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Unformatted text preview: MasteringPhysics: Assignment Print View http://session.masteringphysics.coanchassigmnentPrint?assignmen.. Assignment Display Mode: lspifl.’ PHYSICSBCF08 tit? {Etrcoéts east Magnetism 33% at t2:i§*§;§m at: Monday, fievemher 1?“, 29% View Grading Details ..i..v,IMM,A.itWat....a,Mii.aaM,.i.ttcAawWWthawwmwwmamwwww.WWd wquWWWWWWWmeswaWW, Charging and Discharging a Capacitor in an R-C Circuit Description: Contains derivation of the expressions for charge and current in a series R-C circuit and a number of related conceptual questions. A long problem but it tray be useful to tackle in one sitting Learning Goal: To understand the dynamics of a series RvC circuit. onsider a series circuit containing a resistor of resistance R and a capacitor of capacitance C? connected to a sauce of EMF 5; with negligible internal resistance. The wires are also assumed to have zero resistance. Initially, the 3 i switch is open and the capacitor discharged. E E Let us try to mderstand the processes that take place alter the switch is closed. The charge of the capacitor, the current inthe circuit, and, correspondingly, the voltages R across the resistor and the capacitor, will be changing Note that at any moment in time during the life of our circuit, Kirchhofl‘s loop rule holds and indeed, it is helpful: at w V8 w ts; m t}, where 12;; is the voltage across the resistor, and it}; is the voltage across the capacitor. 3 i i 2 g i ; Part A immediately after the switch is closed, what is the voltage across the capacitor? ANSWER: s it: e RIO *9 Part B Immediately alter the switch is closed, what is the voltage across the resistor? ANSWER: V a t e ZEN) Part C Immediately alter the switch is closed, what is the direction of the current in the circuit? ANSWER: . clockwise counterclockwise There is no current because the capacitor does not allow the current to pass through. While no charge can physically pass through the gap betweenthe capacitor plates, it can flow in the rest of the circuit. The current in the capacitor can be thought of as a difierent sort of current, not involved with the flow of barge, but with an electric field that is increasing with time. This current is called the displacement current. You will learn more about this later. Of course, when the charge of the capacitor is not changing, then there is no merit. ; 3- : Part 1) Afier the switch is closed, which plate of the capacitor eventually becomes positively charged? ANSWER: : the [OP plat: the bottom plate both plates neither plate because electrons are negatively charged Part E Eventually, the process approaches a steady smte. In that steady state, the charge of the capacitor is not changing. What is the current in the circuit in the steady state? Hint E.1 Charge and current 3%? {if If the charge of the capacitor is not changing, then the time derivative of charge, is zero. 1 0f16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmen... ANSWER: Part F In the steady state, what is the charge {I of the capacitor? 3 i § § Hint El Voltage in the steady state a If the current is zero, the voltage across the resistor is zero. That should allow you to determine the voltage across the capacitor. Express your awer in terns of any or all of £1, R, and (f. ANSWER: E q, = {35 Part G How much work W is done by the voltage source by the time the steady state is reached? H .. . .. Chagami .. H . . . .. .. H . i ; By definin‘on, the EMF of the source is defined as the ratio of the work done by the source "pushing" a charge q through the circuit and the magnitude of that charge: 8 we , Express your answer in terlm any or all of S, R, and . . ANSWER= : it! = or“ apacitor stores an armunt of energy {3 m “5“. This is only half the work done by the EMF source. The retraining gt???“ was dissipated in the resistor. So such a simple charging circuit has a high loss percentage, ndependent of the value of the resistance of the circuit. ven though energy is dissipated across the resistor as the capacitor charges, note that the work done depends on (2', but not on I}! This is because it is the capacitor that determines the amount of charge flow through the i E i i i In order to charge the capacitor, a total amount of clmge 2; {26 had to move across the potential difference 63 of the EMF source. The source did work to move this charge equal to W g8 C52. Recall that a charged i ircuit. Charge flow stops when qt} m 8'. The resistance does however affect the rate of charge flow i.e. the current. You will calculate this effect in the parts that follow i Now that we have a feel for the state of the Circuit in its steady state, let obtain expressions for the charge of the capacitor the cmrerit in the resistor as functions of time. We start with‘the loop rule: W V3; V2: Note éthat Vggt} m $838, $38} as and K13“ {£23}. Using these equations, we obtain % then, ‘ ‘ Part H . . . . , , . Integrate both sides of the equation to obtain an expression for g i Hint [-1.1 Constant