ÐÏà¡±á>þÿ )?AGþÿÿÿ:;<=>€BÑu÷€€€Û€ñM Î t €\ÙE Õ €€NñVRüoûÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿì¥Á#` ø¿4pbjbj¡¡ 8bÃÃaSÿÿÿÿÿÿ¤œœœœtttˆ°o°o°o8èoìÔw„ˆ9Ldy†>ê·:$¸$¸$¸Å¹Å¹Å¹˜šššššš$…híT¾ t»£¹"Å¹»»¾œœ$¸$¸Ç6þÎþÎþÎ»(œ$¸t$¸˜þÎ»˜þÎþÎ~´ÿ8¨ Ìt$¸Xy ôf¡®Ç°o<Á†ìX$tý<9DÄAÂÌÞA°AtÅ¹0õ¹"þÎº3ºáÅ¹Å¹Å¹¾¾ Î^Å¹Å¹Å¹9»»»»ˆˆˆDZÌeä ˆˆˆÌeˆˆˆœœœœœœÿÿÿÿInductance 30.1. Identify and Set Up: Apply Eq.(30.4). Execute: (a) EMBED Equation.DSMT4 yes, it is constant. (b) EMBED Equation.DSMT4 M is a property of the pair of coils so is the same as in part (a). Thus EMBED Equation.DSMT4 Evaluate: The induced emf is the same in either case. A constant EMBED Equation.DSMT4 produces a constant emf. 30.2. Identify: EMBED Equation.DSMT4 and EMBED Equation.DSMT4 EMBED Equation.DSMT4 , where EMBED Equation.DSMT4 is the flux through one turn of the second coil. Set Up: M is the same whether we consider an emf induced in coil 1 or in coil 2. Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 Evaluate: We can express M either in terms of the total flux through one coil produced by a current in the other coil, or in terms of the emf induced in one coil by a changing current in the other coil. 30.3. Identify: Replace units of Wb, A and EMBED Equation.DSMT4 by their equivalents. Set Up: EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 Evaluate: We may use whichever equivalent unit is the most convenient in a particular problem. 30.4. Identify: Changing flux from one object induces an emf in another object. (a) Set Up: The magnetic field due to a solenoid is EMBED Equation.DSMT4 Execute: The above formula gives EMBED Equation.DSMT4 The average flux through each turn of the inner solenoid is therefore EMBED Equation.DSMT4 (b) Set Up: The flux is the same through each turn of both solenoids due to the geometry, so EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (c) Set Up: The induced emf is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 Evaluate: A mutual inductance around EMBED Equation.DSMT4 H is not unreasonable. 30.5. Identify and Set Up: Apply Eq.(30.5). Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 Evaluate: M relates the current in one coil to the flux through the other coil. Eq.(30.5) shows that M is the same for a pair of coils, no matter which one has the current and which one has the flux. 30.6. Identify: A changing current in an inductor induces an emf in it. (a) Set Up: The self-inductance of a toroidal solenoid is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (b) Set Up: The magnitude of the induced emf is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (c) The current is decreasing, so the induced emf will be in the same direction as the current, which is from a to b, making b at a higher potential than a. Evaluate: This is a reasonable value for self-inductance, in the range of a mH. 30.7. Identify: EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Set Up: EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 (b) The average flux through each turn is EMBED Equation.DSMT4 Evaluate: The self-induced emf depends on the rate of change of flux and therefore on the rate of change of the current, not on the value of the current. 30.8. Identify: Combine the two expressions for L: EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Set Up: EMBED Equation.DSMT4 is the average flux through one turn of the solenoid. Execute: Solving for EMBED Equation.DSMT4 we have EMBED Equation.DSMT4 Evaluate: The induced emf depends on the time rate of change of the total flux through the solenoid. 30.9. Identify and Set Up: Apply EMBED Equation.DSMT4 Apply Lenz s law to determine the direction of the induced emf in the coil. Execute: (a) EMBED Equation.DSMT4 (b) Terminal EMBED Equation.DSMT4 is at a higher potential since the coil pushes current through from EMBED Equation.DSMT4 to EMBED Equation.DSMT4 and if replaced by a battery it would have the EMBED Equation.DSMT4 terminal at EMBED Equation.DSMT4 Evaluate: The induced emf is directed so as to oppose the decrease in the current. 