Physics PHY101 Handout Electrostatics Summary PDF

PHYSICS –PHY101 VU Summary of Lecture 23 – ELECTROSTATICS II 1. In the last lecture we learned how to calculate the electric field if there are any number of point charges. But how to calcula...

PHYSICS –PHY101 VU Summary of Lecture 23 – ELECTROSTATICS II 1. In the last lecture we learned how to calculate the electric field if there are any number of point charges. But how to calculate this when charges are continuously distributed over some region of space? For this, we need to break up the region into little pieces so that     each piece is small enough to be like a point charge. So, E  E1  E2  E3    , or    E   Ei is the total electric field. Remember that E is a vector that can be resolved  into components, E  Exiˆ  E y ˆj  Ez kˆ. In the limit where the pieces are small enough,   we can write it as an integral, E   dE (or Ex   dEx , E y   dE y , Ez   dEz ) 2. Charge Density: when the charges are continuously distributed over a region - a line, the surface of a material, or inside a sphere - we must specify the charge density. Depending upon how many dimensions the region has, we define: (a) For linear charge distribution: dq   ds (b) For surface charge distribution: dq   dA (c) For volume charge distribution: dq   dV The dimensions of  ,  ,  are determined from the above definitions. 3. As an example of how we work out the electric field coming from a continuous charge distribution, let us work out the electric field from a uniform ring of charge at the point P.  dE  dE cos  P  z y R  ds x The small amount of charge  ds gives rise to an electric field whose magnitude is 1  ds  ds dE   4 0 r 2  4 0 z 2  R 2  z z The component in the z direction is dEz  dE cos with cos  .   1/ 2 r z2  R2 © Copyright Virtual University of Pakistan 65 PHYSICS –PHY101 VU z ds So dEz . Since s, which is the arc length, does not depend upon z or R,   3/ 2 4 0 z 2  R 2 z z  2 R  qz Ez   ds  4 . Answer!!   2 3/ 2   2 3/ 2   3/ 2 4 0 z 2  R 0 z2  R 4 0 z 2  R 2 1 q Note that if you are very far away, the ring looks like a point: Ez  ,  z  R . 4 0 z 2 4. As another example, consider a continuous distribution of charges along a wire that lies along the z -axis, as shown below. We want to know the electric field at a distance x from the wire. By symmetry, the only non-cancelling component lies along the y -axis.  z dE z  dE y  y   z P dE y r dz dq x Applying Coulomb's law to the small amount of charge  dz along the z axis gives, 1 dq 1  dz dE   4 0 r 2 4 0 y  z 2 2 the component along the y direction is dE y  dE cos. Integrating this gives, z  z  z   dz E y   dE   cos dE  2  cos dE   cos. z  z 0 2 0 z 0 y  z2 2 The rest is just technical: to solve the integral, put z  y tan   dz  y sec 2  d. And so,    / 2  E  2 0 y   0 cos d  2 0 y. Now, we could have equally well taken the x axis. The only thing that matters is the distance from the wire, and so the answer is better written as:  E. 2 0 r 5. The flux of any vector field is a particularly important concept. It is the measure of the "flow" or penetration of the field vectors through an imaginary fixed surface. So, if there is a uniform electric field that is normal to a surface of area A, the flux is   EA. More generally, for any surface, we divide the surface up into little pieces and take the © Copyright Virtual University of Pakistan 66 PHYSICS –PHY101 VU   component of the electric field normal to each little piece,  E   Ei  Ai. If the pieces   are made small enough, then in this limit,    E  dA. 5. Let us apply the above concept of flux to calculate the flux leaving a sphere which has a charge at its centre. The electric field at any point on the sphere has magnitude equal 1 q to and it is directed radially outwards. Let us now divide up the surface of the 4 0 r 2 sphere into small areas. Then    E A  E  A  1 q  4 0 r 2  4 r 2. So we end up with q the important result that the flux leaving this closed surface is  . 0 6. Gauss's Law: the total electric flux leaving a closed surface is equal to the charge enclosed by the surface divided by  0. We can express this directly in terms of the   q mathematics we have learned,    E  dA  enclosed. Actually, we have already seen 0 why this law is equivalent to Coulomb's Law in point 5 above, but let's see it again. So,   applying Gauss's Law to a sphere containing charge,  0  E  dA   0  EdA  qenclosed. If the surface is a sphere, then E is constant on the surface and  0 E  dA  q and from this  0 E  4 r 2   q  E  1 q. This is Coulomb's Law again, but the power of Gauss's 4 0 r 2 law is that it holds for any shape of the (closed) surface and for any distribution of charge. 7. Let us apply Gauss's Law to a hollow sphere that has charges only on the surface. At any   distance r from the centre, Gauss's Law is  0 E 4 r 2  qenclosed. Now, if we are inside the sphere then qenclosed  0 and there is no electric field. But if we are outside, then the total 1 q charge is qenclosed  q and E  , which is as if all the charge was concentrated at the 4 0 r 2 centre. 8. Unfortunately, it will not be possible for me to prove Gauss's Law in the short amount of time and space available bu t the general method can be outlined as follows: take any volume and divide it up into little cubes. Each little cube may contain some small amount of charge. Then show that for each little cube, Gauss's Law follows from Coulomb's Law. Finally, add up the results. For details, consult any good book on electromagnetism. © Copyright Virtual University of Pakistan 67 PHYSICS –PHY101 VU Summary of Lecture 24 – ELECTRIC POTENTIAL ENERGY 1. You are already familiar with the concept of gravitational potential energy. When you lift a weight, you have to do work against the downwards pull of the Earth. That work   is stored as potential energy. Suppose a force F acts on something and displaces it by ds.   Then the work done is F  ds. The work done in going from point a to point b (call it Wab ) b   is then got by adding together the little bits of work, Wab =  F  ds. The change in a potential energy is defined as U  U b  U a  Wab. Remember always that we can only define the potential U at a point if the force is conservative. 2. The electrostatic force is conservative and can be represented by a potential. Let us see how to calculate the potential. So consider two charges separated by a distance as below. q q 1 r Let us take the point a very far from the fixed charge q, and the unit charge at the point b to be at a distance R from q. Then the work you did in bringing the unit charge from infinity 1 1 R R 1 q 1 q 1 to R is, Wab   ( qE )dr    dr 2   q    . Since the charges  4 0  r 4 0  R   4 0 R repel each other, it is clear that you had to do work in pushing the two charges closer together. So where did the negative sign come from? Answer: the force you exert on the unit charge is directed towards the charge q, i.e. is in the negative direction. This is why   q 1 F  ds  (qE )dr. Now U  U ( R) - U () . If we take the potential at  to be 4 0 R q 1 zero, then the electric potential due to a charge q at the point r is U (r ) . 