PHY1004W 2010 Electricity and Magnetism Part 3 Prof Andy Buffler Room 503 RW James andy.buffler@uct.ac.za
M&I Chapter 19 Capacitors, Resistors and Batteries
M&I 19.1 Capacitors in circuits Surface charge rearrangements are rapid nanoseconds! a steady state is soon reached. capacitors can be used to cause a slower change of the current in a circuit Capacitor: or A capacitor is made of a sandwich of two metal foils separated by an insulating material and coiled up into a small package. no conducting path through the capacitor charges cannot jump between the plates. insulating material external connections
A charged capacitor (charge Q on each plate): Plate area A Radius, R E fringe Plate separation s There is a fringe field just outside the charged capacitor plates, where... E fringe Q A s R 0 2
M&I 19.2 Charging a capacitor Connect a 6V battery, a light bulb and a capacitor in series... At t = 0.1 s after starting, the bulb shines brightly... At t = 1 s after starting, the bulb is down to medium brightness... At t > 10 s after starting, the bulb is not glowing at all...
Charging a capacitor... 2 After some time, there is significant positive and negative charge on the capacitor plates...... the fringe field of the capacitor acts in a direction opposite to the conventional current, thus decreasing the net field and the current decreases. Eventually the current stops when there is enough charge on the capacitor to create large enough fringe field to counteract the field made by the other charges. The initial establishment of surface charges takes a few nanoseconds, while it takes several seconds to reach static equilibrium (zero current).
Discharging a capacitor Now remove the battery completely... t = 0.1 s t = 1 s t > 10 s The light bulb initially burns brightly again and decays in brightness until the current in the circuit is once again zero... The fringe field of the capacitor plus the electric field of the charges on the surfaces of the wires drives the current so as to reduce the charge on the capacitor plates... and on the wire surfaces... the electric field is reduced and so thus the current.
M&I 19.5 R 1 Applications E a. Different plate radius R: Same charge Q distributed on plates after 0.01 s of charging in each case... Q A s Q s fringe,1 2 0 2R1 0 R1 2R1 R 2 > R 1 E fringe,2 Q A s Q s 2 0 2R2 0 R2 2R2 E fringe,1 Smaller fringe field means a larger net field, hence current in circuit with larger capacitor has decreased less than smaller capacitor.
Applications...2 s 1 b. Different plate separation s: Same charge Q distributed on plates after 0.01 s of charging in each case... E fringe,1 Q A s R 1 0 2 E fringe,2 Q A s R 2 0 2 E fringe,1 s 2 < s 1 Again, a smaller fringe field means a larger net field, hence current in circuit with larger capacitor has decreased less than smaller capacitor.
Applications...3 c. Different dielectric materials between the plates: Same charge Q distributed on plates after 0.01 s of charging in each case... E fringe,1 E plastic E fringe,1 The field of the polarized plastic is in opposite direction to the fringe field, so field due to battery and surface charges is reduced by a smaller amount... net field in wire is larger than it is without the plastic... and hence current flowing has decreased less than without the dielectric material.
Capacitors in parallel Two capacitors in parallel can be thought of as one capacitor with plates that have the sum of the areas of the individual capacitors. An isolated light bulb Why does the light bulb light up?
What are capacitors used for? What will happen to the light bulb if we break the circuit (briefly) here??... capacitors are used in all kinds of circuits to even out rapid changes in voltage...... for example, in a power supply, parallel capacitors filter out high frequency (AC) voltage changes that are riding on top of a constant DC voltage... the circuit is caused to behave sluggishly... unable to change voltage conditions rapidly.
M&I 19.6 Energy considerations Consider the charging and discharging of a capacitor... In terms of +E or E, what was the energy change of the battery, capacitor, light bulb and surroundings during each phase?... fill in the missing six spaces below... Battery: Light bulb: Capacitor: Surroundings: Charging + E Discharging + E
M&I 19.7 The current node rule in a capacitor circuit In a steady state, the electron current entering a node in a circuit is equal to the electron current leaving that node.... consequence of the principle of conservation of charge. Also known as the Kirchhoff node rule I 1 I 2 I 3 I 1 = I 2 = (I 3 +I 4 ) I 4 But I 3 need not be equal to I 4 But what about a capacitor? The current node rule cannot apply to one of the plates of the capacitor in isolation, only to the capacitor as a whole, since there is not a steady state situation on each plate. I I No current here!