30.10. Identify: Apply EMBED Equation.DSMT4 Set Up: The induced emf points from low potential to high potential across the inductor. Execute: (a) The induced emf points from b to a, in the direction of the current. Therefore, the current is decreasing and the induced emf is directed to oppose this decrease. (b) EMBED Equation.DSMT4 so EMBED Equation.DSMT4 In 2.00 s the decrease in i is 8.00 A and the current at 2.00 s is EMBED Equation.DSMT4 Evaluate: When the current is decreasing the end of the inductor where the current enters is at the lower potential. This agrees with our result and with Figure 30.6d in the textbook. 30.11. Identify and Set Up: Use Eq.(30.6) to relate L to the flux through each turn of the solenoid. Use Eq.(28.23) for the magnetic field through the solenoid. Execute: EMBED Equation.DSMT4 If the magnetic field is uniform inside the solenoid EMBED Equation.DSMT4 From Eq.(28.23), EMBED Equation.DSMT4 Then EMBED Equation.DSMT4 Evaluate: Our result is the same as L for a torodial solenoid calculated in Example 30.3, except that the average circumference EMBED Equation.DSMT4 of the toroid is replaced by the length l of the straight solenoid. 30.12. Identify and Set Up: The stored energy is EMBED Equation.DSMT4 The rate at which thermal energy is developed is EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 Evaluate: (c) No. If I is constant then the stored energy U is constant. The energy being consumed by the resistance of the inductor comes from the emf source that maintains the current; it does not come from the energy stored in the inductor. 30.13. Identify and Set Up: Use Eq.(30.9) to relate the energy stored to the inductance. Example 30.3 gives the inductance of a toroidal solenoid to be EMBED Equation.DSMT4 so once we know L we can solve for N. Execute: EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: L and hence U increase according to the square of N. 30.14. Identify: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy. (a) Set Up: The magnetic field inside a toroidal solenoid is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (b) Set Up: The self-inductance of a toroidal solenoid is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (c) Set Up: The energy stored in an inductor is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (d) Set Up: The energy density in a magnetic field is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (e) EMBED Equation.DSMT4 Evaluate: An inductor stores its energy in the magnetic field inside of it. 30.15. Identify: A current-carrying inductor has a magnetic field inside of itself and hence stores magnetic energy. (a) Set Up: The magnetic field inside a solenoid is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (b) Set Up: The energy density in a magnetic field is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (c) Set Up: The total stored energy is U = uV. Execute: EMBED Equation.DSMT4 (d) Set Up: The energy stored in an inductor is EMBED Equation.DSMT4 Execute: Solving for L and putting in the numbers gives EMBED Equation.DSMT4 Evaluate: An inductor stores its energy in the magnetic field inside of it. 30.16. Identify: EMBED Equation.DSMT4 . EMBED Equation.DSMT4 Set Up: EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 Evaluate: A large value of L and a large current would be required, just for one light bulb. Also, the resistance of the inductor would have to be very small, to avoid a large EMBED Equation.DSMT4 rate of electrical energy loss. 30.17. Identify and Set Up: Starting with Eq. (30.9), follow exactly the same steps as in the text except that the magnetic permeability EMBED Equation.DSMT4 is used in place of EMBED Equation.DSMT4 Execute: Using EMBED Equation.DSMT4 and EMBED Equation.DSMT4 gives EMBED Equation.DSMT4 Evaluate: For a given value of B, the energy density is less when EMBED Equation.DSMT4 is larger than EMBED Equation.DSMT4 . 