March 9 Physics 54 Lecture Professor Henry Greenside

1. March 9 Physics 54 LectureProfessor Henry Greenside

2. Reminder: Answers to PRS Questions in Notes Section of PowerPoint Slides B field has same magnitude everywhere but different orientation in the three horizontal strips as shown. Want to do line integral CCW about square of side L, path ABCDA in that order. Answer: (4), the line integral is zero. To evaluate the line integral, you need to break the integral up into a sum of separate contiguous pieces that correspond to each region of space where the magnetic field changes or where the wire changes its orientation with respect to the magnetic field. Thus the line integral along the square ABCDA needs to be broken up into the line integrals over segments AE, EF, FB, BC, CG, GH, HD, and DA. The line integral from point A to point E gives a value of xB since the magnetic field B is parallel to the direction of the path, the magnetic field has constant strength B along the path, and the length of the path is x. The line integral from point E to point F is 0 since the magnetic field is everywhere perpendicular to the path in this region (remember that the dot product of perpendicular vectors is zero). The total line integral from A to F so far is thus xB + 0 = xB. The line integral from point F to point B is -xB since now the magnetic field points in the opposite direction to the direction we are traversing along the path (which is always counterclockwise). This the value of the line integral from A to E cancels the value of the line integral from F to B and, so far, the total line integral from A to B is xB + 0 + (-xB) = 0. The line integral from point B to point C is zero since B is everywhere perpendicular to the path we are following. The total line integral from A to C is therefore xB + 0 + (-xB) + 0 = 0. The line integral from point C to point G is xB since the magnetic field is constant along this part of the path and is parallel to the direction of the path. The total line integral from point A to point G is therefore xB + 0 + (-xB) + 0 + xB = xB. The line integral from point G to point H is zero since the magnetic field is perpendicular to the path GH in this region. The total line integral from A to H is therefore still xB. The line integral from point H to point D is �xB since B now points opposite to the direction we are following along the path. This contribution cancels the contribution from segment GC and so the total line integral from point A to B to C to D is zero. Finally, the line integral from point D back the beginning of the path, point A, is again zero since B is perpendicular to the path in this region. Adding all these separate pieces up, we get zero for the value of the total line integral around the closed loop ABCDA. This means in turn that the total current flowing through the loop must be zero (by Ampere�s law), but this does not mean that there are no currents flowing through the loop itself, just that all the currents, plus or minus, must add to zero. It might be the case, for example, that a huge current is flowing out of the page through this loop and this is balanced by many tiny negative currents passing through the loop in the opposite direction. A challenge: can you think of a way to arrange wires and currents to create this strange three-part magnetic field (the pink arrows), i.e., is this a magnetic field that could occur in nature? If you can, please email me your solution!B field has same magnitude everywhere but different orientation in the three horizontal strips as shown. Want to do line integral CCW about square of side L, path ABCDA in that order. Answer: (4), the line integral is zero. To evaluate the line integral, you need to break the integral up into a sum of separate contiguous pieces that correspond to each region of space where the magnetic field changes or where the wire changes its orientation with respect to the magnetic field. Thus the line integral along the square ABCDA needs to be broken up into the line integrals over segments AE, EF, FB, BC, CG, GH, HD, and DA. The line integral from point A to point E gives a value of xB since the magnetic field B is parallel to the direction of the path, the magnetic field has constant strength B along the path, and the length of the path is x. The line integral from point E to point F is 0 since the magnetic field is everywhere perpendicular to the path in this region (remember that the dot product of perpendicular vectors is zero). The total line integral from A to F so far is thus xB + 0 = xB. The line integral from point F to point B is -xB since now the magnetic field points in the opposite direction to the direction we are traversing along the path (which is always counterclockwise). This the value of the line integral from A to E cancels the value of the line integral from F to B and, so far, the total line integral from A to B is xB + 0 + (-xB) = 0. The line integral from point B to point C is zero since B is everywhere perpendicular to the path we are following. The total line integral from A to C is therefore xB + 0 + (-xB) + 0 = 0. The line integral from point C to point G is xB since the magnetic field is constant along this part of the path and is parallel to the direction of the path. The total line integral from point A to point G is therefore xB + 0 + (-xB) + 0 + xB = xB. The line integral from point G to point H is zero since the magnetic field is perpendicular to the path GH in this region. The total line integral from A to H is therefore still xB. The line integral from point H to point D is �xB since B now points opposite to the direction we are following along the path. This contribution cancels the contribution from segment GC and so the total line integral from point A to B to C to D is zero. Finally, the line integral from point D back the beginning of the path, point A, is again zero since B is perpendicular to the path in this region. Adding all these separate pieces up, we get zero for the value of the total line integral around the closed loop ABCDA. This means in turn that the total current flowing through the loop must be zero (by Ampere�s law), but this does not mean that there are no currents flowing through the loop itself, just that all the currents, plus or minus, must add to zero. It might be the case, for example, that a huge current is flowing out of the page through this loop and this is balanced by many tiny negative currents passing through the loop in the opposite direction. A challenge: can you think of a way to arrange wires and currents to create this strange three-part magnetic field (the pink arrows), i.e., is this a magnetic field that could occur in nature? If you can, please email me your solution!

