying magnetic field in the space around the circuit which Generate magnetic fields by permanent magnets and electric Magnetostatics: constant currents produce induces an emf. A self induced emf always opises a change in 1.0*10^7 m/s in the xy magnetic fields. currents. plne, at angle 30 degrees current due to Lenz’s law says any induced emf opposes the Magnetic field affect objects by magnetic force on moving with yaxis. A uniform Electromagnetics: timevarying charge change in magnetic flux that would produce it. charges, magnetic force and torque on electric currents, and distributions and currents produce time Selfinduced emf a timevarying current in a coil produces a Ex. 5 torque magnetic field of motion of a charge in magnetic field. magnitude 10 T is in the varying electric and magnetic fields. selfinduced emf, opposing the emf in the coil E=L(ΔI/Δt) Bar Magnets are attracted to metal objects. Electromagnetic induction: production of Selfinductance constant L; depend on size and shape alone. positive y direction. Polesat the ends where objects are most strongly attracted an electromotive force across a conductor Φ=LI SI unit is 1H=1Vs/A (north and south) exposed to time varying magnetic fields. Inductor a coil designed to have large selfinductance in circuit Like repel; unlike attract Electric field is created in around space and in which it is inserted Magnetic poles not able to be isolated thus an emf and a current is induced in a Ideal inductor no resistance & represented in circuits by the Hans Oersted discovered that when a compass place close to a conducting loop when the magnetic field 2 symbol ___________ L=μ N A/l0 [H] wire that carries an electric current, the compass needle will changes Voltage drop across an inductor a timevarying current causes turn. Ex. 3 221 Faraday’s Law Magnetic Fields, Sourcemoving charge; effect is to exert a Induction Experiments changing a voltage drop V =abΔl/Δt) Field has a nearly uniform magnitude of 0.50 T over Ex. 4 find selfinductance of solenoid of length 10cm, w/ 1000 force on other moving charge placed in the field. cylindrical region of diameter of 6.0 cm & bout zero magnetic field can produce a current, turns & cross sectional area of 2.0 cm 2 Moving chargemight be isolated point charge/ electron elsewhere. Fund mag force on wire, carrying a primary coil is connected to a battery 2 7 2 4 2 L=μ N0A/l = (4pi×10 Tm?A)(1000) (2.0×10 m )/.10 m morning through currentcarrying wire or last spinning electron current of 40 A. Wire is perp to field and passes secondary connected to an ammeter. the magnet through the center of the cylindrical region. Magnetic flux CQ.2210 Find the sign of the voltge drop V acabss the Magnetic field lines the magnetic field stands for the symbol Emfactually induced by a change in the inductor just after the switch is opened. And estimate current B and characterized graphically by field lines. magnetic flux rather than simply by a through resistor 10^2 s after the switch is opened. change in the magnetic field. Magnetic field exerts a force on a Particle Moving Φ=BAcosθ B = the strength of uniform moving Ex 5 Coils of wire wrapped around rotor in several magnetic field units T*m^2=Wb F=|q|vBsinθ planes. Commutator & brushes allow current to in an External pass through any coil only when coil is in position Magnetic Field Perp θ=0 Parallel θ=90 θ angle between vectors v and B Faraday’s law of Induction emf induced F perpendicular to the plane of v and B, to achieve max mag torque>xurrent pass through If the particle’s around any loop equals the negative rate of coil only when plane of coil is about parallel to B or velocity not perp in the direction of extended right thumb change of outward magnetic flux through when fingers of right hand rotate from v when mag moment vector is about perp to B. Each to field, path is the surface bounded by the loop. Ε =Δϕ/Δt P.2221 A solenoid