of integration 3 C To obtain the conscint of integration, use the initial condition that at time t w. t), the charge on the capacitor is zero. , Express your answer in terms ofany or all of 23', R, if, and 6?. Enter exp (1:) for s3“. ANSWER: fiét}=0§{1M€1§3 : Part I Now difierentiate cg {t} to obtain an expression for the current 1 Qt}. ANSWER: * V s V i E Express your amwer in terms ofairy or all oft“, It, t, and (3. Enter exp (2:) for r”. i i I {z} = Eerie oretically, the steady state is never reached: The exponential functions approach their limits as i --e <33 asymptotically. However, it does not take very long for the values of z} and I {t} to get very close to their limiting alues. The next few questions illustrate this point Note that the quantity Rat? has dimensions of time and is called the time constant, or the relaxation time. It is ofien denoted by 7. Using 7, one can rewrite the expressions for charge and current as follows: w» CS (3 “exp M {at} m C38 (1 weep 2 of 16 ‘ 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigmnentPrint?assignmen... 193} m "b Iii} m» ifihugieg a capacitor current remiss: time» :3" £3 E t fiffiargias z: capacitor 2 charge some» time Part J " Find the time 3.3 that it would take the clurge of the capacitor to reach 99.99% of its maximum value given that I? m 123} .3“! and if? or 5065 gt“. Part J.1 Find an expression for the time Find the time a that it takes the charge ofthe capacitor to reach 99.99% of its maximum value. Him J.l.a How to approach this part What is the maximum charge on the capacitor? Set :1 {t} to be 99.99% of this value and solve for the corresponding time. lt would‘tahe‘ the same @fii’ériifie' r51 the‘cmem to am}; 50.0mm; imuaikfiaammn) value; cornpare‘theveiipressions for its} and I (2} :5 5e; this for yourself. ' ' " ' 3 Express your answer numerically in seconds. Use three significant figures in your answer. ¢ ‘ Notice how quickly the circuit approaches steady state for these typical values of resistance and capacitance! g E s i 3 § % Let us now consider a different R-C circuit. This time, the capacitor is initially charged (:fiit} “4 (2(3), and there is no source of EMF in the circuit. We will assume that the top plate of the capacitor initially holds positive charge. 41 For this circuit, Kirchhofl’s loop rule gives 13 «t g m 0, 01' equivalently. If? M w § § ; g i Part K Find the current I {(3 as a finiction oftime for this circuit. W Find the charge {13} on the capacitor Find the charge a; {a} on the capacitor as a function of time for this circuit. Hint [Clan The relationship between charge and current The current inthe circuit is given by IR} m fizz—lg. Substitute this expression for into the equation {{fifl m “if”? separate variables, and then integate both sides of the resultling equation. ‘ 9 Express your answerin terns of go, {7, i, and If- Enteraxp (x) for 6"- ANSWER: qe'; = with Express your answer in terms of as. ('3’, i, and R. Enteraxp (x) for e. . i i a 5 ANSWER: 3 of16 ' 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://sessionmasteringphysics.com/myct/assignmentPrint?assignmen... >~ negative ‘of current canvbe explained by the fact positive charge capacitors mpvplat‘evdecrEases. ofthese finictionsrare figure. {a} " Qischurging a capacitor: C9370“! 3’ crests “mt? an»; gmxehazgmg a cagacitm; ’ charge transit» time ,3 E :1 % Problem 25.72 Description: (a) Obtain an expression for the rate (dV/dt) at which the voltage across a charging capacitor increases. (b) Evaluate your result at time F 0. (c) Show that if the capacitor continued charging steadily at this rate, it would be fully charged in... Part A : Obtain an expression for the rate (zfi’grfi) at which the voltage across a charging capacitor increases. Express your answer in tents of 1, fl, and f’. ANSWER: ' 7&2” c“? Part B t Evaluate your result at time t U. ANSWER: Port C ; Show that if the capacitor contian charging steadily at this rate, it would be fully charged in exactly one time constant ANSWER: Answer Key: Elf the capacitor were to charge steadily at this rate (i.e§, if the voltage were a linear function of time, (dV_C/dt)_0*t=(emf)*t I/(tau)), the voltage would reach its final value (V__C(infifnity)=E) in just one time constant (i.e., V_C(infinity)=(emf)*t/(tau) for Problem 25.77 Description: A parallel—plate capacitor has plates of 10 ( cm)A2 area separated by a 0. 