30.18. Identify and Set Up: The energy density (energy per unit volume) in a magnetic field (in vacuum) is given by EMBED Equation.DSMT4 (Eq.30.10). Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: Large-scale energy storage in a magnetic field is not practical. The volume in part (a) is quite large and the field in part (b) would be very difficult to achieve. 30.19. Identify: Apply Kirchhoff s loop rule to the circuit. i(t) is given by Eq.(30.14). Set Up: The circuit is sketched in Figure 30.19. INCLUDEPICTURE "youn.ism.ch.30/FIG30-019.tif" \* MERGEFORMAT EMBED Equation.DSMT4 is positive as the current increases from its initial value of zero. Figure 30.19Execute: EMBED Equation.DSMT4 EMBED Equation.DSMT4 (a) Initially (t = 0), i = 0 so EMBED Equation.DSMT4 EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 (Use this equation rather than Eq.(30.15) since i rather than t is given.) Thus EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 (d) Final steady state means EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: Our results agree with Fig.30.12 in the textbook. The current is initially zero and increases to its final value of EMBED Equation.DSMT4 The slope of the current in the figure, which is di/dt, decreases with t. 30.20. Identify: The current decays exponentially. Set Up: After opening the switch, the current is EMBED Equation.DSMT4 and the time constant is ( = L/R. Execute: (a) The initial current is EMBED Equation.DSMT4 = (6.30 V)/(15.0 ©) = 0.420 A. Now solve for L and put in the numbers. EMBED Equation.DSMT4 (b) ( = L/R = (43.3 mH)/(15.0 ©) = 2.89 ms (c) Solve EMBED Equation.DSMT4 for t, giving EMBED Equation.DSMT4 Evaluate: In less than 5 time constants, the current is only 1% of its initial value. 30.21. Identify: EMBED Equation.DSMT4 with EMBED Equation.DSMT4 The energy stored in the inductor is EMBED Equation.DSMT4 Set Up: The maximum current occurs after a long time and is equal to EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 so EMBED Equation.DSMT4 when EMBED Equation.DSMT4 and EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 EMBED Equation.DSMT4 so EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: EMBED Equation.DSMT4 The time in part (a) is EMBED Equation.DSMT4 and the time in part (b) is EMBED Equation.DSMT4 30.22. Identify: With EMBED Equation.DSMT4 closed and EMBED Equation.DSMT4 open, EMBED Equation.DSMT4 is given by Eq.(30.14). With EMBED Equation.DSMT4 open and EMBED Equation.DSMT4 closed, EMBED Equation.DSMT4 is given by Eq.(30.18). Set Up: EMBED Equation.DSMT4 After EMBED Equation.DSMT4 has been closed a long time, i has reached its final value of EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 and EMBED Equation.DSMT4 EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 and EMBED Equation.DSMT4 EMBED Equation.DSMT4 so EMBED Equation.DSMT4 Evaluate: EMBED Equation.DSMT4 The time in part (b) is EMBED Equation.DSMT4 30.23. Identify: L has units of H and R has units of EMBED Equation.DSMT4 . Set Up: EMBED Equation.DSMT4 Execute: Units of EMBED Equation.DSMT4 units of time. Evaluate: EMBED Equation.DSMT4 is dimensionless. 30.24. Identify: Apply the loop rule. Set Up: In applying the loop rule, go around the circuit in the direction of the current. The voltage across the inductor is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 EMBED Equation.DSMT4 gives EMBED Equation.DSMT4 and EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: EMBED Equation.DSMT4 is negative, so there is a potential rise across the inductor; point c is at higher potential than point b. There is a potential drop across the resistor. 30.25. Identify: Apply the concepts of current decay in an R-L circuit. Apply the loop rule to the circuit. i(t) is given by Eq.