9. M = 2 π R µ 0 n. 3. M = π R 2 µ 0 n N correct. 5. M = π R 2 µ 0 n. 8. M = π r 2 µ 0 n N

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1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering points A coil has an inductance of 4.5 mh, and the current through it changes from 0.22 A to 0.7 A in 0.32 s. Find the magnitude of the average induced emf in the coil during this period. Correct answer: 6.75 mv. = I t ) 0.7 A 0.22 A = H) 0.32 s = V = 6.75 mv. ragsdale zdr82) HW9 ditmire 58335) n 2 = µ 0 A l = N/A H m 2 )0.02 m) = turns/m) 2 Therefore, n = turns/m) 2 = turns/m = turns/cm points A long solenoid has inside a coil of fine wire coaxial with it points A small air-core solenoid has a length of 2 cm and a radius of 0.32 cm. The permeability of free space is N/A 2. If the inductance is to be 0.85 mh, how many turns per centimeter whole turns plus fractional turns) are required? Correct answer: turns/cm. et : l = 2 cm = 0.02 m, µ 0 = N/A 2, r = 0.32 cm = m, = 0.85 mh = 0.02 m. and et n be the number of turns per centimeter. The area is A = π r 2 = π m) 2 = m 2. The inductance of the air-core solenoid is = µ 0 n 2 A l I Inside coil has N total turns Outside solenoid has n turns per meter What is the mutual inductance between the solenoid and the inner coil?. M = 2 π µ 0 n N 2. M = 2 π 2 µ 2 0 n N 3. M = π 2 µ 0 n N correct 4. M = π r 2 µ 0 n 5. M = π 2 µ 0 n 6. M = 2 π r 2 µ 2 0 n N 7. M = 2 π r µ 0 n N 8. M = π r 2 µ 0 n N 9. M = 2 π µ 0 n r

2 ragsdale zdr82) HW9 ditmire 58335) 2 0. M = 2 π r µ 0 n The magnetic field of a solenoid is The magnetic flux is so the emf is B = µ 0 n I. Φ B = B A = µ 0 n I)π 2 ), = N d Φ B = π 2 µ 0 n N d I. Because we are interested in the emf in the inner coil, the area to use is the area of the inner coils rather than the solenoid area.) The emf created through a mutual inductance M is = M d I, so we have keywords: M = π 2 µ 0 n N. 004 part of 5) 0.0 points A circuit is set up as shown in the figure. I I 2 S I 2 The switch is closed at t = 0. The current I through the inductor takes the form I = ) e t/τx x where x and τ x are to be determined. Find I immediately after the circuit is closed.. I = 2. I = 3. I = I = 0 correct Before the circuit is closed, no current is flowing. When we have just closed the circuit we are at t = 0 +, a mathematical notation meaning a very short time ǫ after t = 0. Nothing happens in the circuit at t = 0, only immediately after when the switch is, indeed, closed. However, this is just a mathematical detail.) There are two loops in the problem, one with,, 2 and one with,,. So at t = 0 +, the battery wants to drive a current through both loops. The first loop presents no problem; since there is no inductance working against us, a current will immediately be set up. The second loop, however, has an inductor which tries to prevent any change in the current going through it, and so goes up smoothly from I = 0, as can be seen in the given solution just put t = 0 to find I = 0). Therefore, at this instant, the inductor carries no current, and we can neglect it when we find the current through 2. The equivalent resistance is eq = + 2 so, from = eq I we find I 2 = part 2 of 5) 0.0 points Find I 2 immediately after the circuit is closed.. I 2 = 2. I 2 = 2 3. I 2 = correct I 2 = 0 See previous explanation.