3. Key Points From Previous Lecture

4. Getting Ready for Spring Break I Ohm on the range (i.e., the song �Home on the Range�).Ohm on the range (i.e., the song �Home on the Range�).

5. Getting Ready for Spring Break II A mobile ohm (mobile home).A mobile ohm (mobile home).

6. Getting Ready for Spring Break III: Why did the chicken cross the Moebius strip? Answer: To get to the same side. A Moebius strip is a most unusual object in that it has only one surface, unlike say a piece of paper that has two surfaces. You can verify this yourself by mentally tracing a pen along the middle of the above strip. You will see that, because of the half-turn of the strip, you end up back where you started without lifting up your pen, i.e., there is only one side to this object. A challenge: Make a Mobius strip by cutting a long rectangular piece of paper, giving it a half-turn, and then taping the ends together. Now get a scissors and cut the strip down its middle. What do you get?!Answer: To get to the same side. A Moebius strip is a most unusual object in that it has only one surface, unlike say a piece of paper that has two surfaces. You can verify this yourself by mentally tracing a pen along the middle of the above strip. You will see that, because of the half-turn of the strip, you end up back where you started without lifting up your pen, i.e., there is only one side to this object. A challenge: Make a Mobius strip by cutting a long rectangular piece of paper, giving it a half-turn, and then taping the ends together. Now get a scissors and cut the strip down its middle. What do you get?!

7. At the Whiteboard� Finale of Ampere�s law: the toroid. Application of solenoids, toroids, and Faraday�s law to magnetic confinement of a thermonuclear fusion plasma: the ultimate environmentally and politically acceptable power source for the human race? Faraday�s law of induction: an electromotive force emf is produced by a time varying magnetic flux. Mnemonic: First letter of Faraday sounds like �flux�, which helps you remember when to use Faraday�s law versus Ampere�s law. Several worked examples of Faraday�s law such as a wire falling down vertical conducting rails under gravity.

8. Magnetic Field in A Toroid:Another Application of Ampere�s Law

9. Easiest Fusion Reaction on Earth:Hydrogen Isotopes Deuterium and Tritium Tritium T has a half-life of about 12.3 years. Do you see why hydrogen isotopes are the easiest nuclei to fuse? Smallest charge means smallest repulsive force. Nuclei have to touch (be within about 10-15m) for the strong and weak interactions to become active. For subtle reasons having to do with quantum mechanics, a D-T reaction is millions of times more probable than a p-p reaction, we can not achieve fusion the way the Sun does.Tritium T has a half-life of about 12.3 years. Do you see why hydrogen isotopes are the easiest nuclei to fuse? Smallest charge means smallest repulsive force. Nuclei have to touch (be within about 10-15m) for the strong and weak interactions to become active. For subtle reasons having to do with quantum mechanics, a D-T reaction is millions of times more probable than a p-p reaction, we can not achieve fusion the way the Sun does.

10. A Day in the Life of a Gigawatt Reactor Nitrous oxide Nitrous oxide

11. Where to Get Deuterium?

12. Big insight: magnetic confinement

14. ITER: International Thermonuclear Experimental Reactor

15. Transition to Chapter 29

16. PRS Review Question: Direction of Current? Answer: (2) Imagine a mobile positive charge in the ring. As the ring falls, the positive charge moves downward with the same speed as the ring itself. Thus you have a positive charge moving downward in the presence of a magnetic field that points to the upper left on the left side of the disk. The right hand rule for the cross-product of two vectors, V cross B, tells us that a Lorentz force appears that pushes the positive charge into the page on the left side of the ring, and similarly out of the page on the right side of the ring. Thus the answer is (2).Answer: (2) Imagine a mobile positive charge in the ring. As the ring falls, the positive charge moves downward with the same speed as the ring itself. Thus you have a positive charge moving downward in the presence of a magnetic field that points to the upper left on the left side of the disk. The right hand rule for the cross-product of two vectors, V cross B, tells us that a Lorentz force appears that pushes the positive charge into the page on the left side of the ring, and similarly out of the page on the right side of the ring. Thus the answer is (2).