4 0 r dU Remember that we know how to calculate the force given the potential: F  . Apply dr q 1 this here and you see that F  , (which also has the correct repulsive sign). 4 0 r 2 3. From the above, it is quite obvious that the potential energy of two charges q1 , q2 is, 1 q1q2 mm U r  . Compare this with the formula for gravitational energy, U  r   G 1 2. 4 0 r r What is the difference? From here, you can see that the gravitational force is always negative (which means attractive), whereas the electrostatic force can be both attractive © Copyright Virtual University of Pakistan 68 PHYSICS –PHY101 VU or repulsive because we have both + and - charges in nature. 3. The electric potential (or simply potential) is the energy of a unit charge in an electric field. Joule So, in our MKS units, the unit of potential is 1  1 Volt. Another useful unit is Coulomb "electron volt" or eV. The definition is: One electron - volt = energy gained by moving one electron charge through one Volt    1.6  1019 C  1V  1.6  10 19 J It is useful to note that 1 Kev = 103eV (kilo-electron-volt) 1 Mev = 106eV (million-electron-volt) 1 Gev = 109eV (giga-electron-volt) 1 Tev = 1012eV (tera-electron-volt) 4. Every system seeks to minimize its potential energy (that is why a stone falls down!). So, positive charges accelerate toward regions of lower potential, but negative charges accelerate toward regions of higher potential. Note that only the potential difference matters - even if a charge is placed in a region where there is a high potential, it will not want to move unless there is some other place where the potential is higher/lower. 5. Given a system of charges, we can always compute the force - and hence the potential - that arises from them. Here are some important general statements: a)Potentials are more positive in regions which have more positive charge. b)The electric potential is a scalar quantity (a scalar field, actually). dU c)The electric potential determines the force through F   , and hence the electric dr field because F  qE. d)The electric potential exists only because the electrostatic force is conservative. 6. To compute the potential at a point, the potentials arising from charges 1, 2,    N must be added up: N 1 N qi V  V1  V2     VN  Vi  . Here ri is q1 i 1 4 0 i 1 ri the distance of the i'th charge from the point where r12 q2 the potential is being calculated or measured. As an r13 example, the potential from the three charges is: r23 1 q1q2 1 q1q3 1 q2 q3 Vr    . 4 0 r12 4 0 r13 4 0 r23 q3 © Copyright Virtual University of Pakistan 69 PHYSICS –PHY101 VU 7. Let us apply these concepts to the dipole system considered earlier. With two charges, 1  q q  q r2  r1 VP  V1  V2    . We are particularly interested in the situation 4 0  r1 r2  4 0 r1r2 where r  d. from the diagram you can see that r2  r1  d cos and that r1r2  r 2. Hence, q d cos 1 p cos V . So we have calculated the potential at any  with such 4 0 r 2 4 0 r 2  little difficulty. Note that V  0 at  . 2 q r1 P  d r r2 q r2  r1  d cos 8. Now let us calculate the potential which comes from charges that are uniformly spread over a ring. This is the same problem as in the previous P lecture, but simpler. Give the small amount of potential coming from the small amount of charge  1 dq z dq   ds some name, dV . Then obviously 4 0 r 2 y 1 dq 1 q R V   dV   . 4 0 r 2 4 0 R2  z2 ds x © Copyright Virtual University of Pakistan 70 PHYSICS –PHY101 VU Summary of Lecture 25 – CAPACITORS AND CURRENTS 1. Two conductors isolated from one another and from their surroundings, form a capacitor. These conductors may be of any shape and size, and at any distance from each other. If a potential difference is created between the conductors (say, by connecting the terminals of a battery to them), then there is an electric field in the space between them. The electric field comes from the charges that have been pushed to the plates by the battery. The amount of charge pushed on to the conductors is proportional to the potential difference between the battery terminals (which is the same as between the capacitor plates). Hence, Q  V. To convert this into an equality, we write Q  CV. This provides the definition of Q capacitance, C . V 2. Using the above definition, let us calculate the capacitance of two parallel plates separated by a distance d as in the figure below.    q      + + + + + + Gaussian surface d          q   q Recall Gauss's Law:    E  dA  enclosed. Draw any Gaussian surface. Since the electric 0 Q field is zero above the top plate, the flux through the area A of the plate is   EA  , 0 Q where Q is the total charge on the plate. Thus, E  is the electric field in the gap 0 A E Q  A between the plates. The potential difference is V  , and so C   0. You can see d V d that the capacitance will be large if the plates are close to each other, and if the plates have a large area. We have simplified the calculation here by assuming that the electric field is strictly directed downwards. This is only true if the plates are infinitely long. But we can usually neglect the side effects. Note that any arrangement with two plates forms a capacitor: plane, cylindrical, spherical, etc. The capacitance depends upon the geometry, the size of plates and the gap between them. © Copyright Virtual University of Pakistan 71 PHYSICS –PHY101 VU 3. One can take two (or more) capacitors in various ways and thus change the amount of charge they can contain. Consider first two capacitors connected in parallel with each other. The same voltage exists across both. For each capacitor, q1  C1V , q2  C2V where V is the potential between terminals a and b. The total charge is: Q  q1  q2  C1V  C2V  (C1  C2 )V Q Now, let us define an "effective" or "equivalent" capacitance as Ceq . Then we can V immediately see that for 2 capacitors Ceq  C1  C2 , and Ceq   Cn  for n capacitors . C1 Ceq V C2 4. We can repeat the analysis above when the capacitors are put in series. Here the difference is that now we must start with V  V1  V2 , where V1 and V2 are the voltages across the two. Clearly the same charge had to cross both the capacitors. Hence, Q Q 1 1 V  V1  V2    Q(  ). C1 C2 C1 C2 Q 1 1 1 From our definition, Ceq  , it follows that  . The total capacitance is now V Ceq C1 C2 1 1 less than if they were in parallel. In general,   for n capacitors . Ceq Cn V1 V2 Ceq C1 C2 V V 5. When a battery is connected to a capacitor, positive and negative charges appear on the opposite plates. Some energy has been transferred from the battery to the capacitor, and now been stored in it. When the capacitor is discharged, the energy is recovered. Now let us calculate the energy required to charge a capacitor from zero to V volts. Begin: the amount of energy required to transfer a small charge dq to the plates is dU  vdq, where v is the voltage at a time when the charge is q  Cv. As time goes on, the total charge increases until it reaches the final charge Q (at which point the voltage becomes V ). So, © Copyright Virtual University of Pakistan 72 PHYSICS –PHY101 VU Q q q Q2 1 dU  vdq  dq  U   dU   dq   CV 2. C 0 C 2C 2 But where in the capacitor is the energy stored? Answer, it is present in the electric field in the volume between the two plates. We can calculate the energy density: 1 CV 2  V 2 1 energy stored in capacitor U   u   2  0     0 E 2. volume of capacitor Ad Ad 2 d  2  A In the above we have used C  0 , derived earlier. The important result here is that d u  E. Turning it around, wherever there is an electric field, there is energy available. 