with 200 turns and crosssectional area of coil > 100 turns of wire, area of 2.00*10^2 m^2. spiral>>helix 1.0 cm is 3.00 cm long. So how much current should the to B. Current > 2.00 A, magnitude is .500 T. Find the solenoid carry so that the flux produced by its own magnetic Max value –perpendicular θ =90 degrees Trajectory of a Lenz’ Law –the induced current produces a Force is zero when v is parallel to B θ =0 magnetic torque assume m and B are perp to each moving charge magnetic field that opposes the change in field is equivalent to the flux produced by earth’s magnetic field other. of magnitude of .500 G, directed along the solenoid’s axis. Right Hand Rule 214 BiotSavart Law magnetic field in anonuniform magnetic flux that produced it Direction of magnetic field induced current resulting from the e,f is Direction of instantaneous force act on a (+) produced by a small wire segment of High B>small r found from this law. The () in Faraday’s charge is found by this rule Place right hand on the length change in l carrying current I n law is included to indicte the polarity of the v so fingertips point in direction of v and fingers direction determined if you rotate fingers induced emf. can rotate from v to B. When hand is in that positi on right hand from the direction of the |ΔB| 2 on, extended right thumb points in force direction =μ 04pi(IΔlsinθ/r ) Ex. 1 The ring has a radius of 4.00 cm and a current to the direction of the vector B=∑(ΔB) resistance of 1.00*10 Ω. The magnetic locating the field point, with magnitude. Ex. 6 a beam of electrons move at Magnitude> field is increasing at a constant rte from .200 Magnetic Fields 1.00*10^7 m/s describing a circular rcuit containing a resistor R and an idB=μ /0pi(l/r) T to 400 Tin a time interval of 1.00*!0 s. Find the current in ring. Define in terms of the magnetic path in a uniform magnetic field of and L are initially connected in series force exerted on a test charge magnitude 1.00 10^3 T. Find direction f, current isn’t quickly established in te circuit due to the electrical inertia of the inductor. I=E/R(1e ) τ moving in the field with of motion, period, radius. Ex.9 long straight wire directed perp to page and velocity v time constant = τ=L/R Find Rate of change of current in circuit, t=0 L=2.00mH and d mag field B= F/ |q| v sin θ units: [T] at field point 4.00 cm to right of wire. B= SI unit Telsa (T) E=6.00V. 1T=1 N/(Cm/s)= 1 N/Am Δl/Δt=E/L=6.00V/(2.00*10^3H)=3.00*10^3 A/s7 Tm/A)/2pi)(100A/.04 m) Circular current loopmag field produced by a Cgs unit is a Gauss (G) 1 G=10 T earth’s magnetic circular current loop at a point on the axis of the Ex. 5. A very large coil selfinductance 5 loop is given by: B=μ /0(Ia /r ) or B=μ0/2pi(m/r ) 3 of 2.0 H and resistance of 10 ohms. 12 field = .5G=5*10 T Ex.8 Find magnetic field at the center of circular V battery t=0, Find the steady state 212 Magnetic forces on loop of radius 2.00 cm, carrying a current of 10.0 A current and find the current through the current carrying clockwise 2 3 coil at t=.50 s. conductors. CQ.2113A long, straight wire B=μ 0a /2r r=a B=(4pi*10^7 Tm/A)) Wire of length l a current I (10.0A)/2(2.00*10^2 m) carries a positive current I into Motional emfemf induced by the motion of in a uniform, external the page and find direction of magnetic field B the magnetic field at each Magnetic force between parallel wires a conductor through a magnetic field. |E|=Blv experiences a magnetic F/l=μ I I /(2pi*r) Ex. 2left side of the Ushaped conductor has point. 