(30.18). The voltage across the resistor depends on i and the voltage across the inductor depends on di/dt. Set Up: The circuit with EMBED Equation.DSMT4 closed and EMBED Equation.DSMT4 open is sketched in Figure 30.25a. INCLUDEPICTURE "youn.ism.ch.30/FIG30-025a.tif" \* MERGEFORMAT EMBED Equation.DSMT4 Figure 30.25aConstant current established means EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (a) Set Up: The circuit with EMBED Equation.DSMT4 closed and EMBED Equation.DSMT4 open is shown in Figure 30.25b. INCLUDEPICTURE "youn.ism.ch.30/FIG30-025b.tif" \* MERGEFORMAT EMBED Equation.DSMT4 At EMBED Equation.DSMT4 Figure 30.25bThe inductor prevents an instantaneous change in the current; the current in the inductor just after EMBED Equation.DSMT4 is closed and EMBED Equation.DSMT4 is opened equals the current in the inductor just before this is done. (b) Execute: EMBED Equation.DSMT4 (c) Set Up: See Figure 30.25c. INCLUDEPICTURE "FIG30-025c.tif" \* MERGEFORMAT Figure 30.25c Execute: If we trace around the loop in the direction of the current the potential falls as we travel through the resistor so it must rise as we pass through the inductor: EMBED Equation.DSMT4 So point c is at higher potential than point b. EMBED Equation.DSMT4 Or, EMBED Equation.DSMT4 (d) EMBED Equation.DSMT4 EMBED Equation.DSMT4 Taking natural logs of both sides of this equation gives EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: The current decays, as shown in Fig. 30.13 in the textbook. The time constant is EMBED Equation.DSMT4 The values of t in the problem are less than one time constant. At any instant the potential drop across the resistor (in the direction of the current) equals the potential rise across the inductor. 30.26. Identify: Apply Eq.(30.14). Set Up: EMBED Equation.DSMT4 EMBED Equation.DSMT4 The current is increasing, so EMBED Equation.DSMT4 is positive. Execute: (a) At EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 (b) As EMBED Equation.DSMT4 EMBED Equation.DSMT4 and EMBED Equation.DSMT4 EMBED Equation.DSMT4 (c) When EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: At all times, EMBED Equation.DSMT4 as required by the loop rule. 30.27. Identify: EMBED Equation.DSMT4 is given by Eq.(30.14). Set Up: The power input from the battery is EMBED Equation.DSMT4 The rate of dissipation of energy in the resistance is EMBED Equation.DSMT4 The voltage across the inductor has magnitude EMBED Equation.DSMT4 so the rate at which energy is being stored in the inductor is EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: (d) Note that if we expand the square in part (b), then parts (b) and (c) add to give part (a), and the total power delivered is dissipated in the resistor and inductor. Conservation of energy requires that this be so. 30.28. Identify: An L-C circuit oscillates, with the energy going back and forth between the inductor and capacitor. (a) Set Up: The frequency is EMBED Equation.DSMT4 and EMBED Equation.DSMT4 giving EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (b) Set Up: The energy stored in a capacitor is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 (c) Set Up: The current in the circuit is EMBED Equation.DSMT4 , and the energy stored in the inductor is EMBED Equation.DSMT4 Execute: First find ( and Q. ( = 2Àf = 1.336 ( 104 rad/s. Q = CV = (20.0 ( 10 6 F)(150.0 V) = 3.00 ( 10 3 C Now calculate the current: i = EMBED Equation.DSMT4 (1.336 ( 104 rad/s)(3.00 ( 10–3 C) sin[(1.336 ( 104 rad/s)(1.30 ( 10–3 s)] Notice that the argument of the sine is in radians, so convert it to degrees if necessary. The result is i = EMBED Equation.DSMT4 39.92 A Now find the energy in the inductor: EMBED Equation.DSMT4 Evaluate: At the end of 1.30 ms, nearly all the energy is now in the inductor, leaving very little in the capacitor. 