2 1 Q 6. Dielectrics. Consider a free charge  Q. Around it is an electric field, E . 4 0 r r 2 Now suppose this charge is placed among water molecules. These molecules will polarise, i.e. the centre of positive charge and centre of negative charge will be slightly displaced. The negative part of the water molecule will be attracted toward the positive charge  Q. 1 1 Q So, in effect, the electric field is weakened by and becomes,. Here I have r 4 0 r r 2 introduced a new quantity  r called "dielectric constant". This is a number that is usually bigger than one and measures the strength of the polarization induced in the material. For air,  r  1.0003 while  r  80 for pure water. The effect of a dielectric is to increase the capacitance of a capacitor: if the air between the plates of a capacitor is replaced by a  A  A dielectric, C  0  r 0. d d © Copyright Virtual University of Pakistan 73 PHYSICS –PHY101 VU Summary of Lecture 26 – ELECTRIC POTENTIAL ENERGY 1. Electric current is the flow of electrical charge. If a small amount of charge dq flows in dq time dt , then the current is i . If the current is constant in time, then in time t , the dt current that flows is q  i  t. The unit of charge is ampere, which is define as: 1 coulomb 1 ampere  second A car's battery supplies upto 50 amperes when starting the car, but often we need to deal with smaller values: 1 milliampere  1 ma  103 A 1 microampere  1  A  10 6 A 1 nanoampere  1 nA  109 A 1 picoampere  1 pA  1012 A 2. The direction of current flow is the direction in which positive charges move. However, in a typical wire, the positive charges are fixed to the atoms and it is really the negative charges (electrons) that move. In that case the direction of current flow is reversed. electron flow conventional current flow device   3. Current flows because something forces it around a circuit. That "something" is EMF, electromotive force. But remember that we are using bad terminology and that EMF is not a force - it is actually the difference in electric potentials between two parts of a circuit. So, in the figure below, V  Va  Vb is the EMF which causes current to flow in the resistor. How much current? Generally, the larger V is , the more current will flow and we expect I  V. In general this relation will not be completely accurate but when it holds, we say V V Ohm's Law applies: I . Here, R  is called the resistance. R I R I I V © Copyright Virtual University of Pakistan 74 PHYSICS –PHY101 VU 4. Be careful in understanding Ohm's Law. In general the current may depend upon the applied voltage in a complicated way. Another way of saying this is that the resistance may depend upon the current. Example: when current passes through a resistor, it gets hot and its resistance increases. Only when the graph of current versus voltage is a straight line does Ohm's Law hold. Else, we can only define the "incremental resistance". I  Amps  R i V does not obey Ohm's Law obeys Ohm's Law V  volts  5. Charge is always conserved, and therefore current is conserved as well. This means that when a current splits into two currents the sum remains constant, i1  i2  i3. i2 i1 i3 6. When resistors are put in series with each other, the same current flows through both. So, V1  iR1 and V2  iR2. The total potential drop across the pair is V  V1  V2  i ( R1  R2 ).  Req  R1  R2. So resistors in series add up. R1 R2 V1 V2 7. Resistors can also be put in parallel. This means that the same V V voltage V is across both. So the currents are i1  , i2 . R1 R2 V V V Since i  i1  i2 it follows that i    a R1 b Req R1 R2 R2 1 1 1 RR    or Req  1 2. a Req b Req R1 R2 R1  R2 This makes sense: with two possible paths the current will find less resistance than if only one was present. 8. When current flows in a circuit work is done. Suppose a small amount of charge dq is moved through a potential difference V. Then the work done is dW  Vdq  V idt. Hence © Copyright Virtual University of Pakistan 75 PHYSICS –PHY101 VU V dW Vidt  i 2 Rdt ( because i  ). The rate of doing work, i.e. power, is P   i 2 R. R dt V2 This is an important formula. It can also be written as P  , or as P  iV. The unit of R joule coulomb joule power is: 1 volt-ampere  1  1  1 watt. coulomb second second 9. Kirchoff's Law: The sum of the potential differences encountered in moving around a closed circuit is zero. This law is easy to prove: since the electric field is conservative, therefore no work is done in taking a charge all around a circuit and putting it back where it was. However, it is very useful in solving problems. As a trivial example, consider the circuit below. The statement that, starting from any point a we get back to the same potential after going around is: Va  iR    Va. This says  iR    0. a b  i  i R  i d c 10. We can apply Kirchoff's Law to a circuit that consists of a resistor and capacitor in order to see how current flows q through it. Since q  VC , we can see that  iR  0. Now C 1 dq di differentiate with respect to time to find  R  0,   C dt dt  di 1  t C i or  i. This equation has solution: i  i0e RC. The dt RC R product RC is called the time constant  , and it gives the time by which the current has fallen to 1/ e  1/ 2.7 of the initial value. x 2 x3 x 4 A reminder about the exponential function, e x  1  x      2! 3! 4! d x 2 x 3x 2 4 x3 d x From this, e 1        e x Similarly, e  e  x dx 2! 3! 4! dx 11. Circuits often have two or more loops. To find the voltages and currents in such situations, it is best to apply Kirchoff's Law. In the figure below, you see that there are 3 loops and you can see that: © Copyright Virtual University of Pakistan 76 PHYSICS –PHY101 VU 1 2 i1  i3  i2 1  i1R1  i3 R3  0 i i3 R3  i2 R2   2  0 R1 i1 R3 i3 R2 i2 Make sure that you understand each of these, and then check that the solution is: 1  R2  R3    2 R3 1R3   2  R1  R3  1R2   2 R1 i1  i2  i3  R1 R2  R2 R3  R1 R3 R1 R2  R2 R3  R1 R3 R1 R2  R2 R3  R1R3 12. A charge inside a wire moves under the influence of the applied electric field and suffers many collisions that cause it to move on a highly irregular, jagged path as shown below. Nevertheless, it moves on the average to the right at the "drift velocity" (or speed). 