104nm layer of glass insulation with resistivity rho = 1.2 * 10Al3 Omey * m and dielectric constant kappa = 5.6. Because of the finite resistivity, current can leak." 4 of16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmen... i Part B 1 Find the time constant for this capacitor to discharge through its insulation % Express your amwer using two significant figures. ANSWER: z = . *3 Part c Show that it depends only on the properties of the insulating material and not on its dimensions. ANSWER: : Answer Key: :We will use the resistance of the insulation material, R=(rho)*d/A, where d is the thickness of the material and A is the area of the iplates. We will also use the equation for parallel—plate capacitance, C=(kappa)*(epsilon)_0*A/d, where kappa=5.6 is the dielectric constant of glass. The time constant we are seeking is tau=R*C=[(rho)*d/Al*[(epsilon)_0*A/d]=(rho)*(kappa)*(epsilon)_0. This is independent of the geometrical terms d and A, and depends only on the material properties. i i g i 2 ; é i. i z Charge Moving in a Cyclotron Orbit Description: General problem, which goes through charged-particle motion perpendicular to a magnetic field; reviews cyclotron frequency derivation Goes through kinematics and fiequency invariance. Learning Goal: To understand why charged particles move in circles perpendicular to a magnetic field and why the frequency is an invariant. ‘ A particle of charge 11 and rmss m, moves in a region of space where there is a uniform magnetic field I? w 863 (i.e., a magnetic field of magnitude {fit in the +2 direction). In this problem neglect any forces on the particle other than the magnetic force. oeeooonee coco-coon eeoomoeooe y erowofi‘ooo L coceeoeea z *‘ coarseness §§Q¢0§¢le .6 oooooeeaoflgfiz #OtOOtkfiié senescence Part A At a given moment the particle is moving in the +x direction (and the magnetic field is always in the +2 direction). If r; is positive, what is the direction of the force on the particle due to the magnetic field? : Hint A.l The right-hand nine for magnetic force 3 I A charged particle moving through a region of magnetic field experiences a magnetic force. This force is directed perpendicular to boththe velocity vector and the magnetic field vector at the point of interaction The requirement that the force be perpendicular to both of the other vectors specifies the direction ofthe force to within an algebraic sign. This algebraic sign is determined by the right-hand rule. To employ the right-hand rule: 1. Spread your right thumb and index finger apart by 90 degrees. 2. Bend your middle finger so that it is perpendicular to your thumb and index finger. 3. Orient your hand so that your thumb points in the direction of the velocity and your index finger in the direction of the magnetic field. 1f the charge is positive, your middle finger is now pointing in the direction of the force as shown in the figure If the elnrge is negative, the force is in the direction opposite your middle finger. + x direction E. E 5 of 16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.corn/myct/assigmnentPrint?assignmen... - ' -x direction 4» y direction g --»-y direction -3» z direction m z direction 3!? Part B This force will cause the path of the particle to Curve. Therefore, at a later time, the direction of the force will ANSWER: have a component along the direction of motion ' rermin perpendicular to the direction of motion have a component against the direction of motion first have a component along the direction of motion; then ayinst it; then along it; etc. is? Part c The fact that the magnetic field generates a force perpendicular to the instantaneous velocity of the particle Ins implications for the work that the field does on the particle As a consequence, if only the magnetic field acts on the C particle, its kinetic energy will . ANSWER: - increase over time ' decrease over time i retrain constant oscillate ‘ 3% Part D The particle moves in a plane perpendicular to the magnetic field direction as shown in the figure. What is w, the angular fi'equency of the circular motion? ; attenuates: eceeeeoac g accomocaco y cue fleet L g ecekeeo ; x escapee co canon» no a i. coon». «reggae: cannot to cocaine )0 ‘ Hint D.l How to approach the problem i This is a circular dynamics problem Set PM“: we to solve the problem Note that angular speed and angular frequency are the same physical quantity. Part D.2 Determine the magnetic force If the particle is moving With velocity of magnitude is, what is Fm, the magnitude of the rmgnetic force on the particle? Express EMS in terms of u and other givenvariahles. ANSWER: gm = W336 V ‘Part D3 Determine the acceleration of the particle If the particle moves in a circular orbit of radius 1? with uniform speed {$2, what is the radial component of the particle's acceleration 6,? Express at, in terms of a: and other given variables ANSWER: V Part D.4 Express the angular speed in terns of the linear speed What is the angular speed a: in terms of t), the tangential speed? Hint D.4.a Relationship between 3; and a; xpress 15in term of a: and R, the radim of the circle. ANSWER: t, = R Express “J in term of I}, m, and Be. 6 of16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?aSsignmen.L. it appeared in the equations of force and motion, it canceled This implies that the fiequency (but not the linear speed) of the The first particle accelerator built, the cyclotron, was based on the fact that the fi'equency of a charged particle orbiting in a uniform field is independent of the radius. in the cyclotron, radio fi‘equency voltage is applied across gap between the two sides of the conducting vacuum chamber in which the