3 ragsdale zdr82) HW9 ditmire 58335) part 3 of 5) 0.0 points Find I after the circuit is closed for a long time.. I = 2 2. I = 0 3. I = I = correct When we wait a long time, the current I though the inductor levels out, i.e., d I 0. The voltage over the inductor is V = d I, at all times. Thus V 0 as t, and we can replace the inductor in the figure by a straight wire. When the current now comes from to the junction where I splits into I 2 and I, there is no resistance in the I path but a nonzero resistance 2 in the other. Naturally, the current takes the path with no resistance. Since we do pass through in any case, the equivalent resistance is now. At t = I = I 2 = part 4 of 5) 0.0 points Find I 2 after the circuit is closed for a long time.. I 2 = 2 2. I 2 = I 2 = 4. I 2 = 0 correct See previous explanation. 008 part 5 of 5) 0.0 points Determine the time constant τ x ) ) correct 2 To find τ x, we set up the loop equation through I di = 0 and the loop through 2 I 2 I I) = 0 since I 2 = I I. We see from the second equation that I = + I If we substitute this into the first equation, we find ) + I 2 di + 2 = 0 or ) ) I di = 0 meaning the circuit is equivalent to a one-loop circuit with a battery ) 2 = + 2

4 ragsdale zdr82) HW9 ditmire 58335) 4 and a resistance ) 2 =. + 2 With these notations we see so the time constant is I di = 0, τ x = = τ x = + ) 2 Alternative solution to Part 3: The mathematically minded might realize that an equation of the form di I = 0 has a solution which is composed of a homogeneous solution and a particular solution. The particular solution corresponds to the first term in the I in the problem statement, the homogeneous corresponds to the second term. The homogeneous solution is therefore the only one of relevance to the time constant τ x, and we can disregard any batteries. This provides a simple solution to Part 3: Just ignore the battery and find the equivalent resistance connected to, in this case eq = + 2, so the time constant is τ = = + ) eq 2 as before. Nobody seems to have proven that ignoring the batteries always works, so the previous solution is more reliable. 009 part of 4) 0.0 points Consider the figure shown below. The switch is initially set at position b. There is no charge nor current around the upper loop while at position b. At t = 0 the switch is set to position a. c C S b What is the potential difference V ca V a V c at time t = 0?. V ca = correct 2. V ca = e ) 3. V ca = e 4. V ca = 5. V ca = 6. V ca = e 7. V ca = 0 8. V ca = 9. V ca = e ) 0. V ca = Since at time t = 0 the current is zero, we can consider the potential difference V ca V a V c across the battery, which immediately yields V a V c =, V c V a = +, where V c > V a, therefore V c V a < 0. Another way of approaching this problem is by using the formula V ca V a V c = d I or a t=0.

5 ragsdale zdr82) HW9 ditmire 58335) 5 00 part 2 of 4) 0.0 points The switch is still in position a. What is the power consumption through the resistor at time t = 2 τ, where τ is the time constant of the loop?. P ca = 2 2. P ca = 2 3. P ca = 2 4. P ca = 2 5. P ca = 2 6. P ca = 2 + e /2) 2 e +/2) 2 e /2) 2 correct + e +/2) 2 e +/2) 2 e /2) 2 7. P ca = 2 The equation for the power dissipated is given by Pt ) = It) 2 = 2 ) 2 e t/τ At t = τ, we then have 2 P = 2 e /2) 2. 0 part 3 of 4) 0.0 points After a long time at position a, the switch is set back to position b at time t 2 = 0. What is maximum charge on the capacitor while the switch is at position b? 2. Q max = 2. Q max = 3. Q max = 4. Q max = 2 C 5. Q max = C 6. Q max = 2 7. Q max = 8. Q max = 2 C 9. Q max = C 0. Q max = C correct By conservation of energy we have Q max 2 Solving for Q max yields 2 C = 2 I max 2. Q max = I max C = C. 02 part 4 of 4) 0.0 points At time t 2 = 3 T, where T is the period 4 of the C circuit, the relationship between V b and V c, and the direction of the current through the inductor are given by. V c > V b ; flows from c through to b 2. V c > V b ; flows from b through to c 3. V b = V c ; no current flow 4. V b = V c ; flows from b through to c 5. V b > V c ; no current flow 6. V b > V c ; flows from b through to c 7. V b = V c ; flows from c through to b 8. V b > V c ; flows from c through to b 9. V c > V b ; no current flow correct