13. Consider a wire through which charge is flowing. Suppose that the number of charges per unit volume is n. If we multiply n by the crossectional area of the wire A and the length L, then the charge in this section of the wire is q   nAL  e. If the drift velocity of the charges is vd , then the time taken for the charge to move through the wire is L q nALe t. hence the current is i    nAevd. From this we can calculate the drift vd t L / vd i velocity of the charges in terms of the measured current, v d . The current density, nAe i which is the current per unit crossectional area is defined as j  = nev d. If j varies A   inside a volume, then we can easily generalize and write, i   j  dA. L I A I V © Copyright Virtual University of Pakistan 77 PHYSICS –PHY101 VU Summary of Lecture 27 – THE MAGNETIC FIELD 1. The magnetic field exerts a force upon any charge that moves in the field. The greater the size of the charge, and the faster    F  qv  B it moves, the larger the force. The direction of the force is perpendicular to both the direction of motion and the magnetic    B field. If  is the angle between v and B, then F  qvB sin  is the     magnitude of the force. This vanishes when v and B are q  parallel (  0), and is maximum when they are perpendicular. v 2. The unit of magnetic field that is used most commonly is the tesla. A charge of one coulomb moving at 1 metre per second perpendicularly to a field of one tesla experiences a force of 1 newton. Equivalently, newton newton 1 tesla  1 1  104 gauss (CGS unit) coulomb  meter/second ampere  meter In order to have an appreciation for how much a tesla is, here are some typical values of the magnetic field in these units: Earth's surface 10-4 T Bar magnet 10-2 T Powerful electromagnet 1T Superconducting magnet 5T 3. When both magnetic and electric fields are present at a point, the total force acting upon     a charge is the vector sum of the electric and magnetic forces, F  qE  qv  B. This is known as the Lorentz Force. Note that the electric force and magnetic force are very different. The electric force is non-zero even if the charge is stationary, and it is in the  same direction as E. 4. The Lorentz Force can be used to select charged particles of whichever velocity we want. In the diagram below, particles enter from the left with velocity v. They experience a force due to the perpendicular magnetic field, as well as force downwards because of an electric field. Only particles with speed v  E / B are undeflected and keep going straight.      FB  q v B                    FE  qE E qE  q©vBCopyright  v  Virtualvelocity selector University !! of Pakistan B 78 PHYSICS –PHY101 VU 5. A magnetic field can be strong enough to lift an elephant, but it can never increase or  decrease the energy of a particle. Proof: suppose the magnetic force F moves a particle  through a displacement d r. Then the small amount of work done is,      dW  F  d r  q(v  B)  d r    dr  q (v  B)  dt dt     q (v  B)  v dt  0. Basically the force and direction of force are orthogonal, and hence there can be no work done on the particle or an increase in its energy. 6. A magnetic field bends a charged particle into a circular orbit because the particle feels a force that is directed perpendicular to the magnetic field. As we saw above, the particle cannot change its speed, but it certainly does change direction! So it keeps bending and bending until it makes a full circle. The radius of orbit can be easily calculated: the magnetic v2 and centrifugal forces must balance each other for equilibrium. So, qvB  m and we r mv find that r . A strong B forces the particle into a tighter orbit, as you can see. We qB v qB can also calculate the angular frequency,   . This shows that a strong B makes r m the particle go around many times in unit time. There are a very large number of applications of these facts.      v    F   v2 mv F qv B  m  r.     r qB    F    v v     7. The fact that a magnetic field bends charged particles is responsible for shielding the earth from harmful effects of the "solar wind". A large number of charged particles are released from the sun and reach the earth. These can destroy life. Fortunately the earth's magnetic field deflects these particles, which are then trapped in the "Van Allen" belt around the earth. © Copyright Virtual University of Pakistan 79 PHYSICS –PHY101 VU 8. The mass spectrometer is an extremely important equipment      v  that works on the above principle. Ions are made from atoms    by stripping away one electron. Then they pass through a     mv  velocity selector so that they all have the same speed. In a r qB     beam of many different ions, the heavier ones bend less, and     lighter ones more, when they are passed through a B field.      9. A wire carries current, and current is flowing charges. Since each charge experiences a force when placed in a magnetic field, you might expect the same for the current. Indeed, that is exactly the case, and we can easily calculate the force on a wire from the force on individual charges. Suppose N is the total number of charges and they are moving at the     average (or drift) velocity v d. Then the total force is F  Nevd  B. Now suppose that the wire has length L, crossectional area A, and it has n charges per unit volume. Then clearly    N  nAL, and so F  nALevd  B. Remember that the current is the charge that flows   through the wire per unit time, and so nAev d  I. We get the important result that the    force per unit length on the wire is F  I  B. 10. A current that goes around a loop (any shape) produces a  magnetic field. We define the magnetic moment as the   product of current and area,   IAzˆ. Here A is the area of the loop and I the current flowing around it. The direction I is perpendicular to the plane of the loop, as shown. 11. Magnetic fields are produced by currents. Every small bit of current produces a small amount of the B field. Ampere's Law, illustrated below, says that if one goes around a loop (of any shape or size) then the integral of the B field around the loop is equal to the enclosed current. In the loop below I  I1  I 2. Here I 3 is excluded as it lies outside.   i1   B  ds  0 I enclosed i3 Amperian loop   i2  ds  B © Copyright Virtual University of Pakistan 80 PHYSICS –PHY101 VU 12. Let us apply Ampere's Law to a circular loop of radius r outside an infinitely long wire   carrying current I through it. The magnetic field goes around in circles, and so B and ds   are both in the same direction. Hence,  B  ds  B  ds  B  2 r   0 I. We get the 0 I important result that B . 2 r 13. Assuming that the current flows uniformly over the crossection, we can use Ampere's Law to calculate the magnetic field at distance r , where r now lies inside the wire.   r2   Ir B  2 r    0 I  2  B 0 2 R  2 R Here is a sketch of the B field inside and outside the wire as a function of distance r.   