M&I 19.8 Capacitance As more and more charge congregates on the plates, the potential difference ΔV between the plates increases. Define capacitance: Q C V Plate area A Plate separation s + + + + + E +Q Q Units: C V -1, or farads, F The magnitude of the electric field between the plates is Q A And since V Es s 0 0 Therefore C A for a parallel plate capacitor s E Q A 0
If there is a dielectric (insulating) material between the plates, then C K s 0 A where K is the dielectric constant Material Dielectric constant K Vacuum 1.00000 Air (dry) 1.00059 Silicone oil 2.5 Polystyrene 2.56 Teflon 2.1 Nylon 3.4 Paper 3.7 Pyrex glass 5.6 Water 80
M&I 19.10 A macroscopic analysis of circuits A microscopic view has been very useful to understand some fundamental processes, but it is not easily to measure electric field, surface charge, electron drift speed or mobility. On the other hand, it is easy to measure conventional current (rather than electric current), potential difference (rather than electric field) and resistance (rather than mobility) so, then, a macroscopic view might be useful
M&I 19.11 Resistance Conventional current: I q nav q naue Where q is the absolute value of the charge on each carrier n is the number of charge carriers per m 3 u is the mobility of the charge carriers Grouping the material properties together: I ( q nu) AE I And define J ( q nu) E E A Where J is the current density (A m -2 ) and σ is the conductivity q nu (A V -1 m -1 ) We consider J to be a vector pointing in the direction of the conventional current, which is in the direction of the electric field E + E J E
Conductivity with two kinds of charge carrier Apply a potential difference across a solution of NaCl The flow of Na + ions to the right constitutes a conventional current to the right and the flow of Cl ions to the left also constitutes a conventional current to the right. The total current is I q n Av q n Av 1 1 1 2 2 2 q n Au E q n Au E 1 1 1 2 2 2 Cl E Na + And the current density I J A q n u E q n u E E 1 1 1 2 2 2 Hence the conductivity q1 n1u 1 q2 n2u2
Resistance combines conductivity and geometry Consider the potential difference V E d r ΔV across a length L of wire of constant cross sectional area A and L uniform composition In the steady state, the electric field is uniform everywhere in the wire, hence V EL I Since J E this gives A I V A L V V Rearranging: I L A R L Resistance: R Units: ohms, Ω A a quantity which depends on the properties of the material (σ) and the geometry of the resistor (L and A) A
Summary this should all make sense to you Starting with v ue I q nav q na ue V V EL for a straight wire, so E L V V I q nau L L q nau V L I where R with q nu R A of course V IR but never V IR Why?
Microscopic view Macroscopic view v ue J E i nav naue 1 I q nav V R *** Do the Ex.(ercises) in the M&I book!
Constant and varying conductivity By experiment we can determine that the current through one light bulb is less than twice the current through two light bulbs in series since the mobility u is lower at higher temperatures the conductivity depends on u and hence the resistance R changes for different amounts of current. Ohmic resistors Ohmic materials: the conductivity remains constant, independent of the current through the material. I V R for an ohmic resistor Nichrome wire and carbon resistors are nearly always ohmic, while most metals are only ohmic at low temperatures I for a metal V
Semiconductors Some materials and elements such as silicon and germanium are non-ohmic in these semiconductor materials, the density of mobile electrons n and hence the conductivity σ depends exponentially on the potential difference. in a metal, n is a fixed number (the number of free electrons (or holes) per unit volume). in a semiconductor, at low temperatures there are no free electrons or holes and the material behaves very much like an insulator for small applied fields need a large enough electric field to free up (some) electrons, so n is variable and depends on potential difference. Raising the temperature can also free up electrons. I for a semi conductor V
Other non-ohmic circuit elements Capacitors are not ohmic V Q C for a capacitor + Batteries are not ohmic if you double the current through a battery then the potential difference should not change much at all, in fact will decrease slightly. I V R has limited validity, and should only be applied to resistors
M&I 19.12 Loop equation: emf R I R I R I 0 1 2 3 Series resistance emf = ( R R R ) I R I 1 2 3 equivalent R = R R R equivalent 1 2 3 emf R 1 R 3 R 2 Assuming that: R L A the conductivity σ is independent of the current for that material the magnitude of the sum of all the resistor potential drops is equal to the potential rise across the battery the steady state current I is the same in every element
M&I 19.12 ΔV is the same across each resistor and is numerically equal to the emf of the battery I I1 I2 I3 Parallel resistance emf I I 1 I 2 I 3 R 1 R 2 R 3 emf emf emf I R R R 1 2 3 0 1 1 1 emf I emf R R R R 1 2 3 equivalent 1 1 1 1 R R R R equivalent 1 2 3