protons circulate owing to an external magnetic field. Particles in phase with this voltage are accelerated each time they cross the gap (because the eld reverses while they make half a circle) and reach energies of millions of electron volts afier several thousand round trips. Forces between a Charge and a Bar Magnet ‘ Interaction of stationary charge and bar magnet A positive charge is displaced some distance in the « fir. direction from the magnet. Assume that no charges are induced on the magnet. w» Part A Assume that the length of the magnet is much smaller than the separation between it and the charge. As a result of magnetic interaction (i.e., ignore pure Coulomb forces) between the charge and the bar magnet, the magnet will experience which of the following? 3 i z ANSWER: E A torque due to the charge attracting the north pole of the manget I A torque due to the charge attracting the south pole of the magnet A torque only if one magnetic pole is slightly closer to the charge than the other . No torque at all The fact that a stationary charge pro uces no torque on a magnet emphasizes a difference between electric and magnetic forces. You rmy think of the bar magnet as having a positive magnetic charge at the north end and a negative magnetic charge at the south end (even though there exist no nugnetic charges in nature). importantly, these magnetic charges are not the same as electric charges, and they do not interact with stationary electric charges. (This node] of the nagnetic dipole correctly predicts that the magnetic field lines go fiomN to S outside the n'agnet, however.) Interaction of moving charge and bar magnet Consider a second case in which the charge is again some distance in the a direction from the magnet but now it is moving toward the center of the bar magnet. that is, with its velocity along it. u? 3 Due to its motion in the magnetic field of the bar magnet, the charge will experience a force in which direction? 2 Part 3.1 Determine the magnetic field direction near a charge % The bar magnet produces a niagietic field at the position of the moving charge Given the relative location of the charge with respect to the bar magnet, and the orientation of the bar magnet, in which direction will this field point? Hint B.l.a Dipole naignetic field The negnetic field lines outside of a magnetic dipole go fi-omthe dipole's north pole to its south pole. Answer in terms of at, £1, E, or a linear combination thereof. ANSWER: a a»: ’ 7 of16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrint?assignmen.. Determine the direction of force on a charge moving in a magnetic field A charge .3 moving in a rmgnelic feld feels a force given by the Lorena force equation: 15" of? it 3, 1 where y; is positive here but could be positive or negtive in general. The direction of the force is determined by the vector product if X {37 and may be found by applying the right-hand rule: 1. Align your right hand so the fingers point along the direction of . 2i Rotate your wrist so that the fingers will point along the direction of when they are closed. 3. Your right thumb now points along the direction of the force. What two vectors should fill the blanks in these statements to correctly determine the direction of the force? ANSWER: 3%; t (2/1? ANSWER: t l»: - W! hir‘ on in tea; .ii :0 'i This is the only type of magnetic fierce: A moving charge will experience a vector force due to a magnetic field. The force between two permanent mgnets arises because of the (perpetually) circulating charge of the electrons in the magnets. . in reality, electric forces onthe bar magnet might well dominate the magnetic forces discussed here. In practice, substantial rmgnetic forces arise only for pemianently magnetized objects and for current-carrying wires that have significant numbers of moving charges in spite of being essentially neutral electrically. ‘ i i Interaction of mm and bar magnet i i Now the charge is replaced by an electrically neutral piece of initially magnetized sofi iron (for example, a nail) that is not moving. n tron Nail ,9 W 5» . n) As a result of the magnetic interaction betweenthe sofi iron and the bar magnet, which of the following will occur? > C.1 Magnetic induction A piece of sofi iron will contain magnetic domains (srmll dipoles) that have the following properties: A They align themselves along the magnetic field beacuse of the torque exerted on them This is called induction of a magnetic dipole moment. B. They experience a force due to the magnetic field of the bar magnet C. They produce a magnetic field that causes the mgnet to experience a force. d'thelfi that"fai the nail Is the force attractive or repulsive? ANSWER: ; The nagnet will experience a torque due to the iron attracting its north pole. ' The magnet will experience a torque due to the iron attracting its south pole. Q Whichever pole of the magnet is closest to the iron will be attracted to the iron. Whichever pole of the magnet is closest to the iron will be repelled fi-om the iron. The