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- Arizona
- Arizona State University - Tempe
- Physics
- Physics 131
- Adams/menendez
- [SOLUTIONS] Mastering Physics HW33.pdf

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HW33 Due: 11:59pm on Monday, November 30, 2009 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View] The Magnetic Field between Capacitor Plates A capacitor consists of two parallel circular plates of radius . The capacitor has capacitance and is being charged in a simple circuit loop. The circuit has an initial current and consists of the capacitor, a battery with voltage , and a resistor with resistance . Part A What is the magnetic field in the middle of the capacitor plates at a distance from the center, as a function of time ? Hint A.1 How to approach the problem Hint not displayed Hint A.2 What is the current in Ampère's law? Hint not displayed Hint A.3 The displacement current of a charging capacitor Hint not displayed Give your answer in terms of , , , , , , and any necessary constants. ANSWER: = Correct Part B Assume that the capacitor has been charging for a long, long time ( ). What is the approximate value of the magnetic field between the plates now? Hint B.1Current as Hint not displayed ANSWER: = 0 Correct T LR Circuit Theory Learning Goal: To review the procedure for setting up and solving equations for determining current growth in an LR circuit connected to a battery Consider an LR circuit as shown in the figure. The battery provides an EMF . The inductor has inductance , and the resistor has resistance . The switch is initially open as shown. At time , the switch is closed. Part A What is the differential equation governing the growth of current in the circuit as a function of time after ? Hint A.1 Find the voltage change across the resistor Hint not displayed Hint A.2 Find the voltage drop across the inductor Hint not displayed Hint A.3 Use Kirchhoff's loop law to sum the voltages Hint not displayed Express the right-hand side of the differential equation for in terms of , , , and . ANSWER: = Correct Part B We know that the current in the circuit is growing. It will approach a steady state after a long time (as tends to infinity), which implies that the differential term in our circuit equation will tend to zero, hence simplifying our equation. The current around the circuit will tend to an asymptotic value . What is ? Express your answer in terms of and and other constants and variables given in the introduction. ANSWER: = Correct Part C Solve the differential equation obtained in Part A for the current as a function of time after the switch is closed at . Hint C.1 Substituting into the differential equation Hint not displayed Hint C.2 A helpful integral Hint not displayed Hint C.3 Solving the differential equation 1: a substitution Hint not displayed Hint C.4 Solving the differential equation 2: new limits Hint not displayed Express your answer in terms of , , and . Use the notation exp(x) for . ANSWER: = Correct The current approaches its asymptotic/steady-state value exponentially, i.e., usually very fast. However, the time constant depends on the actual values of the inductance and resistance. Exercise 26.43 An emf source with a magnitude of = 120 , a resistor with a resistance of = 78.0 , and a capacitor with a capacitance of = 5.10 are connected in series. Part A As the capacitor charges, when the current in the resistor is 0.850 , what is the magnitude of the charge on each plate of the capacitor? ANSWER: = 2.74×10 −4 Correct A Growing Current in an LR Circuit An inductor with an inductance of 2.40 and a resistance of 7.20 is connected to the terminals of a battery with an emf of 6.30 and negligible internal resistance. Part A Find the initial rate of increase of current in the circuit. ANSWER: 2.63 Correct Part B Find the inductor voltage at the instant when the current is 0.470 . ANSWER: 2.92 Correct Part C Find the rate of increase of current at the instant when the current is 0.470 . ANSWER: 1.22 Correct Part D Find the current 0.200 after the circuit is closed. ANSWER: 0.395 Correct Part E Find the final steady-state current. ANSWER: 0.875 Correct Problem 30.58 An L-C circuit consists of an inductor with an inductance of 55.0 and a capacitor with a capacitance of 250 . The initial charge on the capacitor is 6.50 , and the initial current in the inductor is zero. Part A What is the maximum voltage across the capacitor? ANSWER: = 2.60×10 −2 Correct Part B What is the maximum current in the inductor? ANSWER: = 1.75×10 −3 Correct Part C What is the maximum energy stored in the inductor? ANSWER: = 8.45×10 −8 Correct Part D When the current in the inductor has half its maximum value, what is the charge on the capacitor? ANSWER: = 5.63×10 −6 Correct Part E When the current in the inductor has half its maximum value, what is the energy stored in the inductor? ANSWER: = 2.11×10 −8 Correct Problem 30.58 At in an L-C circuit, the capacitor has its maximum charge. It is easy to show that the electric and magnetic energies are equal at where T is the period of the oscillations. Part A What is the next time after when the electric and magnetic energies again are equal? ANSWER: Correct Problem 30.68 In the circuit shown in the figure , the capacitor is originally uncharged. The switch starts in the open position and is then flipped to position 1 for . It is then flipped to position 2 and left there. Part A If the resistance is very small, what is the upper limit for the amount of charge the capacitor could receive? ANSWER: 8.94×10 −4 Correct C Part B Even if is very small, how much electrical energy will be dissipated in it? ANSWER: 2.00×10 −2 Correct J Score Summary: Your score on this assignment is 98.7%. You received 49.36 out of a possible total of 50 points. clockwork MasteringPhysics: Assignment Print View

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