M&I 19.13 Work and power in a circuit Moving a small amount of charge q from one location to another in a circuit requires potential energy U e where U e qv If this work is done in time t then the power P U t e qv t But since q I the current in the circuit, t Power P IV In a resistor, V IR then P 2 2 V I R In a capacitor, V q C then the work done W to charge a 2 capacitor from no charge to charge Q : 1 Q W 2 C R
M&I 19.14 Internal resistance s Recall our mechanical battery... The electrons on the motor-driven belt are acted on by both the non-coulomb force exerted by the motor F NC and the Coulomb force exerted by the charges on the plates F C = ee c. F C E C F NC In general, FNC FC because of the finite mobility of the charges...... leads to an average drift speed of the ions in the battery, which in turn is proportional to F F... NC I FNC Then J (force per unit charge) EC A e C
Internal resistance...2 J F E NC NC If C then EC e F e I A If the short length of the battery is s, the electric field is nearly uniform, and the potential difference across the battery is V E s then F s s NC V e A I C work per unit charge, or emf of the battery call s A the internal resistance r int of the battery emf V emf r I r int Write battery int
M&I 19.15 The ammeter A... used to measure current in a circuit Ammeter sign convention: If conventional current flows into the socket marked + then the ammeter indicates a positive current. Ammeters are connected in series and therefore should have as small a resistance as possible so as not to change the current that you are trying to measure. In order to measure large currents, a shunt resistor is placed in parallel with the ammeter to shunt part of the current around the ammeter. A
M&I 19.15 The voltmeter V... used to measure potential difference An ammeter with a resistor in series forms a voltmeter Voltmeter sign convention: If the potential is higher at the socket marked + then the ammeter indicates a positive potential difference. Voltmeters are connected in parallel and therefore should have as large a resistance as possible so that as small a current as possible flows through it.
Two useful examples with meters 1. The ammeter. Say we have a galvanometer which has a full scale defection current of 50 ma and internal resistance 20 Ω, and want to use it as an ammeter to measure a (full scale) current of 5 A, what shunt resistance R S must be connected in parallel? G RS r int 2. The voltmeter. If we now want to use the same galvanometer as a voltmeter with a full scale deflection at 10 volts, what shunt resistance must be connected in series? G r int R S
M&I Quantitative analysis of an RC circuit 19.16 Charging a capacitor through resistor I R Suppose the capacitor is uncharged at t = 0. +Q At some time t > 0, a ε current I is established and Q a charge Q builds up on the capacitor. Q Energy equation for this circuit: Vround trip RI 0 C See RCcircuit.py or Q dq Q RI R C dt C Differential equation Solution: Q( t) C 1e t RC C
Q(t) εc 0.63 εc I(t) ε/r 0.63 ε/r V C (t) ε V R (t) ε τ τ Charging a capacitor through resistor 2 t t t t t RC Q( t) C 1e RC = time constant It () C dq() t dt e R t RC e R t V V IR V R R t (1 e ) V t e C
Discharging a capacitor through resistor Suppose the capacitor has charge Q 0 at t = 0. At some time t > 0, a current I is established as the charge drains off the capacitor. Energy equation for this circuit: Q Vround trip RI 0 C or Q dq Q 0 RI R C dt C dq 1 Write dt 1 giving ln Q t K Q RC RC I ε R +Q Q C or Q( t) Q0e RC t K ln Q 0
Q(t) Q 0 0.63 Q 0 I(t) e Q 0 /RC Q 0 /RC V C (t) Q 0 /C τ Discharging a capacitor through resistor 2 t t Q( t) 0 t Q e RC = time constant dq() t It () dt Q e RC V C Q C Q t 0 RC 0 Q 0 C e t RC e t V R (t) Q 0 /C t t t 0 VR IR e t Q 0 e C RQ RC
1 2 3 4 5 Which graph shows current versus time while discharging?
1 2 3 4 5 The capacitor is initially charged (left plate negative) When the circuit is connected and there is a current: 1. Charge on the plates stays constant. 2. Left plate gets less negative. 3. Left plate gets more negative.
1 2 3 4 5 Which graph shows current versus time while charging?
1 2 3 4 5 The capacitor is initially uncharged. When the circuit is connected: 1. Electrons jump across the gap between the plates. 2. Electrons pile up on the left plate. 3. Electrons pile up on the right plate.
1 2 3 4 5 Which graph shows magnitude of charge on the right plate versus time while charging?
1 2 3 4 5 The capacitors are initially uncharged. Q A s E fringe 2R 0 After 0.01 s of charging: 1. The fringe field of each capacitor is the same 2. The smaller capacitor (#1) has a larger fringe field 3. The larger capacitor (#2) has a larger fringe field
1 2 3 4 5 The capacitor is initially uncharged. Which graph shows the magnitude of the fringe field of the capacitor at location A while charging?
1 2 3 4 5 The capacitor is initially uncharged. Which graph shows the magnitude of the net field at location A while charging?
