Matter & Interactions Chapter 18 Solutions

Transcription

1 Q12: To quote from pg. 769 of the textbook, It is important to keep in mind that although the units of emf are volts, the emf is not a potential difference. Potential difference is a path integral of the electric field made by real charges. The emf is the energy input per unit charge and might in principle be gravitational or nuclear in nature. In the case of a battery, the emf of the battery is produced by chemical energy input per unit charge. Q13: The battery maintains positive surface charge on the positive terminal and negative surface charge on the negative terminal. The electric field within the battery points from the positive terminal toward the negative terminal. Within the wires, the electric field also points away from the positive terminal and toward the negative terminal. However, in a sketch of the circuit, the electric field just inside the battery will be opposite the electric field within the wire just outside the battery. Q14: (a According to conservation of charge (the node rule, i 1 i 2. Thus, i 1 i 2 nau 1 nau 2 E 2 u 1 u 2 E 2 E 2 ( u1 u (b Neglect the potential difference (energy loss across the connecting wires. Apply the loop rule to the circuit with the single bulb (bulb 1, which we will call 0 to distinguish this case from the previous case in part (a. emf E 0 0 The current is i 0 nau 1 E ( 0 emf nau 1 Now apply the loop rule to the circuit with two bulbs. 843

2 emf E 2 0 ( 1 emf E emf emf 4 The current through bulb 1 is i 1 nau 1 nau 1 ( 3 4 emf ut we know that ( emf i 0 nau 1 So, substitution gives i i 0 (c Evidently replacing ulb 2 with a wire increased the current through the battery by a factor of 4/3. Or, to say it in another way, if you originally have a battery and bulb 1 and you connect bulb 2 in series with bulb 1, the current through the battery and bulbs will be 3/4 of the original current. i Cu i filament n Cu A Cu u Cu E Cu n fil A fil u fil E fil The cross-sectional area of a wire is at least 100 times larger than the area of a filament. The mobility of copper is about 10 times larger than tungsten. Thus, the electric field in the copper wire is at least 1000 times smaller than in the tungsten. The electric fields in all the copper wires are the same because i naue and i (and n, A, and u is the same in all the copper wires. Thus, E must also be the same. (d The graph is shown in the figure below. Gray lines are for the circuit with two bulbs (part a. 844

3 Q15: (a No current will flow through the wire(s. The reason is that if you apply the loop rule to the entire circuit, you get emf 1 emf 2 V wire 0 where V wire is the potential difference across the entire long wire, measured from the negative terminal of battery 1 to the negative terminal of battery 2. Since the batteries are identical emf 1 emf 2, so V wire 0. As a result, E 0 and i naue 0 everywhere inside the wire(s. This means that the surface charge density must be uniform along the wires. There is no gradient (i.e. change in the surface charge density. A sketch is shown in the figure below

4 (b The electric field is zero at all locations inside the wire. (c The brightness of the bulb in this case is zero. There is no current through the bulb. If one of the batteries was not connected backwards, then there would be current through the bulb and the bulb would light. (d It is worth doing the experiment to see the results for yourself. Q16: Note that the battery remains the same in all experiments, and according to the oop Rule, emf V where V is the potential difference across the wire. Since V E, then emf E where is the length of the wire. This can be a useful relationship. Answers are shown in Table??. Experiment Effect on current Parameter that changed Double i changes by a factor of 1/2 E; E changes by a factor of 1/2 because increased by 2 and i A. Double A i changes by a factor of 2 A; A changes by a factor of 1/2 and i A. Double i changes by a factor of 1/2 E; Two bulbs in series is like having a single filament that is twice as long. Doubling changes E by 1/2 and i E. Double emf i changes by factor of 2 E; 2emf results in 2E which results in 2i since i E. Q17: The resistor (thin wire and thick connecting wires are in series. Applying conservation of charge (i.e. the Current Node Rule to the resistor and wires leads to the conclusion that the electron current through the thin wire must equal the electron current through the thick wire. Use i neaue and the fact that n and u are properties of the material in order to compare the quantities given in Table??. resistor <,, or > thick wires i R i w n R n w A R < A w u R u w E R > E w v R > v w Q18: (a The electric field is uniform throughout the wire and points away from the higher potential (i.e. positive terminal of the battery and toward the lower potential (i.e. negative terminal of the battery. A sketch of E is shown in the figure below. 846