6 ragsdale zdr82) HW9 ditmire 58335) 6 et s track the oscillation. At t 2 = 0 the current is at its maximum value flowing leftto-right through the inductor; the capacitor is uncharged. At) one quarter of the pe-, the right plate is fully riod i.e., t 2 = T 4 charged positive. At t 2 = T the current 2 is flowing right-to-left through the inductor while at time t 2 = 3 T, the left plate of the 4 capacitor is fully charged positive. Hence at time t 2 = 3 4 T, I = 0 and V c > V b. 03 part of 3) 0.0 points In the C circuit below the capacitor is charged to its maximum 23.8 µc while the switch S is open. Then the switch is closed at t = 0. µf 2 H Q S Find the energy stored in the inductor at the quater of a period of oscillations in the circuit. Correct answer: µj. I = d q = ω q max sinω t + δ) with ω = 2 π T =, where T is the period C of oscillation. Solution: When the switch closes, initially there is zero current in the circuit. Thus the solution has the correct phase at t = 0 Now at t = T 4, Thus I = d Q = ω q max sinω t). 2 π I = q 0 ω sin = q 0 ω = A. T T 4 U = 2 I2 = 2 q 0 ω) 2 = U max = U Cmax = 2 C q2 0 = F) C) 2 06 µj J = µj. ) et : t = T 4, C = µf = 0 6 F, = 2 H, and Q 0 = 23.8 µc = C. C Q S Basic concept: Oscillation in C circuit: q = q max cosω t + δ) Observe that at t = T, the magnitude of 4 the current flow through the inductor is at a maximum, and the charge q = q max cos ω t is zero at t = T 4 = π. Thus, at this moment 2 ω there is no energy stored in the capacitor and all the energy is stored in the inductor. At any other time, one can show that for an ideal C circuit where there is no dissipation from resistance, the total energy stored is the sum of the energy in the capacitor and the energy in the inductor, and this energy is constant in time although the energy in each one oscillates). 04 part 2 of 3) 0.0 points

7 ragsdale zdr82) HW9 ditmire 58335) 7 Find total energy at t = T 2. Correct answer: µj. The total energy does not change with respect to time, so U total = U max = U Cmax = µj. 05 part 3 of 3) 0.0 points Consider a new initial condition at t = 0: q 0 = 0, I 0 = +.0 A. et q = q max cosω t+δ), I = d q, where q max is assumed to be positive. Find the new phase angle δ for this new situation.. δ = δ = δ = δ = δ = 270 correct 6. δ = δ = δ = 0 We find the correct δ by fitting the choices into the formulas below: q = q max cosω t + δ) I = d q = ω q max sinω t + δ) At t = 0, q should be 0, while I is positive. emember q max is assumed to be positive. Then only δ = 270 fits all the requirement; i.e., at t = 0 and q = q max cos 270 = 0, keywords: points In an series circuit, an inductor of 4.7 H and a resistor of 7.5 Ω are connected to a 23.5 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. ventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Correct answer: J. et : = 4.7 H, = 7.5 Ω, and = 23.5 V. The current in an circuit is I = e t/). The final equilibrium value of the current, which occurs as t, is I 0 = = 23.5 V 7.5 Ω = A. The energy stored in the inductor carrying a current A is U = 2 I2 = H) A)2 = J points In the circuit below, initially the switch S is in position a and switch S 2 is closed. C 2 C S 2 S b a I = ω q max sin 270 = ω q max > 0.