17. Relativity of Motion Implies Creation of E Without Charge Motion!

18. Demo: Magnet, Coil, and Galvanometer

19. Demo of currents being generated by changing magnetic fields: magnet falling down a conducting tube

20. Demo of currents generated by changing magnetic field: a ring launcher

21. Transcranial Magnetic Stimulation: Changing MagneticField Causes Currents to Flow in Brain Do NOT try this on your own, can cause epilepsy if not careful. Magnet over Broca�s region can shut down speech center, person can�t talk. Professor at Stanford claims big jumps in artistic and mathematical ability when certain parts of brain temporarily deactivated, controversial but extremely interesting.Do NOT try this on your own, can cause epilepsy if not careful. Magnet over Broca�s region can shut down speech center, person can�t talk. Professor at Stanford claims big jumps in artistic and mathematical ability when certain parts of brain temporarily deactivated, controversial but extremely interesting.

22. The Jewel in the Crown: Faraday�s Law

23. Magnetic Flux F

24. PRS Question: Flux Through Solenoid Answer: 2. For a non-uniform magnetic field, the magnetic flux Phi through some surface like the flat disk spanning the circular loop of wire is obtained by adding up all the little fluxes arising from small areas of the disk times the local magnetic field. For the loop a, the magnetic field is zero everywhere except inside the solenoid (to a good approximation if the solenoid is tightly wound and quite long) so the flux through loop a is simply the flux through a cross section of the solenoid itself, namely (pi R^2)B where R is the radius of the solenoid and B = mu_o n I is the magnetic field in a solenoid associated with a current I. This flux is independent of the radius of the loop as long as the loop is bigger than the solenoid. From this, we immediately see that Phi_a = Phi_b since both of these loops entirely contain the solenoid. The flux through loop c is smaller than the fluxes through a and b since the radius of c is smaller than the radius of the solenoid, i.e., AB is smaller since A is smaller and B is the same. Finally, the flux through loop d is the smallest, namely zero, since no field lines of B pass through the disk spanning d, because the loop is turned parallel to the field lines.Answer: 2. For a non-uniform magnetic field, the magnetic flux Phi through some surface like the flat disk spanning the circular loop of wire is obtained by adding up all the little fluxes arising from small areas of the disk times the local magnetic field. For the loop a, the magnetic field is zero everywhere except inside the solenoid (to a good approximation if the solenoid is tightly wound and quite long) so the flux through loop a is simply the flux through a cross section of the solenoid itself, namely (pi R^2)B where R is the radius of the solenoid and B = mu_o n I is the magnetic field in a solenoid associated with a current I. This flux is independent of the radius of the loop as long as the loop is bigger than the solenoid. From this, we immediately see that Phi_a = Phi_b since both of these loops entirely contain the solenoid. The flux through loop c is smaller than the fluxes through a and b since the radius of c is smaller than the radius of the solenoid, i.e., AB is smaller since A is smaller and B is the same. Finally, the flux through loop d is the smallest, namely zero, since no field lines of B pass through the disk spanning d, because the loop is turned parallel to the field lines.