forces here result fi'om magnetic domains (small dipoles) contained within the soft iron that are aligned by the mgietic field of the bar magnet (These donnins my be thought of as srmll perpetual current loops, and thei tendency to be aligned is called perme ' 'ty.) Whi ole of the bar gnet is do 'll dominate this alignment and will align the dipoles to attract that pole of the bar magnet. WWM Magnetic Field from Two Wir Description: Calculate, at several different locations, the net magnetic field due to two straight infinite wires carrying anti-parallel currents. Students are asked to look for a pattern in the results as the points of imerest become more and more remote from the wires. i E Learning Goal: To understand how to use the principle of superposition in conjunction withthe Biot-Savart (or Ampere's) law. Fromthe Biot—Savart law, it can be calculated that the magnitude of the magnetic field due to a long straight wire is given by i 8 of16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View i equidistant fi‘om the wires. g i i z t i i E .PartA I The same result can be obtained from Ampere's law as wellr As you answer the questions posed here, try to look for a pattern in your answers. writing sometimes automatically use their "pencil-flee hand" to detemiine the direction of {31) http://session.masteringphysics.com/myct/assignmentPrint?assignmen... of the page and that in wire 2 is directed into the page. The distance between the wires is 241. The x axis is perpendicular to the line connecting the wires and is where 159 ( may 3, 333'"? - mfA) is the permeability constant, I is the current in the wire, and (i is the distance fromthe wire to the location at which the magnmde of the magnetic field is being calculated E In this problem, you will be asked to calculate the nagnetic field due to a set of two wires with antiparallel currents as shown in the diagram. Each of the wires carries a current of magnitude 1. The current in wire 1 is directed out I Which of the vectors best represents the direction of the rmgnetic field created at point K (see the diagram in the problem introduction) by wire 1 alone? Part B ANSWER: 3 -: Part C 9of16 Enter the number of the vector with the appropriate direction. Enter the number of the vector with the appropriate direction. Enter the number of the vector with the appropriate direction. Which of the vectors best represents the direction of the rmgnetic field created at point K by wire 2 alone? Which of these vectors best represents the direction of the net mgnetic field created at point K by both wires? i The direction of vector can be found using the right-hand rule. (Take care in applying the right-hand rule. Many students mistakenly use their lefi hand while applying the right-hand rule since those who use their right hand for i i 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmentPrinflassigmnerL.. ' Part D Find the magnitude of the magnetic field 3m created at point K by wire 1. Express your answer in term of I, (i, and appropriate constants. ANSWER: Of course, Part E 5 Find the mag-rimde of the net magnetic field I?“ created at point K by both wires. g Express your answer in term of I , (i, and appropriate constants. ANSWER: Q _ W1 1V Part F Point L is located a distance n’v’tfi fi'omthe midpoint betweenthe two wires. Find the magnitude of the magnetic field 13“, created at point L by wire 1. Hint El How to approach the problem Use the distances provided and the Pythagorean Theorem to fird the distance between wire 1 and point L. Express your answer in terns of .f , d, and appropriate constants. ANSWER: 3 Mr at, = “WW Extix/g § § § i z z 2 g !. e g M Part G g Point L is located a distance (33% from the midpoint between the two wires. Find the magnitude of the net magnetic field 3L created at point L by both wires. i 1 G.l How to approach the problem Sketch a detailed diagram with all angles marked; draw vectors I?» and 13%,»; then add them using the parallelogram rule. Part G.2 Find the direction of the magnetic field due to wire 1 Which of the vectors best represents the direction of the magIetic field created at point L (see the diagram in the problem introduction) by wire 1 alone? Enter the number of the vector with the appropriate direction m c: Find the direction at the magnetic field due to wire 2 Which of the vectors best represents the direction of the magnetic field created at point L by wire 2 alone? nter the number of the vector with the appropriate direction § g: g; i, t 10 0f16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignrnentPrint?assignmen... Part GA ‘ Find the direction of the net magnetic field Which of the vectors best represents the direction of the net magnetic field created at point L by both wires? Enter the number of the vector with the appropriate direction ANSWER: 3 §Now thatthe directions ofthe magne ' fi ld Lamithe ; Use the distances provided and your knowledge of right angle triangle trigonometry to find the angle between the magnetic field due to Part G.6 “ Find the angle between magnetic field due to Wire 1 and the x axis i t 33 j Hint (1.5 Angle between magnetic field due to wire 1 and the x axis 3 Use the distances provided and your knowledge of right angle triangle trigonometry to find the angle fig between the magnetic field due to wire 1 at point L and the x axis. Express your answer numerically, in degrees. ANSWER: g}; = 3&3 Q Him 6.7 Net magnetic field i Corsider the symmetry of the magnetic field at point L due to wire 1 and the nugtetic field due to wire 2. You should note that the y components of these two vectors are of equal magnitude but are opposite in direction. Therefore they will cancel when added together, leaving you only to worry about the x components. Find the x component of the magnetic field at point L due to wire l by using the magnitude of the vector (found in Part F) and the angle * : between the x axis and the magnetic field vector (found in the previous hint). Because of synmetry, the x component of the magnetic field at point L due to wire 2 is the same size. To find the net magnetic field at point L you need , to add together the x components of the magnetic field at point L due to wire 1 and of the magnetic field due to wire 2. Express your answer in tern of I , (i, and appropriate constants. x‘ 23m! *3 Part H Point M is located a distance 2‘; fi'omthe midpoint between the two wires. Find the magnitude of the magnetic field {Em created at point M by wire 1. Express your answer in terms of 1, xi, and appropriate constants. ANSWER: : g.“ = my " sway/’3“? “‘11 Part 1 Find the magnitude of the net magnetic field [fig created at point M by both wires. Hint 1.1 ketch a detailed diagram with all angles marked; draw vectors Em and 33“; then add them using the parallelogram rule. How to approach the problem Find the direction of the magnetic field due to wu-e l 9 § 11 of16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://sessionmasteringphysics.com/myct/assigmnentPrint?assignmen... , Which of the vectors best represents the direction of the magnetic field created at point M by wire 1 alone? ; Enter the number of the vector with the appropriate direction. -_ Part 1.3 Find the direction of the net magnetic field ‘ Which ofthe vectors best represents the direction of the net rmgnetic field created at point M by both wires? Enter the number of the vector with the appropriate direction. Ll: Angle between magnetic field due to wire 1 and the x axis U the d‘ tances provided and our knowledge of right angle triangle trigonometry to find the angle between the magnetic field due to wire 1 at point M and the x axis. g Part 1.5 Find the angle between magnetic field due to wire 1 and the x axis Use the distances provided and your knowledge of right angle triangle trigonometry to find the angle {33¢ between the magnetic field due to wire 1 at point M and the x axis. Express your answer numerically, in degrees. ANSWER: g” : {33A 6 Hint 1.6 Net magnetic field Consider the symmetry of the magnetic field at point M due to wire l and the magnetic field due to wire 2. You should note that the y componems of these two vectors are of equal magnitude but are opposite in direction. 'l‘herefore they will cancel when added together, leaving you only to worry about the x components. Find the x component of the magnetic field at point M due to wire 1 by using the magnitude of the vector (fem-id in Part H) and angle between the x axis and the rmgnetic field vector (found inthe previous him). Because of symmetry, the x component of the mgneh‘c field at point M due to wire 2 is the same size. To find the net magnetic field at point you need to add together the x components of the magnetic field at point M due to wire 1 and of the magnetic field due to wire 2. fi—W ANSWER: S in: and E 3 Express your answer in terms of 2", xi, and appropriate constants. E z it»! ran .1 Fimlly, consider point X (not shown inthe diagram) located on the x axis very far away in the positive x direction. Which of the vectors best represents the direction of the magnetic field created at point X by wire 1 alone? Enter the number of the vector with the appropriate direction. 4:- \N 12 0f16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.conchassiglmentPrint?assignmen... ANSWER: f 1 iii Part K Which of the vectors best represents the direction of the magnetic field created at point X by wire 2 alone? Enter the number of the vector with the appropriate direction. As you can see, at a very large distance, the individual mgnetic fields (and the corresponding magnetic field lines) created by the wires are directed nearly opposite to each other, thus ensuring tint the net magnetic field is very, very small even compared to the magnitudes of the individual magnetic fields, which are also relatively small at a large distance Eromthe wires. Thus, at a large distance, the magnetic fields due to the two wires almost ancel each other out! (That is, if point X is very far from each wire, thenthe ratio Emilia is very close to zero) Another