30.29. Identify: The energy moves back and forth between the inductor and capacitor. (a) Set Up: The period is EMBED Equation.DSMT4 Execute: Solving for L gives EMBED Equation.DSMT4 (b) Set Up: The charge on a capacitor is Q = CV. Execute: Q = CV = (7.50 ( 10 9 F)(12.0 V) = 9.00 ( 10 8 C (c) Set Up: The stored energy is U = Q2/2C. Execute: EMBED Equation.DSMT4 (d) Set Up: The maximum current occurs when the capacitor is discharged, so the inductor has all the initial energy. EMBED Equation.DSMT4 EMBED Equation.DSMT4 Execute: Solve for the current: EMBED Equation.DSMT4 Evaluate: The energy oscillates back and forth forever. However if there is any resistance in the circuit, no matter how small, all this energy will eventually be dissipated as heat in the resistor. 30.30. Identify: The circuit is described in Figure 30.14 of the textbook. Set Up: The energy stored in the inductor is EMBED Equation.DSMT4 and the energy stored in the capacitor is EMBED Equation.DSMT4 Initially, EMBED Equation.DSMT4 with EMBED Equation.DSMT4 The period of oscillation is EMBED Equation.DSMT4 Execute: (a) Energy conservation says EMBED Equation.DSMT4 and EMBED Equation.DSMT4 EMBED Equation.DSMT4 The charge on the capacitor is zero because all the energy is in the inductor. (b) From Figure 30.14 in the textbook, EMBED Equation.DSMT4 at EMBED Equation.DSMT4 and at EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 is the maximum charge on the plates. The graphs are sketched in Figure 30.30. q refers to the charge on one plate and the sign of i indicates the direction of the current. Evaluate: If the capacitor is fully charged at EMBED Equation.DSMT4 it is fully charged again at EMBED Equation.DSMT4 but with the opposite polarity. INCLUDEPICTURE "youn.ism.ch.30/FIG30-030.tif" \* MERGEFORMAT Figure 30.30 30.31. Identify and Set Up: The angular frequency is given by Eq.(30.22). q(t) and i(t) are given by Eqs.(30.21) and (30.23). The energy stored in the capacitor is EMBED Equation.DSMT4 The energy stored in the inductor is EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 which rounds to 105 rad/s. The period is given by EMBED Equation.DSMT4 (b) The circuit containing the battery and capacitor is sketched in Figure 30.31. INCLUDEPICTURE "FIG30-031.tif" \* MERGEFORMAT EMBED Equation.DSMT4 EMBED Equation.DSMT4 Figure 30.31(c) EMBED Equation.DSMT4 (d) EMBED Equation.DSMT4 (Eq.30.21) EMBED Equation.DSMT4 EMBED Equation.DSMT4 The minus sign means that the capacitor has discharged fully and then partially charged again by the current maintained by the inductor; the plate that initially had positive charge now has negative charge and the plate that initially had negative charge now has positive charge. (e) EMBED Equation.DSMT4 (Eq.30.23) EMBED Equation.DSMT4 The negative sign means the current is counterclockwise in Figure 30.15 in the textbook. or EMBED Equation.DSMT4 (Eq.30.26) EMBED Equation.DSMT4 which checks. (f) EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: Note that EMBED Equation.DSMT4 This agrees with the total energy initially stored in the capacitor, EMBED Equation.DSMT4 Energy is conserved. At some times there is energy stored in both the capacitor and the inductor. When i = 0 all the energy is stored in the capacitor and when q = 0 all the energy is stored in the inductor. But at all times the total energy stored is the same. 30.32. Identify: EMBED Equation.DSMT4 Set Up: EMBED Equation.DSMT4 is the angular frequency in rad/s and f is the corresponding frequency in Hz. Execute: (a) EMBED Equation.DSMT4 (b) The maximum capacitance corresponds to the minimum frequency. EMBED Equation.DSMT4 Evaluate: To vary f by a factor of three (approximately the range in this problem), C must be varied by a factor of nine. 30.33. Identify: Apply energy conservation and Eqs. (30.22) and (30.23). Set Up: If I is the maximum current, EMBED Equation.DSMT4 . For the inductor, EMBED Equation.DSMT4 . Execute: (a) EMBED Equation.DSMT4 gives EMBED Equation.DSMT4 . (b) EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . (c) EMBED Equation.DSMT4 at EMBED Equation.DSMT4 means EMBED Equation.DSMT4 . EMBED Equation.DSMT4 , so EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . Evaluate: The total energy of the system is EMBED Equation.DSMT4 . At EMBED Equation.DSMT4 , the current is close to its maximum value and most of the system s energy is stored in the inductor. 