5 i T N I q T N T N q I ( ( 1 2 (6 mol ions/mol C/ion 0.3 C/s s 44.4 h P33: True: 1 False: 2, 3, 4, 5 In steady state, the currents at and D must be equal so i i D e/s. Use the properties of the thick wire to calculate the electric field inside it. i n A u E E i n A u electrons/s ( electrons/m 3 ( π ( m 2 ( N/C 4 m/s N/C P34: (a Electrons move to the right. Their magnetic field must point toward the top of the page. (b The current divides equally between bulbs 2 and 3 since they are identical. Thus the current through each of them is e/s. (c ( 2 3 < 1 (d i nau E E i nau electrons/s ( electrons/m 3 ( m 2 ( N/C 4 m/s N/C The electric field points to the left. 852

6 P35: (a The needle deflects downward. (b The needle deflects 13 upward. (c i nau electrons/s ( electrons/m 3 ( m 2 ( V/m 4 m/s N/C points to the left. In bulb 3, the electric field is to the right. P36: (a Treat the battery as if it is a capacitor made of two oppositely charged discs separated a distance 4.5 cm. To make a rough estimation, assume that the electric field is nearly uniform between the plates and given by that of a capacitor. E Q/A ε 0 If the electric field is uniform, then the potential difference is V E where is the distance between the terminals and V is equal to the emf of the battery. V Q/A ε 0 Q ε 0A V ( C 2 /(N m 2 π(0.01 m 2 (1.7 V m C This is a very small amount of charge. It is also a very rough approximation because the equation for the electric field within a capacitor was derived for the case of small separation between the discs (compared to the radius of the discs. (b The charge on a piece of tape is on the order of nanocoulombs (10 9 C. The charge on the terminal of a battery is about 10,000 times smaller than the charge on a piece of tape. Thus, the coulomb force between the terminal of the battery and the tape would be 10,000 times smaller than the coulomb force between two identical pieces of tape. This is too small to be noticeable. 853

7 Use this constant to find i 1 in this new circuit. i 1 na 1 u 2 emf ( s s 1 P50: It makes sense that this current is greater than with two identical bulbs because it is easier to push current through a thick filament than a thin filament. (a Apply the loop rule. Call the thin wire 1 and the thick wire 2. 2emf 1 E Apply the node rule. i 1 i 2 na 1 u na 2 ue 2 A 1 A 2 E 2 A 2 A 1 E 2 (0.352 ( E E 2 2E 2 Substitute into the loop equation. 2emf 2E 2 1 E emf E 2 ( E 2 2emf (1.5 V 2(0.5 m m 2.6 V m So, 2E V m. 865

8 (b v ue, so calculate drift speed and t for each wire seperately since v is different for each wire. 5 m/s v 1 (7 10 V/m (5.2 V m m/s t 1 1 v m m/s 1374 s For wire 2, v 2 ue m/s t m m/s 824 s t t 1 + t s s 2200 s ( 1 mm (2200 s 60 s 36.6 min (c The minimum time required to reach steady state is given by the speed of light. Using m, c t t c 0.65 m m/s s This is 2 ns, which is much smaller than the time interval for a mobile electron to travel around the circuit. 866

9 (d The electron current is i 1 na 1 u ( ( m m 5 m/s π ( V/m (5.2 V m s 1 Thus, electrons pass from the thin wire to the thick wire every second. P51: (a The loop rule gives emf E 0 E emf 1.7 V 0.75 m 2.27 V m (b The electric field remains the same. It only depends on V and. However, the current will change. (c i naue. i u so 4u gives 4i. As a result, (3 is true. P52: (a Conventional current flows from the positive terminal of one battery to the negative terminal of the other battery. For the upper compass in the first circuit, conventional current flows south which creates a magnetic field beneath the wire that is east. Thus, the compass points NE. For the lower compass in the first circuit, conventional current flows north which creates a magnetic field beneath the wire that is west. Thus, the compass points NW. For the compass with bulb in the second circuit, the current flows south so the compass is deflected to the NE. A sketch is shown in the figure below. For clarity, angles are shown greater than what they would appear on an actual compass. 867