B B ds r inside outside R r B rR © Copyright Virtual University of Pakistan 81 PHYSICS –PHY101 VU Summary of Lecture 28 – ELECTROMAGNETIC INDUCTION 1. Earlier we had defined the flux of any vector field. For a magnetic field, this means that flux of a uniform magnetic field (see figure) is   B A  BA cos. If the field is not A constant over the area then we must add up all the little   pieces of flux:  B   B  dA. The dimension of flux is  B magnetic field  area, and the unit is called weber, where  1 weber  1 tesla  metre2. 2. A fundamental law of magnetism states that the net flux through a closed surface is always   zero,  B   B  dA  0. Note that this is very different from what you learned earlier in electrostatics where the flux is essentially the electric charge. There is no such thing as a magnetic charge! What we call the magnetic north (or south) pole of a magnet are actually due to the particular electronic currents, not magnetic charges. In the bar magnet below, no matter which closed surface you draw, the amount of flux leaving the surface is equal to that entering it. N S Example : A sphere of radius R is placed near a long, straight wire that carries a steady current I. The magnetic field generated by the current is B. Find the total magnetic flux passing through the sphere. © Copyright Virtual University of Pakistan 82 PHYSICS –PHY101 VU Answer: zero, of course! 3. Faraday's Law for Induced EMF: when the magnetic flux changes in a circuit, an electro- motive force is induced which is proportional to the rate of change of flux. Mathematically,    dB where  is the induced emf. If the coil consists of N turns, then    N dB. dt dt How does the flux through a coil change? Consider a coil and magnet. We can: a) move the magnet, b) change the size and shape of the coil by squeezing it, c) move the coil. dB In all cases, the flux through the coil changes and is non-zero leading to an induced dt emf. Example: A flexible loop has a radius of 12cm and is in a magnetic field of strength 0.15T. The loop is grasped at points A and B and stretched until it closes. If it takes 0.20s to close the loop, find the magnitude of the average induced emf in it during this time. Solution: Here the loop area changes, hence the flux. So the induced emf is:  0   (0.12) 2  0.15     dB    final flux  initial flux        0.034 Volts. dt  time taken  0.2  Example : A wire loop of radius 0.30m lies so that an external magnetic field of +0.30T is perpendicular to the loop. The field changes to -0.20T in 1.5s. Find the magnitude of the average induced EMF in the loop during this time.  B Solution: Again, we will find the initial and final fluxes first and then divide by the time taken for the change. Use   BA  B r 2 to calculate the flux.  i  0.30    (0.30) 2  0.085 Tm 2  f  0.20    (0.30) 2  0.057 Tm 2   f   i 0.085  0.057     0.095 V t t 1.5 © Copyright Virtual University of Pakistan 83 PHYSICS –PHY101 VU 4. Remember that the electromotive force is not really a force but the difference in electric potentials between two points. In going around a circular wire, where the electric field is constant as a function of angle, the emf is   E  2 r . More generally, for any size or     dB shape of a closed circuit,    E  ds. So Faraday's Law reads:  E  ds  . dt Example : A conducting wire rests upon two parallel rails and is pulled towards the right with speed v. A magnetic field B is perpendicular to the plain of the rails as shown below. Find the current that flows in the circuit.  x F2              D v             F3  B (into paper) Solution: As the wire is pulled to the right, the area of the circuit increases and so the flux increases. Measure x as above so that x  vt , i.e. x keeps increasing as we pull. The flux at any value of x is  B  BDx, and so, d  BDx     dB    BD dx  BDv. dt dt dt  BD v To calculate the current, we simply use Ohm's Law: I  . Here R is the R R resistance of the circuit. Example : Find the power dissipated in the above circuit using I 2 R, and then by directly calculating the work you do by pulling the wire. B2 D2 v2 Solution: Clearly I 2 R  is the power dissipated, as per usual formula. Now let us R calculate the force acting upon the piece of wire (of length D ) that you are pulling. From    B2 D2v the formula for the force on a wire, F  I L  B, the magnitude is F  IBD . So, R B2 D 2v2 the power is P  Fv . This is exactly the value calculated above! R © Copyright Virtual University of Pakistan 84 PHYSICS –PHY101 VU 5. Lenz's Law : The direction of any magnetic induction effect is such as to oppose the cause of the effect. Imagine a coil wound with wire of finite resistance. If the magnetic field decreases, the induced EMF is positive. This produces a positive current. The magnetic field produced by the current opposes the decrease in flux. Of course, because of finite resistance in loop, the induced current cannot completely oppose the change in flux. © Copyright Virtual University of Pakistan 85 PHYSICS –PHY101 VU Summary of Lecture 29 – ALTERNATING CURRENT 1. Alternating current (AC) is current that flows first in one direction along a wire, and then in the reverse direction. The most common AC is sinuisoidal in which the current (and voltage) follow a sine function, as in the graph below. The average value is zero because the current flows for the same time in one direction as in the other. m    m sin  t   average      0 t However, the square of any AC wave is always positive. Thus its average is not zero. As we shall see, upon averaging the square we get half the square of the peak value.  rms  02  0 t The calculation follows: if there is a sine wave of amplitude (height) equal to one and frequency  , (  2 / T , T=time period), then the squared amplitude is sin 2 t and its average is: 1  cos 2 t T T T 1 1 1 1 1  sin 2 t   T0 dt sin 2  t   dt ( T0 2 )   dt  T0 2 2 Taking the square root gives the root mean square value as 1/ 2 of the maximum value. Of course, it does not matter whether this is of the voltage or current:  m2  m I m2 I  rms    0.707 m , and I rms   m  0.707 I m. 2 2 2 2 Exactly the same results are obtained for cosine waves. This is what one expects since the difference between sine and cosine is only that one starts earlier than the other. © Copyright Virtual University of Pakistan 86 PHYSICS –PHY101 VU 2. AC is generated by a coil rotating in a magnetic field. We know from Faraday's Law,    d  B , that a changing magnetic flux gives rise to an emf. Imagine a magnetic dt field and a coil of area A, rotating with frequency  so that the flux through the coil at any instant of time is  B  BA cos t. Then the induced emf is   BA sin  t. 3. AC is particular useful because transformers make it possible to step up or step down voltages. The basic transformer consists of two coils - the primary and secondary - both wrapped around a core (typically iron) that enhances the magnetic field.  