MARCH 11, 2014 REVIEW OF ELECTROMAGNETISM

Reminders about the quiz Remember to print the approved AP equation pages to take to the quiz. Note that Lenz’ Law and the various RHRs aren’t in the equation list. You need to be able to apply these from memory. It’s acceptable to use hand motions for the RHRs during the quiz. Remember to take a hand calculator. If you’re using a laptop, have your AC adapter handy. While taking the quiz, the only application you may have open is WebAssign.

Quiz Advice Review your work, the keys, and my comments for Ch. 22,23 assignments. For additional practice beyond what’s provided in this presentation, see the Chapter Reviews. The practice problems will refresh your memory in using fundamental relationships such as those for the magnetic forces on charged particles and currents and for induced emfs due to changing magnetic flux. The problems will also give you practice in applying RHRs and Lenz’ Law.

More specific advice Review the concept of torque and its application to motors. Torque is the product of a force and its moment arm. The latter is the perpendicular distance from the axis of rotation to the line of action of the force. (See WA E and V17.) Review the function and operation of the electromagnetic devices we’ve studied: motors, generators, transformers. Know what fundamental physical principles they make use of and how. When determining the magnetic force on a moving charge in a magnetic field, start with the general relationship for the magnitude of the force: F mag = |q|vBsin . Note that all symbols in this equation are magnitudes. Use a RHR to determine the direction of the force. In applying Lenz’ Law, remember that this has to do with how a system responds to a change in magnetic flux. In order to have an induced emf, the magnetic flux must change. There are 3 possible changes: change in field strength, change in area covered by the magnetic field, change in angle between the field lines and the normal to the area.

Even more specific WA L15AL, Item 1. This question dealt with the change in current as the magnet was being pushed into the coil. The observation was that as the magnet was pushed in, the current first increased and then decreased to 0. The current was always >= 0 while the magnet was being pushed in. The goal of the problem was to explain the observation. Here is an excellent explanation. “The induced current was proportional to the rate of change of the magnetic flux. The main factor which affected flux in this experiment was the position of the magnet near the coil. The rate of change of the position, which is velocity, was proportional to the rate of change of flux, which relates to the induced current. This shows that the current varied with the velocity, so while the magnet was accelerated into the coil, the induced current rose with the velocity. When the magnet decelerated in the coil, the velocity and current both lowered. When the velocity was once again zero, the current was zero as well.” -B. Dalla Rosa

Follow up on L14 (field of a bar magnet) Now consider the problem in L14 about determining how the net magnetic field changes along the perpendicular bisector of the magnet. To the right is shown two magnetic field lines. First, note how they’re skewed to the right. This is to be expected from the fact that the magnet is oriented perpendicular to the Earth’s magnetic field. At points P and Q, the magnetic field vectors of the earth, B e, and the bar magnet, B bar, are drawn together with the net magnetic field, B net. Note that B e has the same direction and magnitude at both points. This is because there’s negligible change in the Earth’s field over such a small distance (compared to the size of the Earth). On the other hand, B bar decreases significantly with increasing distance from the magnet. As a result, B net not only decreases in magnitude but also gets steeper at increasing distance. One can imagine that at very large distances from the magnet, B bar would be insignificant, and B e = B net.

Follow up on L14, con’t. By the way, the decrease of magnetic field with distance is probably the primary reason why the value of the field that you calculated for the bar magnet was less than the typical value quoted in the text. The way the field of a bar magnet would typically be measured is close to a pole. This field is expected to be strongest here. You can see for yourself that the field lines are dense close to a pole. Of course, another reason for the difference may simply be that your magnet is different than the one the text calls typical.

Follow up on L27 (Genecon) In this lab, the primary observation is that the Genecon becomes harder to turn physically the less electrical resistance there is between the contacts. Here’s an explanation: If you turn at an approximately constant rate, the rate of change of magnetic flux and hence, the induced emf, will be approximately constant. (Faraday’s Law) Since I = V/R with V representing the induced emf, the current will be inversely proportional to the electrical resistance. Now consider the mechanical resistance. With greater current, there will be greater magnetic force on the windings of the armature. (F mag = BILsin  ) This results in a counter torque, which opposes the motion of the armature. Thus, the mechanical resistance to turning depends directly on the current, which is, in turn, inversely proportional to the electrical resistance.