8 ragsdale zdr82) HW9 ditmire 58335) 8 When the switch S is then thrown to b, the voltage will oscillate at a frequency which is It). the same whether switch S 2 is open or closed. 2. zero, since the circuit was not oscillating to begin with. S b a 3. higher when switch S 2 is open. correct 4. lower when switch S 2 is open. 5. zero, since the battery is no longer in the circuit. The oscillation frequency is ω = C = C + C 2 ). The capacitors are parallel C = C + C 2, thus when switch S 2 is opened the capacitance decreases and its inverse increases. Therefore, the frequency ω is higher when only ON of the capacitors is connected. 08 part of 3) 0.0 points The switch in the circuit in the figure below is initially in position a with no connection to position b. After a very long time, the switch is thrown from position a to position b with no connection to position a) at time t. But, while the switch is moving from position a to position b, both positions are connected to the switch. This allows the current through the inductor to be maintained by the battery while the switch is in the process of being thrown until the switch is thrown all the way to position b. If I 0 = It ) is the current in the inductor at time t = t, what is the current It) as a function of time?. It) = I 0 e t t)/) 2. It) = I 0 e t t)/ t/ 3. It) = I 0 e [ 4. It) = I 0 e t/] t/ 5. It) = I 0 e 6. It) = I 0 e t t)/) 7. It) = I 0 e t t)/ correct [ 8. It) = I 0 e t/] Basic Concepts: circuits. Magnetic energy stored in an inductor: U M = 2 I2. Power dissipated in a resistor: P = I V = I 2 = V 2. For an -circuit, the current after the battery is disconnected is given by It) = I 0 e t t)/τ, where τ = and I 0 is the current after the circuit has been connected to the battery for a long time. I 0 =. The current is given by It) = I 0 e t /τ.

9 ragsdale zdr82) HW9 ditmire 58335) 9 If t = t t, = and d I = I 0 ) e t /τ. τ Thus for the power transferred from the inductor to the rest of the circuit, d U M = I0 2 ) e 2 t /τ τ ) = I0 2 e 2t /τ / = I 2 t). Thus, we get the same results as we found using the simple argument from conservation of energy. The minus sign tells us that energy is being lost from the magnetic field as magnetic energy is converted first into electrical energy and then into heat energy in the resistor. 09 part 2 of 3) 0.0 points What is the instantaneous rate at which the magnetic energy stored in the inductor s magnetic field is converted into electrical energy as a function of time after the switch is thrown from position a to position b?. P diss t) = d I 2. P diss t) = It) d I 3. P diss t) = 2 It) d I 4. P diss t) = I 2 t) correct 5. P diss t) = It) 6. P diss t) = It) 7. P diss t) = 2 I2 t) 8. P diss t) = It) The Simple Way: From conservation of energy, all of the energy which is converted from magnetic energy to electrical energy is dissipated in the resistor. There is nothing in the circuit to store the electrical energy i.e., there are no capacitors). This means that the rate at which magnetic energy is converted into electrical energy is equal to the rate at which electrical energy is converted into heat energy. The rate of change of energy is power. This means that the rate at which magnetic energy is converted into electrical energy is equal to the power dissipated in the resistor, so d U M = P diss t) = I 2 t). The Complicated Way: The energy stored in the magnetic field of the inductor is given by U M t) = 2 I2 t). The rate at which this energy is converted into electrical energy is then d U M = d [ ] 2 I2 t) = I d I. 020 part 3 of 3) 0.0 points What is the total amount of electrical energy dissipated in the resistor after the switch is thrown to position b?. diss = I 0 d I 2. diss = I 0 d I 3. diss = 2 I 0 t=t t=t 4. diss = 2 I2 0 correct 5. diss = I diss = I 0 7. diss = d I 8. diss = I 0 t=t

10 The Simple Way: Once again, we use the principle of conservation of energy. The initial energy stored in the magnetic field of the inductor is U Mi = 2 I2 0. The final energy stored in the inductor is U Mf = 2 I2 f = 2 0 = 0. The change in the magnetic energy is thus U M = U Mf U Mi = 2 I2 0. The minus sign again tells us that magnetic energy is decreasing as it is converted into electrical energy. From conservation of energy, all of this energy is dissipated in the resistor, so the amount of energy dissipated in the resistor is diss = 2 I2 0. The Complicated Way: The power dissipated in the resistor is P diss t) = I 2 t). From the definition of power, P diss = d diss. We can solve for the energy dissipated by integrating this with respect to time; diss = = t t = I 2 0 = I 2 0 ragsdale zdr82) HW9 ditmire 58335) 0 P diss t) I 2 t) t [ e t t)/τ] 2 t e 2 t t)/τ. et t = t t ; then so =, and diss = I0 2 e 2 t /τ 0 = I0 2 τ )[e ] 2 t /τ 2 0 = ) 2 I2 0 [0 ] = 2 I2 0. This is the same result we got using conservation of energy directly.

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