Suppose that the flux in the core is  and that its rate of change is. Call the number t of turns in the primary and secondary N p and N s respectively. Then, from Faraday's law,   the primary emf is  p   N p and the secondary emf is  s   N s. The ratio is, t t p N p N N . So, the secondary emf is  s  s  p. If s is less than one, then it is called s Ns Np Np a step-down transformer because the secondary voltage is less than the input voltage. Else, it is a step-up transformer. Both types are used. 4. If this is a lossless transformer (and good transformers are 99% lossless), then the input power must equal the output power,  p I p   s I s. From above, this shows that the ratio Np of currents is I s  I p. Ns 5. Whatever the shape or size of a current carrying loop, the magnetic flux that passes through it is proportional to the current,   I. The inductance L (called the self-inductance if © Copyright Virtual University of Pakistan 87 PHYSICS –PHY101 VU there is only one coil) is the constant of proportionality in the relation   LI. The unit of inductance is called Henry, 1 Henry= 1 Tesla metre2 / Ampere. Note that inductance, like capacitance, is purely geometrical and depends only upon the shape and sizes of wires. It does not depend on the current. 6. Let us calculate the inductance of a long coil wound with n turns per unit length. As calculated earlier, the B field is B  0 nI , and the flux passing through N turns of the coil of length l and area A is N  B   nl  BA   0 n 2 IAl. From the definition, we find: N  B 0 n 2lIA L   0 n 2lA (inductance of long solenoid). I I Example: Find the inductance of a coil with 3500 turns, length 10 cm, and radius 5cm.   0.05m  2 T m T  m2 Solution: L  4  10 7  35002   1.21  1.21H. A 0.10m A 7. When the current changes through an inductor, by Faraday's dB d ( LI ) dI Law it induces an emf equal to    L. dt dt dt Let us use this to calculate the current through the circuit R shown here, where a resistor is present. Then, by using dI dI L Kirchoff's Law,   IR  L or  I  I 2 R  LI. In dt dt words: the power expended by the battery equals energy dissipated in the resistor + work done on inductor. 8. Before we go on to the AC case, suppose that we suddenly connect a battery to an inductor so that the voltage suddenly increases from zero to  across the circuit. Then,  L dI dt   L  IR   has solution: I  t   1  e t /  where  . You can easily see that R R dI  1 t /  this true using  e. The solution shows that the current increases from zero to dt R   a maximum of , i.e. that given by Ohm's Law. We need to pay a little attention to the R "time constant"  , which is the time after which the constant approaches 63% of its final value. Units:     L   henry  volt.second / ampere   volt  second  second.  ampere.ohm   R  ohm ohm   © Copyright Virtual University of Pakistan 88 PHYSICS –PHY101 VU    Note that when t= , then I  R 1  e 1   1  0.37   0.63 R R 9. When we pass current through an inductor a changing magnetic field is produced. This, by Faraday's Law, induces an emf across the coil. So work has to be done to force the current through. How much work? The power, or rate of doing work, is emf  current. dU B  dI  dI Let U B be the work done in passing current I. Then,   L  I  LI. Let us dt  dt  dt UB I 1 2 integrate dU B  LI dI. Then,  0 dU B   LIdI  U B  0 2 LI. This is an important 1 2 result. It tells us that an inductor L carrying current I requires work LI. By conservation 2 of energy, this is also the energy stored in the inductor. Compare this result with the result 1 U E  CV 2 for a capacitor. Notice that C  L and V  I. 2 10. Let us use the result derived earlier for the inductance of a solenoid and the magnetic field 1 1 in it, L  0 n 2lA and B  0 nI. Putting this into U B  LI 2 gives U B  ( 0 n 2lA)( B / 0 n)2. 2 2 2 UB B Divide the energy by the volume of the solenoid, . This directly gives the volume 20 energy density (energy per unit volume) contained in a magnetic field. 11. Electromagnetic oscillations in an LC circuit. We know that energy can be stored in a capacitor as well C as in an inductor. What happens when we connect them up L together and put some charge on the capacitor? As it discharges, it creates a current that transfers energy to the inductor. The total energy remains constant, of course. 1 2 1 q2 This means that the sum U  U B  U E  LI  is constant. Differentiate this: 2 2C dU d  1 2 1 q 2  dq dI d 2 q d 2q 1 0   LI  . Since I  , and  2 ,  2  q0 dt dt  2 2C dt dt dt dt LC 1 d 2q To make this look nicer, put  2   2   2 q  0. We have seen this equation LC dt many times earlier. The solution is: q  qm cos  t. The important result here is that the © Copyright Virtual University of Pakistan 89 PHYSICS –PHY101 VU 1 charge, current, and voltage will oscillate with frequency  . This oscillation will LC go on for forever if there is no resistance in the circuit. © Copyright Virtual University of Pakistan 90 PHYSICS –PHY101 VU Summary of Lecture 30 – ELECTROMAGNETIC WAVES 1. Before the investigations of James Clerk Maxwell around 1865, the known laws of electromagnetism were:   q a) Gauss' law of electricity:  E  dA  (integral is over any closed surface) 0   b) Gauss' law of magnetism:  B  dA  0 (integral is over any closed surface)   dB c) Faraday's law of induction:  E  ds   (integral is over any closed loop) dt   d) Ampere's law:  B  ds  0 I (integral is over any closed loop) 2. But Maxwell realized that the above 4 laws were not consistent with the conservation of charge, which is a fundamental principle. He argued that if you take the space between two capacitors (see below) and take different surfaces 1,2,3,4 then applying Ampere's     Law gives an inconsistency:   B  ds     B  ds  because obviously charge cannot 1,2,4  3 flow in the gap between plates. So Ampere's Law gives different results depending upon which surface is bounded by the loop shown! circuit   Maxwell modified Ampere's law as follows:  B  ds    I  I  where the "displacement 0 d dE current" is I d   0. Let's look at the reasoning that led to Maxwell's discovery of the dt dQ displacement current. The current that flows in the circuit is I . But the charge on the dt d d ( EA) dE capacitor plate is Q   0 EA. Hence, I    0 EA   0  0  I D. In words, the dt dt dt changing electric field in the gap acts as source of the magnetic field in just the same way as the current in the outside wires. This is really the most important point - a magnetic field may have two separate reasons for existence - flowing charges or changing electric fields. © Copyright Virtual University of Pakistan 91 PHYSICS –PHY101 VU 3. The famous Maxwell's equations are as follows:   Q a)  E  dS  0   dB b)  E  d    dt   c)  B  dS  0    dE  d)  B  d   0  I   0   dt        Together with the Lorentz Force F  q E  v  B they provide a complete description of all electromagnetic phenomena, including waves. 