Circuit problems 1 What is the current through each resistor in the circuit below?
Circuit problems 2 What is the current through each resistor in the circuit below?
Circuit problems 2
M&I Chapter 20 Magnetic Force Remember the magnetic field of a moving charge... Biot-Savart law: B ˆ 0 qvr 2 4 r for a point charge and B 0 I 4 lrˆ 2 r for a small current element
M&I 20.1 Magnetic force on a moving charge Magnetic force on a moving charge: Fmagnetic qv B where: q is the charge (including the sign) of the moving charge v is the velocity of the moving charge B is the applied magnetic field (in tesla) F magnetic... does not change the speed of the charge, only its direction, since F B magnetic Magnitude of : F magnetic e v v B F ( e) vb B out the page F magnetic dp dt qvbsin for v c
F qv B magnetic A charge follows a circular path when traveling in a magnetic field... v e B F ev B Recall that for any rotating vector X : Momentum principle: dx dt X The momentum p of the particle above is a rotating vector... See helix_in_b.py dp dt p q vbsin90 mv 1 v c 2 2 qb 1 v m q vb c 2 2
Circular motion at low speeds: qb if v c m... can also get this result starting from the centripetal 2 acceleration of the particle: v R 2 Then: mv q vb R v qb and since : R m Time T for one complete circular orbit: 2 m since, T 2 T qb Momentum p: dp v since p p q vb dt R... momentum: p q BR... valid even for relativistic speeds.
The magnetic force on a charge moving in a magnetic field For each situation below, decide if the magnetic force on the charge is zero. If not zero, then indicate the direction of the force. (a) + v B (b) + v B (c) - v B (d) - v into paper B
The magnetic force on a charge moving in a magnetic field 2 (e) (f) B - v 0 B + v (g) (h).... + B into paper. v... - B out of paper....
1 2 3 4 5 What is the direction of the magnetic force on the proton? 1. +x 2. x 3. +y 4. y 5. +z 6. z 7. zero magnitude
1 2 3 4 5 An electron is traveling in the y direction. At its location there is a magnetic field in the z direction. What is the direction of the magnetic force on the electron? 1. +x 2. x 3. +y 4. y 5. +z 6. z 7. zero magnitude
Applications: The cyclotron E = 0 in dees B out the page in dees E accelerates charged particles across the gap We need to flip the electric field rapidly in the gap in order to synchronize the acceleration of the particles.
ithemba LABS Faure, South Africa 200 MeV separated sector cyclotron See http://www.tlabs.ac.za
M&I 20.2 Magnetic force on a current-carrying wire Consider a bunch of positive charges contained in a small volume with length Δl and cross sectional area A, moving with drift velocity v. If there are n moving charges per unit volume, then there are nal moving charges in this small volume. l B v Then the force on this wire due to an external magnetic field is Fmagnetic ( nal) qvb ( qnav ) l B B F Il B magnetic
Magnetic force on a current-carrying wire 2 Consider this situation F Il B magnetic All the way around the loop, F magnetic points outwards. The resultant force on the side pieces is zero (although the bottom wire is stretched!?) I + - h L B out the page The magnitude of F magnetic F magnetic IlBsin ILBsin90 ILB on the bottom wire is with the direction of F magnetic downwards.
Magnetic force on a current-carrying wire 2 Why does the wire move at all if the magnetic force only acts on the drifting electrons?? The answer relies on the Hall effect excess charge builds up on the top and bottom surfaces on the wire and an electric field builds up until E vb and the resultant force on the drifting electrons is zero. E E However also acts on the stationary positive atomic cores which experience a net downwards force I E B v
Magnetic force between two parallel wires d F 21 L I 1 Consider two very long parallel straight wires of length L, a distance d apart. The upper wire carries conventional current I 1 and the lower wire conventional current I 2 in the same direction. No external magnetic fields present. 0 2I1 Magnetic field at bottom wire due to top wire: B1 4 d 0 2I1 Then force on lower wire due to B 1 : F21 I2LB1 sin 90 I2L 4 d in a direction upwards. Similarly the force on the upper wire due to B 2 from lower wire: 0 2LI1I2 F 21 in a direction downwards 4 d B 1 If I 1 and I 2 were in opposite directions, then the two wires will repel each other. I 2
1 2 3 4 5 What is the direction of the magnetic force on the wire? 1. +x 2. x 3. +y 4. y 5. +z 6. z 7. zero magnitude
1 2 3 4 5 A proton moving in the +y direction experiences a magnetic force in the x direction. What is the direction of the conventional current in the wire? 1. +y 2. y 3. I = 0 4. Not enough information
M&I 20.3 The Lorentz force If a charge moves in region where there are both electric and magnetic fields present, then the net force acting on the particle is given by F F F resultant electric magnetic F qe qv B Lorentz force In each of the situations below, what is the direction of the Lorentz force on a proton at this instant? [and for an electron?] y B v 0 y B v y B v y B z E x z E x z E x v E z x v in z direction
1 2 3 4 5 What are the directions of the forces on the moving positive charge? F elec F mag 1 up up 2 down down 3 up down 4 down up
1 2 3 4 5 What are the directions of the forces on the moving negative charge? F elec F mag 1 up up 2 down down 3 up down 4 down up
1 2 3 4 5 A particle of charge +q travels in a straight line at constant speed. Which is true? 1. qe = qvb 2. E = B 3. E = q / (vb) 4. qe = qb
E = 1000 V/m, B = 0.5 T, q = 2 nc A particle travels in a straight line at constant speed. What is the speed v of the particle? 1 2 3 4 5 1. 2000 m/s 2. 0.5 m/s 3. 500 m/s 4. 1000 m/s 5. 1.0 m/s
Applications: The velocity selector A positively charged particle moves through a hole in a plate into a region with an electric field and a magnetic field. The two fields and the velocity of the particle are all mutually perpendicular. B into paper + v E 1. Draw arrows to indicate the directions of the electric and magnetic forces acting on the particle. (Are there any other forces acting on the particle?)