25. Faraday Law Example: Loop Moving from High to Low Magnetic Field Note: dx/dt=v and d(L-x)/dt=-v. Since the total flux through the loop is Phi = h(L-x)B1 + hxB2, dPhi/dt=hv(B2-B1), by Faraday�s law, the emf of the loop is �dPhi/dt = hv(B1-B2). Note that this emf vanishes if B1=B2, which makes sense since then there is no change in flux if the magnetic field is the same everywhere as the loop moves to the right. We can see that the expression hv(B1-B2) for the emf justifies a previous statement made in lecture, that a straight conducting bar of length h moving with speed v in a magnetic field of strength B (with h, v, and B mutually perpendicular) acts like a battery of emf hvB. One �battery� corresponds to the vertical left side of the loop, with emf hvB1 combines in series with a second battery of smaller emf hvB2 on the right (with positive terminal to positive terminal, negative terminal to negative terminal) so that the total emf is the difference of the two batteries, as it should be by the Kirchoff loop rule. Since the loop is moving, we can immediately deduce the direction of the current around the loop by considering a positive charge in the left vertical side of the loop. This charge is moving with speed v to the right in a magnetic field coming out of the page so feels a Lorentz force qvB1 downward, which implies a CCW current. A positive charge in the right vertical side also feels a downward force but this is weaker since the magnetic field strength B2 < B1, so the direction of the current is that of the stronger force qvB1. We can get the direction of the current also by Lenz�s law. As the loop moves to the right, the number of dots in the loop decreases (more of the loop is in the low field region) which means that an induced current must appear whose new magnetic field increases the number of dots in the loop. By the right hand rule, this has to be a CCW current. Once Faraday�s law tells us that the emf in the loop is hv(B1-B2), we immediately know the current I in the loop from Ohm�s law, I = emf/R where R is the total resistance of the loop. Then the electrical power dissipated is P = emf I = emf^2/R = h^2 v^2 (B1-B2)^2/R. By energy conservation, this must be the power supplied by pulling the loop with a force F to the right with speed v, mechanical power = Fv, so equating the mechanical power to the electrical power gives us an expression for the force needed to pull this loop at a constant speed to the right: Fv = (emf)^2/R or F = (emf)^2/(vR) = h^2 (B1-B2)^2 v/R. This expression is none other than the force IhB1 needed to pull the left vertical bar of the loop carrying current I through the field B1 minus the force IhB2 needed to pull the right vertical bar of the loop also carrying current I through the field B2 to the right. The net result is that we see that Faraday�s law based on a time-varying flux gives exactly the same answer for the current and total force that we would get if we didn�t know about fluxes and just looked at forces acting directly on charges in the loops that are moving with speed v in the presence of two magnetic fields. The advantage of Faraday�s law is that it generalizes to cases where the magnetic field is changing without any motion of the loop or charges and gives a unifying picture that takes the relative motion of charges and magnetic fields into account.Note: dx/dt=v and d(L-x)/dt=-v. Since the total flux through the loop is Phi = h(L-x)B1 + hxB2, dPhi/dt=hv(B2-B1), by Faraday�s law, the emf of the loop is �dPhi/dt = hv(B1-B2). Note that this emf vanishes if B1=B2, which makes sense since then there is no change in flux if the magnetic field is the same everywhere as the loop moves to the right. We can see that the expression hv(B1-B2) for the emf justifies a previous statement made in lecture, that a straight conducting bar of length h moving with speed v in a magnetic field of strength B (with h, v, and B mutually perpendicular) acts like a battery of emf hvB. One �battery� corresponds to the vertical left side of the loop, with emf hvB1 combines in series with a second battery of smaller emf hvB2 on the right (with positive terminal to positive terminal, negative terminal to negative terminal) so that the total emf is the difference of the two batteries, as it should be by the Kirchoff loop rule. Since the loop is moving, we can immediately deduce the direction of the current around the loop by considering a positive charge in the left vertical side of the loop. This charge is moving with speed v to the right in a magnetic field coming out of the page so feels a Lorentz force qvB1 downward, which implies a CCW current. A positive charge in the right vertical side also feels a downward force but this is weaker since the magnetic field strength B2 < B1, so the direction of the current is that of the stronger force qvB1. We can get the direction of the current also by Lenz�s law. As the loop moves to the right, the number of dots in the loop decreases (more of the loop is in the low field region) which means that an induced current must appear whose new magnetic field increases the number of dots in the loop. By the right hand rule, this has to be a CCW current. Once Faraday�s law tells us that the emf in the loop is hv(B1-B2), we immediately know the current I in the loop from Ohm�s law, I = emf/R where R is the total resistance of the loop. Then the electrical power dissipated is P = emf I = emf^2/R = h^2 v^2 (B1-B2)^2/R. By energy conservation, this must be the power supplied by pulling the loop with a force F to the right with speed v, mechanical power = Fv, so equating the mechanical power to the electrical power gives us an expression for the force needed to pull this loop at a constant speed to the right: Fv = (emf)^2/R or F = (emf)^2/(vR) = h^2 (B1-B2)^2 v/R. This expression is none other than the force IhB1 needed to pull the left vertical bar of the loop carrying current I through the field B1 minus the force IhB2 needed to pull the right vertical bar of the loop also carrying current I through the field B2 to the right. The net result is that we see that Faraday�s law based on a time-varying flux gives exactly the same answer for the current and total force that we would get if we didn�t know about fluxes and just looked at forces acting directly on charges in the loops that are moving with speed v in the presence of two magnetic fields. The advantage of Faraday�s law is that it generalizes to cases where the magnetic field is changing without any motion of the loop or charges and gives a unifying picture that takes the relative motion of charges and magnetic fields into account.