4. Electromagnetic waves were predicted by Maxwell and experimentally discovered many years later by Hertz. Note that for these waves: a) Absolutely no medium is required - they travel through vacuum. b) The speed of propagation is c for all waves in the vacuum. c) There is no limit to the amplitude or frequency. A wave is characterized by the amplitude and frequency, as illustrated below. wavelength amplitude wavelength node Example: Red light has  = 700 nm. The frequency is calculated as follows: 3.0  108 m / sec  7  4.29  1014 Hertz 7  10 m By comparison, the electromagnetic waves inside a microwave oven have wavelength of 6 cm, radio waves are a few metres long. For visible light, see below. On the other hand, X-rays and gamma-rays have wavelengths of the size of atoms and even much smaller. 400nm 500nm 600nm 700nm © Copyright Virtual University of Pakistan 92 PHYSICS –PHY101 VU 5. We now consider how electromagnetic waves can travel through empty space. Suppose an electric field (due to distant charges in an antenna) has been created. If this changes dE   dE then this creates a changing electric flux which, through  B  ds   0 0 , dt dt dB   dB creates a changing magnetic flux. This, through  E  ds   , creates a changing dt dt electric field. This chain of events in free space then allows a wave to propagate. x z In the diagram above, an electromagnetic wave is moving in the z direction. The electric field is in the x direction, Ex  E0 sin(kz   t ), and the magnetic field is perpendicular to it, By  B0 sin(k z   t ). Here   kc. From Maxwell's equations the amplitudes of the two fields are related by E0  cB0. Note that the two fields are in phase with each other. 6. The production of electromagnetic waves is done by forcing current to vary rapidly in a small piece of wire. Consider your mobile phone, for example. Using the power from the battery, the circuits inside produce a high frequency current that goes into a "dipole antenna" made of two small pieces of conductor. The electric field between the two oppositely charged pieces is rapidly changing and so creates a magnetic field. Both fields propagate outwards, the amplitude falling as 1/ r. sin 2  The power, which is the square of the amplitude, falls off as I ( ) . The sin 2  r2 dependence shows that the power is radiated unequally as a function of direction. The maximum power is a t    / 2 and the least at   0. © Copyright Virtual University of Pakistan 93 PHYSICS –PHY101 VU 7. The reception of electromagnetic waves requires an antenna. The incoming wave has an electric field that forces the electrons to run up and down the antenna wire, i.e. it produces a tiny electric current. This current is then amplified (increased in amplitude) electronically. This is schematically indicated below. Here the variable capacitor is used to tune to different frequencies. 8. As we have seen, the electric field of a wave is perpendicular to the direction of its motion. If this is a fixed direction (say, xˆ ), then we say that wave is polarized in the x direction. Most sources - a candle, the sun, any light bulb - produce light that is unpolarized. In this case, there is no definite direction of the electric field, no definite phase between the orthogonal components, and the atomic or molecular dipoles that emit the light are randomly oriented in the source. But for a typical linearly polarized plane electromagnetic wave E polarized along xˆ , E x  E0 sin( kz   t ), B y  0 sin(kz   t ) with all other components zero. c Of course, it may be that the wave is polarized at an angle  relative to xˆ , in which case E x  E0 cos   sin( kz   t ), E y  E0 sin   sin( k z   t ), E z  0. 9. Electromagnetic waves from an unpolarized source (e.g. a burning candle or microwave oven) can be polarized by passing them through a simple polarizer of the kind below. A met al plate with slits cut into will allow only the electric field component perpendicular to the slits. Thus, it will produce linearly polarized waves from unpolarized ones. microwave oven polarization metal plate © Copyright Virtual University of Pakistan 94 PHYSICS –PHY101 VU Summary of Lecture 31 – LIGHT 1. Light travels very fast but its speed is not infinite. Early attempts to measure the speed using earth based experiments failed. Then in 1675 the astronomer Roemer studied timing of the eclipse of one of Jupiter's moon called Io. In the diagram below Io is observed with Earth at A and then at C. The eclipse is 16.6 minutes late, which is the time taken for light to travel AC. Roemer estimated that c  3  108 metres/sec, a value that is remarkably close to the best modern measurement, c  299792458.6 metres/sec. 2. Light is electromagnetic waves. Different frequencies correspond to different colours. Equivalently, different wavelength  correspond to different colours. Recall that the product    c. In the diagram below you see that visible light is only one small part of the total electromagnetic spectrum. Here nm means nanometres or 10-9 metres. 3. If light contained all frequencies with equal strength, it would appear as white to us. course, most things around us appear coloured. That is because they radiate more strongly in one range of frequency than in others. If there is more intensity in the yellow range than the green range, we will see mostly yellow. The sky appears blue to us on a clear day because tiny dust particles high above in the atmosphere reflect a lot of the blue light coming from the sun. In the figure below you can see the hump at smaller wavelengths. © Copyright Virtual University of Pakistan 95 PHYSICS –PHY101 VU Compare this with the spectrum of light emitted from a tungsten bulb. You can see that this is smoother and yellow dominates. Intensity Intensity Normal Daylight Tungsten Bulb 400nm 500nm 600nm 700nm 400nm 500nm 600nm 700nm 4. What path does light travel upon? If there is no obstruction, it obviously likes to travel on straight line which is the shortest path between any two points, say A and B. Fermat's Principle states that in all situations, light will always take that path for which it takes the least time. As an example, let us apply Fermat's Principle to the case of light reflected from a mirror, as below. B 1 A 1 a  b 1 1 x dx d The total distance travelled by the ray is L  a 2  x 2  b2   d  x . The time taken 2 is t  L / c. To find the smallest time, we must differentiate and then set the derivative dt 1 dL to zero, 0  dx c dx   a 2  x 2   2 x   b 2   d  x    2  d  x  1 1 1/ 2 1 2 1/ 2 2c 2c   x dx From here we immediately see that  From the a 2  x2 b  d  x 2 2 above diagram, sin 1  sin 1. Of course, it is no surprise that the angle of reflection equals the angle of incidence. You knew this from before, and this seems like a very complicated derivation of a simple fact. But it is still nice to see that there is a deeper principle