The velocity selector 2 2. State in words the conditions that are necessary for the particle to move straight ahead at a constant velocity. 3. Say now that the particle entered the region travelling at 2v. What would now be its trajectory between the plates? Explain in terms of the forces acting on the particle. 4. Say now that the particle entered the region travelling at half of v. What would now be its trajectory between the plates? Explain in terms of the forces acting on the particle.
Worked example: the electron and current loop I v 3R +Q If the force on the moving electron is zero, then what is the magnitude and direction of the current I? Fmag Felectric on electron evb B 1 eq 4 3 0 0 R 2 1 Q 4 9R v 2
The electron and current loop 2 The magnetic field of a loop is At the centre of the 2 loops z = 0 Hence for two loops: Direction of I? or 2 4 I 2 I R B loop 0 = 4 0 0 1 Q 36 Rv 0 0 Since F vb mag is to the right, I must be anticlockwise around the loops. Bloop = 4 1 Q 2 4 9R v 2 IR 2 2 2 3 2 R z 0 2 I R
M&I 20.4 The Hall effect I Consider a metal block through which there is a conventional current I the free electrons will tend to drift in the opposite direction to E with speed v. E B v Now apply a strong external magnetic field B perpendicular to E the magnetic force on the electrons will cause them to be deflected towards the bottom of the block, where they will accumulate on the surface the top surface of the block becomes positive, and the bottom surface negative.
The Hall effect 2 more and more electrons pile up on the bottom surface of the block a new electric field E I builds up between the top and bottom surfaces and pile up of electrons will cease when Fmagnetic Felectric (the net vertical force is zero) and the electrons move horizontally through the bar with no vertical deflection. When this happens: ee evbsin90 or E vb (similar to the velocity selector) E The voltmeter will read a value equal to E h vbh (The voltmeter will read zero when B = 0) L B v d h
The Hall effect 3 The appearance of a sideways or transverse potential difference across a current-carrying wire in the presence of a magnetic field is called the Hall effect. If we had positive particles moving to the right in our conductor, instead of electrons moving to the left, then the bottom of the block would have an excess positive charge with excess negative charge on the top... and the voltmeter would read a negative voltage! therefore by measuring the Hall effect for a particular metal, we can determine the sign of the moving particles that make up the current.