way to think of this is as follows: If you are really far fi-omthe wires, then it‘s hard to tell them apart. It would seem as if the current were traveling up and down, almost along the same line, thereby appearing much the same as a single wire with almost no net current (because the up and down currents almost cancel each other), and therefore almost no magnetic field. Note that this only works for points very far from the wires; otherwsie .o v ‘< a g E 5‘ E 5 fi ‘3 s O. a % i s § 3 £5; E (D E 5 "a 5 é a 8' 5 e e R? a E 8 3% a t comes as no surprise then that one way to eliminate unnecessary nagietic fields in electric circuits is to twist together the wires carrying equal currents in opposite directions. Torque on a Current Loop in a Magnetic Field Description: Students calculate the torque for a rectangular current loop tilted in a uniform magnetic field, uses magnetic moment. This problem will show you how to calculate the torque on a nagnetic dipole in a uniform magnetic field. We start with a rectangular current loop, the shape of which allows us to calculate the Lorentz forces explicitly. Then we generalize our result. Even if you already know the general formula to solve this problem, you might find it instructive to discover where it comes from ’ z § § g E g § § E , A current I flows in a plane rectangular current loop with height a and horizontal sides in. The loop is placed into a uniform magnetic field I? in such a way that the sides of length a are perpendicular to £3“ , and there is an angle 8 betweenthe sides oflength 3; and E . Calculate 1;, the magnitude of the torque about the vertical axis of the current loop due to the interaction of the current through the loop with the magnetic field. View from the side pro em F‘rst find the f rces (di ti and magnitude) on each segment of the loo . Then find the tor ue due to each force and add them up. Part A.2 Forces on the parts of the loop that have length a As current is moving through the loop, forces act on its difi‘erent parts. These result from Lorentz forces on the charges that rmve through the wire. What is the direction of the forces on the two pieces of wire of length a? View from above 13 ofl6 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://sessionmasteringphysics.com/myct/assignmentPrint?assignmen... Hint A.2.a Force on a straight current-carrying wire in a magnetic field The force on a straight wire of length a carrying current I ina magnetic field is given by - } Select the correct diagram from the four options. The forces are symbolized by the red arrows. ANSWER: I 5 A- ual and opposite, but do produce; net torque for t}, smce therr lines of action do not pass throughthe center of The forces on these parts 0 cup on cause a motion of the center of mass, since they are eq - mass of the loop. Force on a current-carrying wire in a magnetic field = Find F, the magnitude of the force on each of the vertical wires (those with length a). Hint A.3.n Relevant equation ecall that 3‘5 1;? x 3? and x :3} ubxiufa}, where (2 is the smaller angle between a and S. Express 1? interms of 3’, a, 8, and 3. Note that not all of these may appearin the answer. ANSWER: p = “i”; Part AA Find am, the component of the torque around the vertical axis of the loop (i.e., out of the plane of the top-view figure). Torque on a loop Hint A.4.a Definition of torque The net torque :9 is a vector pointing in the direction of the change of angular momemum It is defined by 3 Express mm in term of ANSWER: Part A.5 Forces on the parts of the loop that have length i: As current is moving through the loop, forces act on its different parts. These result from Lorentz forces onthe charges that move through the wire. What can you say about the forces on the two pieces of wire of length b? lect the most accurate qualitative description fro the following list. ANSWER: The forces point away fromthe center of the loop and cancel each other. The forces act as a torque about an axis through the midpoints of the two two pieces of length b and make the loop turn in the direction of positive 8. The forces point toward the center of the loop and cancel each other. Both forces point upward perpendicular to I? and It. a ince forces on two of loop cancel out. ‘ofacdonvgoesthrough of loop, weuneed not consider‘them in calculation ofthe torque on loop: key‘is that the forces ' l’ the' to ue contributions about any point cancel Express the magnitude of the torque in terms of the given variables. You will need a trigonomeric function [e.g., stage; or emit?” Use 17:? for the magnitude of the magnetic field. ANSWER= ; '5' = mmm £53 Part B Give a more general expression for the magnitude of the torque r. Rewrite the answer found in Part A in term of the magnitude of the magnetic dipole morrem of the current loop at. Define the angle between the vector perpendicular to the plane of the coil and the magnetic field to be a}, noting that this angle is the complement of angle a in Part A. t 3.1 Defmrtron of the magnetic dipole moment The