10 Note that this is the same as for bulb A in the one-bulb circuit. Thus, the compass next to bulb will also deflect 5. Applying the node rule to bulbs and C leads to the conclusion that the current entering the negative terminal of the battery is equal to i i + i C 2i. Since the current through the batteries is twice the current through bulb, then current will create a magnetic field that is twice the magnitude of the magnetic field created by the current through bulb. The compass deflection is given by tan θ wire earth wire, tan(5 ( T T Now, multiply this magnetic field by 2 and solve for θ. This gives tan θ wire earth T T θ 9.9 Note that for small angles tan θ θ, so twice the magnetic field results in approximately twice the deflection angle. (c In general, there is a surface charge gradient that changes from high positive surface charge density at the positive terminal of the battery to high negative surface charge density at the negative terminal of the battery. And at the midpoint of the circuit, the surface charge density will be zero because it is the point of transition from positive surface charge density to negative surface charge density. ecause the electric field in the thin filament of the bulb is very large compared to the connecting wires, then the gradient (i.e. change in surface charge density along the filament will be very large. ecause the electric field in the thick wire is smaller, then the gradient (i.e. change in surface charge density along the thick wire will be very small. For the one-bulb circuit, sketch a few positive charges at the positive side of bulb A and a few negative charges at the negative side of bulb A. Then, sketch a a couple more positive surface charges at the positive terminal of the battery and a couple more negative surface charges at the negative terminal of the other battery. There will be high density positive and negative surface charge the other terminals of the two batteries where they are connected by a single wire. For the two bulb circuit, make a similar sketch. (d Applying the loop rule to the one-bulb circuit gives 2emf E A 0 E A 2emf 2(1.5 V m 750 V/m Since the potential difference across bulb and bulb C is also 3.0 V and since they are identical bulbs, then E E C E A 750 V/m. Since bulbs and C have the same current and potential difference across them as bulb A, then they will have the same brightness. 869

11 A sketch of the compasses is shown in the figure below. The deflection angles in the sketch are exaggerated for greater clarity. Note that conventional current flows from the + terminal of one battery to the terminal of the second battery. As a result, the compass near the + terminal deflects NW since wire is west, and the compass near the terminal deflects NE since wire is east. North 4 deg 19 deg 15 deg 19 deg P54: Since this system takes a very long time to come to static equilibrium, we can consider it to be in a quasi steady state. That is, at any instant we can assume electron current i is the same at all locations in the wire. We are interested in the electron current in the wire a short time t after the connection has been made long enough for the rearrangement of surface charge to have occurred (this takes only nanoseconds, but short enough that the charge on the two spheres has changed very little. Since i naue, and we know n, A, and u, we need to find the electric field E in the wire. There are three sources of charge contributing to E in the wire: the two charged spheres, and the charge on the surface of the wire. Although there is not a lot of charge on the wire, this charge is very close to the wire, and its contribution to E inside the wire is significant. Since we donõt know the exact amount and distribution of charge on the wire, we can t calculate E directly from Coulomb s law. Therefore, we start (as usual from a fundamental principle: V 0 for a round trip path. We choose a path that goes through the wire, where we want to know E, and also through the air, as shown in the figure below. 871

12 Consider the part of the path indicated by a dotted line. Along the sections of the path inside the spheres, the electric field is zero, and hence V is zero. Along the section of the path in the air, if we assume that (1 we are sufficiently far from the wire that the small amount of charge on the wire contributes a negligible electric field, and (2 the charged spheres are sufficiently far from each other that the electric field of one is negligible near the other, then we find that along the dotted line portion of the path: V air V V A 1 ( q 4πε o r Q R So for a round trip path from A back to A, we have: V 0 V wire + V air V wire + 1 ( q 4πε o r Q R Since the system is in a quasi steady state, E must be uniform in the wire, so V wire E (it is positive, since we are going from to A, traveling opposite to the direction of E in the wire. Thus, E 1 4πε o ( q r + Q R and i naue 1 4πε ( q nau o r + Q R Therefore, a number of electrons equal to nau 1 4πε ( q r + Q R o t leave the small sphere in a time t. 872