The Hall effect 4 In a metal bar of length L, height h and width w, carrying a conventional current I q nav : Then: V Hall I VHall vbh Bh A q n So can be measured for different materials, and can be used to confirm that in some metals (e.g. aluminium and zinc) and p-type semiconductors, it is positive holes in the electron sea, rather than the electrons themselves that are the dominant charge carriers. V Hall Turn it around! if is known for a particular material, together with I, A, q, n and h (easy), then we can measure an unknown B ( build an instrument to measure magnetic fields) a Hall probe
What is the direction of E inside the bar? 1 2 3 4 5 1. +x 2. x 3. +y 4. y 5. +z 6. z 7. zero magnitude
1 2 3 4 5 If the mobile charges are negative, what is their direction of motion inside the bar? 1. +x 2. x 3. +y 4. y 5. +z 6. z 7. zero magnitude
1 2 3 4 5 If the mobile charges are negative, then what is the sign of the voltmeter reading? (Voltmeter reads positive if the + lead is connected to a higher potential location) 1. positive 2. negative 3. zero
1 2 3 4 5 If the mobile charges are positive, what is the direction of the magnetic force? 1. +x 2. x 3. +y 4. y 5. +z 6. z 7. zero magnitude
1 2 3 4 5 If the mobile charges are positive, then what is the sign of the voltmeter reading? (Voltmeter reads positive if the + lead is connected to a higher potential location) 1. positive 2. negative 3. zero
M&I Motional emf: Currents due to magnetic forces. 20.5 If we move a wire through a magnetic field, then can we generate a current? Consider a metal bar of length L that is moving through region of uniform magnetic field B, into the page... The mobile electrons inside the bar experience a force Fmagnetic ( e) vb downward with the result that the mobile electron sea shifts downwards and the bar becomes polarized electric field E pol inside the bar. L + + E pol + + + + v v B B
Motional emf: Currents due to magnetic forces 2 At first the only force is Fmagnetic evb downwards, but as more electrons shift downwards, the larger the upward electric force F ( e) E electric Polarization continues until Fmagnetic Felectric pol or evb ee or vb E (no net force on the mobile electrons) Think! we have a non-zero electric field inside a metal in static equilibrium but there is no current! The potential difference between the ends of the bar is simply V EL ebl Since there is no current in the bar, there is no net force on the bar due to B and the polarized bar will coast along with no force required to keep it going.
Motional emf: Currents due to magnetic forces 3 Now let the bar run along frictionless metal rails through the uniform magnetic field, and connect a resistor R across the rails (the bar and rails have negligible resistance.) A current will run in this circuit, driven by a battery (the moving metal bar) electrons are driven the wrong way, moving towards the negative end of the bar. The surface-charge distribution on the circuit looks something like R R + + + + + + + + + + + + + + + + + + + + + + + + + v v B B
The electron current in the resistor is continuously depleting the charges on the ends of the moving bar so the electric field inside the bar is always slightly less than is needed to balance the magnetic force i.e. ee Motional emf: Currents due to magnetic forces 4 evb so the electrons move towards the negative end of the bar. F NC The magnetic force is a non-coulomb force the work done by this force in moving an electron from one end of the bar to the other is W FNCL evbl evbl Then the non-coulomb work per unit charge = e If the resistance of the bar is negligible, then V emf vbl motional emf An external force ILB needs to applied on the bar to the right to keep it moving with constant speed v.
1 2 3 4 5 A neutral iron bar is dragged at speed v through a region with magnetic field B. Which diagram best shows the state of the bar?
1 2 3 4 5 A neutral copper bar is dragged at speed v through a region with magnetic field B. Which diagram best shows the state of the bar?
1 2 3 4 5 In which direction will there be a conventional current? 1. Clockwise 2. Counter-clockwise 3. No current
1 2 3 4 5 There is a current as shown. Is there a magnetic force on the entire bar? 1. F mag on bar is upward 2. F mag on bar is downward 3. F mag on bar is to the right 4. F mag on bar is to the left 5. F mag on bar = 0
1 2 3 4 5 Four identical horizontal bars each start from rest and fall vertically. The bars slide down vertical, frictionless rods that are connected at the top by a resistor. In what order do the bars reach the bottom of the vertical rods? 1 2 3 4................ B B = 0 B into paper B out of paper
M&I 20.6 Magnetic forces in moving reference frames Consider two protons initially traveling parallel to each other with the same speed v, a distance r apart... Proton 1 contributes an electric field E 1 at the position of proton 2 which experiences an electric force F21, e q2e1 2 1 e or F21, e 2 downwards 4 r 0 r proton 2 proton 1 E 1 ˆr v F 21,m B 1 v F 21,e Proton 1 also contributes a magnetic field at the position of proton 2 where 0 ev B into the page. 1 2 4 r Hence there is a magnetic force F 21,m on proton 2, where 2 2 0 ev F21, m upwards 2 4 r B 1