behind it. © Copyright Virtual University of Pakistan 96 PHYSICS –PHY101 VU 5. The speed of light in vacuum is a fixed constant of nature which we usually call c, but in a medium light can travel slower or faster than c. We define the "refractive index" of that Speed of light in vacuum c medium as: Refractive index  (or n  ). Usually the values Speed of light in material v of n are bigger than one (e.g. for glass it is around 1.5) but in some special media, its value can be less than one. The value of n also depends on the wavelength (or frequency) of light. This is called dispersion, and it means that different colours travel at different speeds inside a medium. This is why, as in the diagram below, white light gets separated into different colours. The fact that in a glass prism blue light travels faster than red light is responsible for the many colours we see here. 6. We can apply Fermat's Principle to find the path A followed by a ray of light when it goes from one 1 L1 medium to another. Part of the light is reflected, a and part is "refracted", i.e. it bends away or 1 n1 x dx v1 towards the normal. The total time is, n2 v2 L L c nL n L L  2 L2 t  1  2, n t  1 1 2 2  v1 v 2 v c c b 2 L  n1 L1  n2 L2  n1 a 2  x 2  n2 b 2   d  x  2 d Fermat's Principle says that the time t must be C   a  x 2   2 x   2 b 2   d  x    2  d  x  1 dt 1 dL n1 2 1/ 2 n 2 1/ 2 minimized: 0   dx c dx 2c 2c   x dx And so we get "Snell's Law", n1  n2 or n1 sin 1  n2 sin  2. This a 2  x2 b2   d  x  2 required a little bit of mathematics, but you can see how powerful Fermat's Principle is! 7. Light coming from air into water bends toward the normal. Conversely, light from a source in the water will bend away from the normal. What if you keep increasing the angle with respect to the normal so that the light bends and begins to just follow the surface? This phenomenon is called total internal reflection and  c is called the critical © Copyright Virtual University of Pakistan 97 PHYSICS –PHY101 VU n2 angle. It obeys: n1 sin  c  n2 sin 900 from which  c  sin 1. n1 n1 n2 c 8. Fibre optic cables, which are now common everywhere, make use of the total internal reflection principle to carry light. Here is what a fibre optic cable looks like from inside: Even if the cable is bent, the light will continue to travel along it. The glass inside the cable must have exceedingly good consistency - if it thicker or thinner in any part, the refractive index will become non-uniform and a lot of light will get lost. Optical fibres now carry thousands of telephone calls in a cable whose diameter is only a little bigger than a human hair! © Copyright Virtual University of Pakistan 98 PHYSICS –PHY101 VU Summary of Lecture 32 – INTERACTION OF LIGHT WITH MATTER 1. In this lecture I shall deal with the 4 basic ways in which light interacts with matter: a) Emission - matter releases energy as light. b) Absorption - matter takes energy from light. c) Transmission - matter allows light to pass through it. d) Reflection - matter repels light in another direction. 2. When an object (for example, an iron rod or the filament of a tungsten bulb) is heated, it emits light. When the temperature is around 800o C, it is red hot. Around 2500o C it is yellowish-white. At temperatures lower than 800o C, infrared (IR) light is emitted but our eyes cannot see this. This kind of emission is called blackbody radiation. Blackbody radiation is continuous - all wavelengths are emitted. However most of the energy is radiated close to the peak. As you can see in the graph, the position of the peak goes to smaller wavelengths (or higher frequencies) as the object becomes hotter. The scale of temperature is shown in degrees Kelvin (o K ). To convert from o C to o K , simply add 273. We shall have more to say about the Kelvin scale later. Where exactly does the peak occur? Wien's Law states that maxT  2.90  103 m K. We can derive this in an advanced physics course, but for now you must take this as given. © Copyright Virtual University of Pakistan 99 PHYSICS –PHY101 VU 3. In the lecture on electromagnetic waves you had learnt that these waves are emitted when charges accelerate. Blackbody radiation occurs for exactly this reason as well. When a body is heated up, the electrons, atoms, and molecules which it contains undergo violent random motion. Light may emitted by electrons in one atom and absorbed in another. Even an empty box will be filled with blackbody radiation because the sides of the box are made up of material that has charged constituents that radiate energy when they undergo acceleration during their random motion. 4. When can you use Wien's Law? More generally, when can you expect a body to emit blackbody radiation? Answer: only for objects that emit light, not for those that merely reflect light (e.g. flowers). The Sun and other stars obey Wien's Law since the gases they are composed of emit radiation that is in equilibrium with the other materials. Wien's law allows astronomers to determine the temperature of a star because the wavelength at which a star is brightest is related to its temperature. 5. All heated matter radiates energy, and hotter objects radiate more energy. The famous Stefan-Boltzman Law, which we unfortunately cannot derive in this introductory course, states that the power radiated per unit area of a hot body is P   T 4 , where the Stefan- Boltzman constant is  = 5.67  10-8 W m-2 K -4. 6. Let us apply P   T 4 for finding the temperature of a planet that is at distance R from the sun. The sun has temperature Tsun and radius Rsun. In equilibrium, the energy received from the sun is exactly equal to the energy radiated by the planet. Now, the total energy radiated 1 by the sun is  T04  4 Rsun 2. But on a unit area of the planet, only of this is received. 4 R 2 1 So the energy received per unit area on the planet is  T04  4 Rsun2 . This must be 4 R 2 R equal to  T 4  T  T0 sun. R 7. The above was for blackbody radiation where the emitted light has a continuous spectrum. But if a gas of identical atoms is excited by some mechanism, then only a few discrete wavelengths are emitted. Each chemical element produces a very distinct pattern of colors called an emission spectrum. So, for example, laboratory hydrogen gas lamps emit 3 lines in the visible region, as you can see below. Whenever we see 3 lines spaced apart in this way, we immediately know that hydrogen gas is present. It is as good as the thumbprint of a man! © Copyright Virtual University of Pakistan 100 PHYSICS –PHY101 VU 8. But how do we get atoms excited so that they can start revealing their identity? One way is to simply heat material containing those atoms. You saw in the lecture how different colours come from sprinkling different materials on a flame. 9. Everything that I have said about the emission of light applies exactly to the absorption of light as well. So, for example, when white light (w