Low voltage here a, v coils Problem 1. Considering just the initial acceleration of the electrons between the two vertical plates, determine an equation for v 0 in terms of e, m, and V 1, where m is the mass of an electron. Solutions for P207

Low voltage here a, v P207, Problem 1, con’t. Strategy: This can be done either as a net force problem (combined with a dvat to determined an expression for the acceleration) or a conservation of energy problem. We choose the latter below. System – electron and vertical plates Initial state – electron with 0 velocity at left plate Final state – electron with speed v 0 at right plate Ext forces – none (gravity ignored) W ext =  E sys 0 =  K +  U el = (1/2)m(v f 2 – v i 2 ) + q  V = (1/2)m(v 0 2 – 0) + q(V f – V i ) = (1/2)mv q(V 1 – 0) = (1/2)mv (-e)V 1 (note the substitution of –e for q)

Low voltage here a, v P207, con’t. Solving for v 0 : v 0 = (2eV 1 /m) 1/2 Problem 2. In order for electrons to move from left-to-right at constant velocity between the horizontal plates, what must the direction of the magnetic field be? Explain your answer. We established in the last WebEx session that the electric force was up. Therefore, the magnetic force must be down to achieve equilibrium. Using the left hand rule with thumb pointing down and the fingers pointing in the direction of the velocity, the fingers curl into the screen. This is the magnetic field direction. Problem 3. Assuming that the electrons move with constant velocity v 0 within the region of electric and magnetic fields in the evacuated tube, determine an equation for e/m in terms of B, V 1, V 2, and d only. This is a net force problem. See Problem 7 of M10c for the solution leading to v 0 = E/B. Substituting E = V 2 /d and the result of Problem 1 for v 0, we obtain e/m = V 2 2 /(2B 2 d 2 V 1 )

Low voltage here a, v P207, con’t. Problem 4. Now assume that the electric field of the horizontal plates is turned off, leaving only the magnetic field. The electrons then move in a circular arc of radius R. Determine an equation for e/m in terms of V 1, B, and R only. This is a net force problem with only a magnetic force. See Problem 7 of M10b leading to the result R = mv 0 /(eB). Solving for e/m and substituting the expression for v 0 from Problem 1, the result is e/m = 2V 1 /(BR) 2.

Low voltage here More problems for practice Next is a series of problems from past AP exams. We’ll use these to review concepts and relationships in electromagnetism.

Low voltage here a, v

Low voltage here a, v This is a conservation of energy problem. With the system being the cart and the Earth, the only external force being the normal force of the plane, which does no work, the initial state being the cart at the highest point and the final state just before the cart enters the field, you can show that v = (2gy 0 ) 1/2.

Low voltage here a, v

Low voltage here a, v i.The magnitude of the induced emf =  /  t =  (BAcos  )/  t. The normal to the side of the cart is parallel to the field, hence,  = 0 o. B is constant. Thus the emf becomes B  A/  t. The area is A = hw, and h is constant in the field. Thus, induced emf = Bh(  w/  t) = Bhv = Bh(2gy 0 ) 1/2. ii.The current is I = (induced emf)/R = Bh(2gy 0 ) 1/2 /R.

Low voltage here a, v

Low voltage here a, v Bhw Bhw/R

Low voltage here a, v

Low voltage here a, v Using the RHR where you grasp the wire with your right hand and your thumb pointing in the direction of the current, your fingers trace the magnetic field lines. For I 1, the field is into the screen at point P. For I 2, the field is out of the screen. The field due to I 1 is stronger since r 1 < r 2. Therefore, the net field is in the same direction at point P as the field of I 1, into the screen.

Low voltage here a, v

Low voltage here a, v B net = B 1 – B 2, where the positive direction is into the screen. Substituting into the given formula, Bnet = (  0 /2  )(I 1 /r 1 – I 2 /r 2 )

Low voltage here a, v

Low voltage here a, v The force that the left wire exerts on the right wire is F 12 = B 1 I 2 Lsin . Note that the field used is that of the wire exerting the force, while the current used is that of wire that the force is being exerted on. The angle between the current and the field is always 90 o ; therefore, F 12 /L = B 1 I 2. Check: The force per unit length that the right wire exerts on the left would be F 21 /L = B 2 I 1 by similar reasoning. By Newton’s 3 rd Law, F 12 = F 21. Therefore B 1 I 2 = B 2 I 1. This equality can be shown to be true using the equation for the magnetic field of a long, straight wire: B =  0 I/(2  r). Since r is the same for both wires, B/I is a constant.

Low voltage here a, v

Low voltage here a, v Consider when the particle is in the magnetic field only. Using the RHR at the point of entry into this region, the magnetic force is down (to the center of the circle) so the thumb points down. Fingers curl into the page in the direction of the field into the screen. The extended fingers point to the left. Since this is opposite the direction of motion, the particles must be negative. x F mag

Low voltage here a, v

Low voltage here a, v E The magnetic force is down between the plates. Therefore the electric force must be up in order to have equilibrium. Since the particles are negative, the electric field is opposite the direction of the electric force.

Low voltage here a, v

Low voltage here a, v F el F mag +y F net = F el – F mag = |q|E - |q|vBsin   0 = E – vB  = V/d – vB  V = vBd = (1.9E6 m/s)(0.20 T)(6.0E-3 m)  = 2300 V

Low voltage here a, v

Low voltage here a, v F mag Take + to the left in the direction of the magnetic force shown on the diagram. Do a net force analysis to determine q/m. F net = F mag mv 2 /r = |q|vBsin  |q|/m = v/(rB) = (1.9E6 m/s)/[(0.10 m)(0.20 T) = 9.5E7 C/kg