Magnetic forces in moving reference frames...2 The resultant force on proton 2 due to proton 1 is F21, e F21, Taking the ratio of these two forces: But 1 1 0 0 ev F r v F 2 2 0 2 21, m 2 4 2 0 0 21, e 1 e 2 4 0 r 8.85 10-12 C 2 N m -2 4π 10-7 T m A -1 16 2-2 2 8.99 10 m s c!!! m F F 21, m 21, e v c 2 2 So as v c, F21, m F21, e
Magnetic forces in moving reference frames...3 Now considering the full Lorentz force of proton 1 on proton 2: 1 e e v F F downward 2 2 2 0 21, e 21, m 2 2 4 0 r 4 r The resultant force on proton 2 is downward... and the resultant Lorentz force on proton 1 due to proton 2 is upward. The two protons repel each other and their trajectories are curving line apart. Read Jack and Jill and Einstein in M&I. F 12,resultant F 21,resultant v v
Relativistic field transformations M&I 20.7 Consider two frames of reference: y x z Lab frame y x z Moving frame v (moving at speed v relative to lab frame) How do electric and magnetic fields transform from one reference frame to the other?. Need special relativity ' x x E E ' x x B B ' 2 2 1 y z y E vb E v c ' 2 2 1 z y z E vb E v c 2 ' 2 2 1 y z y v B E c B v c 2 ' 2 2 1 z y z v B E c B v c
Relativistic field transformations 2 Now consider a positively charged particle at rest in the lab frame. 1 q In the lab frame: E ˆ 2 r B 0 4 r 0 In the moving reference frame the particle is seen to be moving with speed v in the x direction In the moving frame above the particle, E y > 0, therefore: B ' z At low speeds, v c : v v Bz E 2 y E 2 y c c 1v c 1v c 2 2 2 2 B ' z v v 1 q E 2 y 2 2 c c 4 0 r
In the moving frame: Relativistic field transformations 3 v v ' 1 q ' ' Above the moving particle Bz E 2 y 2 2, By Bx 0 c c 4 0 r 1 2 and since c ' 0 qv : Bz 2 as predicted by Biot-Savart 0 0 4 r ' v ' ' Similarly, below the particle: Bz E 2 y, By Bx 0 c v z y B x ' v At this position: By E 2 z c, ' v and on the other side: By E 2 z c, B ' x B ' z ' ' Bx Bz magnetic fields are a relativistic consequence of electric fields 0 0
Electric field of a rapidly moving charged particle If the speed of the moving reference frame is very large (approaching c) then the electric field of the particle is altered significantly in the moving frame. If B z = 0: y E ' x E x v + ' E E E ' z ' y E y 1 v E z 1 v c 2 2 c 2 2 ' E y and z Moving frame E increase as v c ' z x
Moving through a region of uniform magnetic field Now consider a case where Bx 0 B 0 B y z B in a rest frame: y B In the moving frame: z rest frame x E ' y 0 vbz v( B) 1v c 1v c 2 2 2 2 vb if v << c y ' B ' E v z moving frame x
Moving through a region of uniform magnetic field 2 If we are in the moving frame and observe a metal bar which is at rest in this frame the metal bar polarizes and a downward ' field E pol is produced, until the net field inside the metal is zero (there is static equilibrium and E vb ). ' pol In the lab frame the bar is observed moving at v in the x-direction, and the bar polarizes due to the magnetic force qvb acting on the mobile charges At equilibrium qepol qvb or E vb pol z z y y ' B ' E ' E pol + + + + + + moving frame B E lab frame + + + + + + pol x x v
Moving through a region of uniform magnetic field 3 The amount of charge buildup is the same in either reference frame, but the explanation that observers in the two frames give for the cause of the charge buildup is different. Principle of relativity: There may be different mechanisms for different observers in different reference frames, but all observers can correctly predict what will happen in their own frames, using the same relativistically correct physical laws. Another example: the velocity selector
Moving through a velocity selector z y E v B x lab frame A particle moves with speed v in the +x direction through a region where there is an electric field of magnitude E in the +y direction and a magnetic field of magnitude B in the +z direction. If E = vb then the particle passes through with no deflection because the electric force of magnitude ee is equal and opposite to the magnetic force of magnitude qvb. If you move through this region with speed v in the +x direction, without the charged particle being present, what electric and magnetic fields would you observe in your (moving) frame?
Moving through a velocity selector 2 The only fields that would be observed in the moving frame are: E ' y Ey vbz E vb vb vb 2 2 2 2 2 2 1v c 1v c 1v c 0 B ' z v v v c c c 1 v c 1 v c 1 v c 1 v c Bz E 2 y B E B vb 2 2 2 2 1 v c B 2 2 2 2 2 2 2 2 A charge at rest is unaffected by the magnetic field and there is no electric field so there is no force on the particle. A particle at rest in the moving frame moves with constant velocity in the rest frame, straight through the velocity selector. a different explanation for each frame for the same phenomenon, each of which makes sense in that frame.