0 1 2 resistance of .10 ohms and that there is force F of magnetic Parallel wires carry current in same direction F=IlBsinθ Solenoids a coil attracting each other. Wires that carry opposing negligible resistance in each of the other three form, in shape of a helix. Produces a strong magnetic sides of the loop which includes the rod, that Magnetic Energy Ex. 6 Find magnetic energy Right hand rule determine currents would repel each other. U =(1/2)LI 2 [J] density inside a direction of F. field in alimited Ex.10 two long, straight, parallel wires 1.00 cm apart has a length of 10 cm nd moves at a constant m inside speed of 4.0 m/s through a magnetic field of Magnetic energy density superconducting coil that Torque on a Current Loop in an External each that carry a current of 1.00 A going in same magnitude .50 T. Calculate the current magnetic energy per unit volume produces a magnetic field 215 Electron Spindirection. Find the magnitude f the force on a 1.00 m Field a spin. But since length of either wire…F=μ /20i(I I 1 2) induced in the loop. inside the inductor may be shown of 10.0T. u m(1/2) A current loop in a uniform, external R=.1 ohm L=10cm=.1m B=.5 T in therms of the field strength as (B /μ0)=(1/2) produces a current and m (meg moment, produces a 2 3 ((10.0T) /4pi*10^7T magnetic field experiences zero net force, magnetic field at position r. B=μ /0pi(m/r )3 V=4.0 m/s |E|=Blv=(.50T)(.10m) u m(1/2)(B /μ )0 [J/m ] but loop experiences a torque tending to Magnetization of permanent magnet is the magnet’s (4.0m/s)=.20 V m/A)=3.98*10^7 J/m^3 I=.20V/.10ohm=2.0 A align its magnetic moment which is m with magnetic moment per unit volume M=m/v=N/V(m ) e direction of the B field. m e9.27*10 Am24 2 Electric Genertors Rotating a coil of wire in m=NIA 6 223 Resistor and an AC source Ex. 13 The max magnetization of iron is 1.7*10 A/m. an external magnetic field at an angular Capacitor and an AC source Torque and Net torque Calculate the number of aligned electron magnetic velocity w results in an induced emf E and a i=V/R=Isinwt the current and voltage are in i=Δq/Δt=Icos wt I=wCV Τ=Frsinθ [Nm] phase moments per unit volume of the iron. terminal potential difference V ab Power loss in a resistor P =I 2 R i=I sin (wtϕ) T=+Fd for counterclockwise N/V=M/m e V ab=E sin0t w=2pif=2pi/T av rms ϕ=90degrees 213 Motion of Point Charge in a Magnetic Magnets and Solenoids V rmssqrt(2) 2 or more forces act on an object, net torque Irms/sqrt(2) capacitor’s reactance: Field Magnetic field of cylindrical permanent magnet w? a Ex.3 An electric generator operating at 60.0 Hz has Xc=1/wC Uniform Magnetic field: by uniform M along its axis is the same as that of a tightly a coil consisting of 100 turns enclosing an area of ohm’s law: I=V/X c ∑τ=τ +1 +…2 2 When a particle of charge q and mass m is wound solenoid with same dimensions as the magnet if 5.00 m in a magnetic field of .100 T. (a) Calculate calculated into a uniform magnetic field B the product of the solenoid’s turns per unit length n A vertical magnetic field of 2.00×10 T, a ring of Loop experience a torque tending to align its magnetic moment m with direction of B to the and current equals M. radius 1.00 cm is flipped in the air as one would field, F=|q|vBsin 90°=|q|vB M=nI flip a coin, so that it begins rotating about a Inductor and an AC Source a=F/m=|q|vB/m=v /r centripetal m and 2 V=Vsinwt ho0rizontal axis at a rate of 50.0 rev/s. find max i=Icoswt Find rms current that resuts from acceleration Electromagnetan iron core is often placed inside a instantaneous emf induced in the ring and max connecting a 2.00muF capacitor directly the charge will move along a circular path solenoid to enhance its magnetic field. Total magnetic P=IE =050.0A)(18800V)=940000 i=Issin(wtϕ) inductive reactance: X =LL across a 120 Vrms source at a frequency of radius r and period T field –sum up the fields produced by soleoid current by W=940kW=.940MW ring has of a 60.H r=mv/|q|B T=2pim/|q|B iron;core produces a strong resultant field. ohms 2153 metal rod of length 20.0 cm and mass 100 g is Find reactance of 2.00mH inductor at a free to slide over two parallel, horizontal metallic rails frequency 60.0 Hz connected to a 10.0 V batter. Internal resitance of .100
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