26. Worked Example of Faraday�s Law: Electrical Power from Falling Bar Bar falling with speed v causes positive charges in the bar to move downward with speed v, which causes the charges to start moving with Lorentz force qvB to the left (by the right hand rule for v cross B), so a CCW current appears in this circuit. As we have discussed before, since the bar is moving through a magnetic field perpendicular to its motion, the bar acts like an emf source of magnitude emf = vLB (where L is the length of the bar between the vertical conducting frictionless rails) so a current with magnitude I = emf/R = vLB/R appears, where R is the total resistance of the circuit (here the light bulb) and the light bulb lights up. The current is CCW as discussed above. The current I flowing through the falling bar in the presence of the magnetic field coming out of the page causes a new force F_mag to appear pointing upward, F_mag=ILB=L^2 B^2 v/R so the total force on the bar is �upward magnetic force� minus �downward gravitational force mg� or F = L^2 B^2 v/R � mg. The total force starts at �mg and increases (becomesmore positive) as the bar falls faster and faster until the magnetic force upward exactly balances the downward gravitational force and the bar then attains a constant terminal speed given by setting the total force to zero: L^2 B^2 v/R = mg or the terminal speed of the falling bar is v = Rmg/(L^2 B^2). It is insightful to solve the entire problem a second time using Faraday�s and Lentz�s laws. If the height of the bar above the ground is denoted by the positive coordinate x, then dx/dt = -v where the instantaneous speed v is considered a positive number. The flux Phi of the magnetic field through the circuit consisting of the bar, vertical rails, and light bulb is BA=B(Lx) so the emf in the circuit, by Faraday�s law, is emf = -dPhi/dt = vLB, exactly the expression we got above so the rest of the analysis regarding currents and terminal speeds holds as before. As the bar falls, the flux Phi decreases because the area of the loop is decreasing (there are fewer and fewer dots of magnetic field line passing through the loop). By Lentz�s law, an induced current in the loop must appear whose own magnetic field must increase the number of dots (increase the flux). The right hand rule with thumb pointing along the direction of the current tells us that this must be a CCW current, in accord with what we deduced by just considering the Lorentz forces acting on moving positive charges. In general, I recommend that you deduce the magnitude of the emf from Faraday�s law without worrying about the sign, then use Lentz�s law or direct arguments based on the motion of microscopic charges to deduce the direction of the current, which then in turn gives the us �direction� of the emf.Bar falling with speed v causes positive charges in the bar to move downward with speed v, which causes the charges to start moving with Lorentz force qvB to the left (by the right hand rule for v cross B), so a CCW current appears in this circuit. As we have discussed before, since the bar is moving through a magnetic field perpendicular to its motion, the bar acts like an emf source of magnitude emf = vLB (where L is the length of the bar between the vertical conducting frictionless rails) so a current with magnitude I = emf/R = vLB/R appears, where R is the total resistance of the circuit (here the light bulb) and the light bulb lights up. The current is CCW as discussed above. The current I flowing through the falling bar in the presence of the magnetic field coming out of the page causes a new force F_mag to appear pointing upward, F_mag=ILB=L^2 B^2 v/R so the total force on the bar is �upward magnetic force� minus �downward gravitational force mg� or F = L^2 B^2 v/R � mg. The total force starts at �mg and increases (becomesmore positive) as the bar falls faster and faster until the magnetic force upward exactly balances the downward gravitational force and the bar then attains a constant terminal speed given by setting the total force to zero: L^2 B^2 v/R = mg or the terminal speed of the falling bar is v = Rmg/(L^2 B^2). It is insightful to solve the entire problem a second time using Faraday�s and Lentz�s laws. If the height of the bar above the ground is denoted by the positive coordinate x, then dx/dt = -v where the instantaneous speed v is considered a positive number. The flux Phi of the magnetic field through the circuit consisting of the bar, vertical rails, and light bulb is BA=B(Lx) so the emf in the circuit, by Faraday�s law, is emf = -dPhi/dt = vLB, exactly the expression we got above so the rest of the analysis regarding currents and terminal speeds holds as before. As the bar falls, the flux Phi decreases because the area of the loop is decreasing (there are fewer and fewer dots of magnetic field line passing through the loop). By Lentz�s law, an induced current in the loop must appear whose own magnetic field must increase the number of dots (increase the flux). The right hand rule with thumb pointing along the direction of the current tells us that this must be a CCW current, in accord with what we deduced by just considering the Lorentz forces acting on moving positive charges. In general, I recommend that you deduce the magnitude of the emf from Faraday�s law without worrying about the sign, then use Lentz�s law or direct arguments based on the motion of microscopic charges to deduce the direction of the current, which then in turn gives the us �direction� of the emf.

27. Noninvasive Brain Studies: SQUID MEGSuperconducting Quantum Interference Device (SQUID)Magnetoencephalography (MEG)