M&I 20.8 Magnetic torque on a magnetic dipole moment Consider a rectangular current-carrying loop of wire in a uniform magnetic field B, free to rotate on an insulating horizontal axle... h w B I... A force F IwB acts on two sides of the loop, causing it to rotate such that its magnetic dipole moment μ lines up with the external magnetic field B. F IwB B I out IA Ihw I in F IwB
Magnetic torque on a magnetic dipole moment...2 F IwB I h F IwBsin B 2 h 2 I F F IwBsin IwB h 2 Total torque on the loop = 2rF sin 2 F hiwbsin Since IA Ihw write torque = B sin or τ μb (for any shaped loop)
M&I 20.9 Potential energy for a magnetic dipole moment When a magnetic dipole moment rotates freely to align with an applied magnetic field, the aligned magnetic dipole moment must be associated with a lower potential energy in the magnetic field. How much work will be done (by us) to move the magnetic dipole moment out of alignment? Work done in rotating a rectangular loop from i to f... r f f f h W U m Fdr 2IwBsin d IwhB sind 2 r i i i U ( IwhB)[cos cos ] m f i write Um Bcos μ B
Force on a magnetic dipole moment Bring one end of a bar magnet near to a current-carrying loop of radius R the magnetic field due to the bar magnet is diverging at some angle θ and is therefore not uniform. Each segment of the ring experiences a magnetic force df Idl B df cos(90 ) B S N 2 df cos(90 ) df df B B I( R ) The horizontal components of the force is df cos(90 ) df sin The vertical components of the forces cancel. F IBsin dl IB(2 R)sin net (2Bsin ) R
Force on a magnetic dipole moment 2 μ S N F mag F by us B 1 B 2 Take a situation where a magnetic dipole moment is aligned with the magnetic field from a bar magnet the magnetic exerts a force F mag to the left and we exert a force F by us which is slightly larger than F mag and we move the magnetic dipole moment from position 1 to 2. Since Um μbb (in this case) then the work done by us = x F x U B B B B B by us m 2 1 2 1 which is greater than zero since B2 B1
Force on a magnetic dipole moment 3 U m B db Fby us a positive value. x x dx du (In general, Fx ) dx db Note that there is no force if the field is uniform 0 dx 0 21 The magnetic field along the axis of a bar magnet: B1 3 4 x Where 1 is the total magnetic dipole moment of all the atomic magnetic dipole moments in the magnet The force that this magnet will exert on a magnetic dipole moment lying along the axis of F 2 1 0 1 0 1 mag 2 2 3 2 4 1 db d 2 2 3 dx dx 4 x 4 x
The magnetic torque on a current-loop dipole What is the direction of the torque on each of the loops below? B (a) (b) (c) (d)
The magnetic torque on a current-loop dipole 2 What is the direction of the torque on each of the loops below? Which way would the magnetic torque tend to rotate the loop? I B B I I (a) The plane of the circular loop is perpendicular to the page and the magnetic field is in the plane of the paper. (b) The plane of the circular loop is perpendicular to the page and the magnetic field is in the plane of the paper. (c) The plane of the circular loop and the magnetic field are both perpendicular to the plane of the paper.
The magnetic torque on a current-loop dipole 3 What is the direction of the torque on each of the loops below? Which way would the magnetic torque tend to rotate the loop? I B I I (d) The circular loop is in the plane of the paper and the magnetic field is into the paper. (e) The circular loop and the magnetic field are both in the plane of the paper. (f) The plane of the circular loop is perpendicular to the paper and the magnetic field is in the plane of the paper.
M&I 20.10 The electric generator Mechanically rotate a loop of dimensions w and h at a constant angular speed d dt (in radians per second) inside a uniform magnetic field. The tangential speed of the left or right wire is v h 2, since the wire is a distance h/2 from the axle. The magnetic forces drive a conventional current clockwise in the loop Magnetic force on each current carrier: F qvbsin h w v B B B v
The electric generator 2 For the right wire, the magnetic force is uniform throughout the wire, so the motional emf (non-coulomb work per unit charge) is given by qvbsin w vbwsin q The left wire contributes the same emf, so the total emf around the loop is: 2vBwsin 2 h 2Bw sin t B( hw)sint BAsint Connect the rotating loop to an external circuit using two commutators. Then the current, I BAsint R I I R I I
The electric generator 3 For our generator of N loops: emf NBAsint Which produces an alternating current in the external circuit this is the basis of nearly all commercial electricity generation emf NBA t In South Africa, frequency of AC is 50 Hz and root mean square voltage is 220V.
The electric generator 4 What power needed to rotate the loop? An external force Fext IwBsin is required at both ends of the loop, perpendicular to the loop, to balance the perpendicular component F IwBsin of the magnetic force of the loop. F IwBsin F ext I F B h 2 IwB F ext I F IwBsin
The electric generator 5 If we rotate the loop through a small angle, we move each end a small distance h 2 and do a small amount of work h W 2IwBsin 2 Divide by t and take the limit as t 0 dw dt h d 2 sin sin sin 2 dt IwB I Bwh I BA and since I BAsint R Power = dw I ( RI ) RI dt 2