magnen'c dipole moment 36% is a vector quantity defined for every current loop by the relation aim at”. The quantity :3: is the vector area of the loop, which is perpendicular to the plane and taken to point up if the current flows around the loop in a counterclockwise direction when looking down on the loop (right-hand rule). Its magnitude is the area of the loop. _ Give your amwer in terms of the magnetic moment m, magnetic field B, and $- ANSWER: 7 = relish! W} 14 of16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assigmnentPrinflassignmen... The more general vector form of this expression is i Part c A current i flows around a plane circular loop ofradim r, giving the loop a magnetic dipole moment of mgr-ritude m. The loop is placed in a uniform magnetic field with an angle e; betweenthe direction of the field lines and the magnetic dipole moment as shown in the figure. Find an expression for the magnitude of the torque r on the current loop. Hint c.1 " Formula for the area ofa circle ‘ The area of a circle with radius 2* is m“. Express the torque explicitly in tenm of r, I, x, {in and I; (where r and a are the magnitudes of the respective vector quantities). Do not use :31. Youwill need a trigonomeric function [e.g.. smite} or meigéfl. ‘ Problem 26.25 i Description: Electrons and protons with the same kinetic energy are moving at right angles to a uniform magnetic field. (a) How do their orbital radii compare? \ Part A How do their orbital radii compare? Express your awer wing two significant figures. . .90 . ANSWER. :3 = #3 5W.me ”E f Problem 26 62 Description: Part of a long wire is bent into a semicircle of radius a, as in the figure . A current] flows in the direction shown. (a) Use the Biot-Savart law to find the magnitude of the magnetic field at the center of the semicircle (point P). (b) Find Part of a long wire is bent into a semicircle of radius a, as in the figure . A current } flows in the direction shown. Use the Biot-Savart law to find the magnitude of the magnetic field at the center of the semicircle (point P). i Express your amwer in terms of the variables I, a, and cotant W ANSWER: Part B Find the direction of the magnetic field at the center of the semicircle (point P). {2: u 15 0f16 11/19/2008 9:45 AM MasteringPhysics: Assignment Print View http://session.masteringphysics.com/myct/assignmerttPrint7assignmen... ANSWER“ f The magnetic field is directed to the right The magnetic field is directed to the lefl. . The magnetic field is directed into the page. The rmgnetic field is directed out of the page. i““scywnvmwwwwmw to Problem 26.87 Description: A straight segment of wire of length L, extending along the x axis from x_0 to x_0+L carries a steady currentl inthe +x direction (fed by other wires not under consideration). It's known that the magnetic field at the aim 0,y in the x-y plane due i g P g A straight segment of wire of length L, extending along the x axis from Jigs to a“; ~§~ 15., carries a steady current I in the “é‘ y direction (fed by other wires not under consideration). It’s known that the magnetic field at the point {i}, g} A 2 § § . . . . m1 its 4} L 13:; s in the gr m 3; plane due to the segment alone is directed out of the page and has a magmmde of mm» WW w y W m + f»? r 2f) «a a? Use the result above to find the magnitude of the magnetic field at the center of a square loop of side a carrying current I . : Express your answer in terms of f, a and cotant w. : E E ANSWER: E Problem 26.91 3 Description: You're taking a class in space weather physics. Space weather deals with the dynamics of the far upper atmosphere and the magnetic regions surrounding the Earth. You're preparing a term paper on the van Allen I r belts, regions where high-enery particles ‘ é You're taking a class in space weather physics. Space weather deals withthe dynamics of the far upper atmosphere and the magnetic regions surrounding the Earth You‘re preparing a term paper on the van Allen belts, regions where E l higlrenergy particles are trapped in the Earth's magnetic field. Your textbook says the magnetic field strength at the belts is 0.1 t3. 3 i t i é Your space weather project is going well, but you must add a section on the generation of the Earth's magnetic field. One highly simplified model suggests the Earth's field arises fi'om a single loop of current at the outer edge of the g planet's liquid core (radius 3000 km) concentric withthe center. What current will give the measured field of 62 3T at the north magnetic pole? i ANSWER= i extra 3 i g . g 1 as. me a E ’ as, it?“ A E . as «M Problem 26.33 r Description: A single-turn wire loop is d in diameter and carries a I current. (a) Find the magnetic field strength at the loop center. (h) Find the [mgnetic field strength on the loop axis, x fi'omthe center. i E i A single-turn wire loop is 2.8 cm in diameter and carries a 620 am current ‘ Summary 0 of 12 dens couplets (0% avg. score 0 of 97 points 16 of16 11/19/2008 9:45 AM ...
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