30.34. Identify: Apply Eq.(30.25). Set Up: EMBED Equation.DSMT4 when EMBED Equation.DSMT4 . EMBED Equation.DSMT4 when EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . Execute: (a) EMBED Equation.DSMT4 . EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 . Evaluate: The value of q calculated in part (b) is less than the maximum value Q calculated in part (a). 30.35. Identify: EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Set Up: EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . Execute: (a) EMBED Equation.DSMT4 EMBED Equation.DSMT4 , since EMBED Equation.DSMT4 . (b) EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 is a constant. Evaluate: Eqs.(30.21) and (30.23) are consistent with conservation of energy in the L-C circuit. 30.36. Identify: Evaluate EMBED Equation.DSMT4 and insert into Eq.(20.20). Set Up: Equation (30.20) is EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 EMBED Equation.DSMT4 . Evaluate: The value of EMBED Equation.DSMT4 depends on the initial conditions, the value of q at EMBED Equation.DSMT4 . 30.37. Identify: The unit of L is H and the unit of C is F. Set Up: EMBED Equation.DSMT4 says EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . Execute: EMBED Equation.DSMT4 . Therefore, LC has units of EMBED Equation.DSMT4 and EMBED Equation.DSMT4 has units of s. Evaluate: Our result shows that EMBED Equation.DSMT4 is dimensionless, since EMBED Equation.DSMT4 . 30.38. Identify: The presence of resistance in an L-R-C circuit affects the frequency of oscillation and causes the amplitude of the oscillations to decrease over time. (a) Set Up: The frequency of damped oscillations is EMBED Equation.DSMT4 . Execute: EMBED Equation.DSMT4 The frequency f is EMBED Equation.DSMT4 . (b) Set Up: The amplitude decreases as A(t) = A0 e (R/2L)t. Execute: Solving for t and putting in the numbers gives: EMBED Equation.DSMT4 (c) Set Up: At critical damping, EMBED Equation.DSMT4 . Execute: EMBED Equation.DSMT4 Evaluate: The frequency with damping is almost the same as the resonance frequency of this circuit ( EMBED Equation.DSMT4 ), which is plausible because the 75-© resistance is considerably less than the 2420 © required for critical damping. 30.39. Identify: Follow the procedure specified in the problem. Set Up: Make the substitutions EMBED Equation.DSMT4 . Execute: (a) Eq. (13.41): EMBED Equation.DSMT4 . This becomes EMBED Equation.DSMT4 , which is Eq.(30.27). (b) Eq. (13.43): EMBED Equation.DSMT4 . This becomes EMBED Equation.DSMT4 , which is Eq.(30.29). (c) Eq. (13.42): EMBED Equation.DSMT4 . This becomes EMBED Equation.DSMT4 , which is Eq.(30.28). Evaluate: Equations for the L-R-C circuit and for a damped harmonic oscillator have the same form. 30.40. Identify: For part (a), evaluate the derivatives as specified in the problem. For part (b) set EMBED Equation.DSMT4 in Eq.(30.28) and set EMBED Equation.DSMT4 in the expression for EMBED Equation.DSMT4 . Set Up: In terms of EMBED Equation.DSMT4 , Eq.(30.28) is EMBED Equation.DSMT4 . Execute: (a) EMBED Equation.DSMT4 . EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 , so EMBED Equation.DSMT4 . (b) At EMBED Equation.DSMT4 , so EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . This gives EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . Evaluate: If EMBED Equation.DSMT4 , then EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . 30.41. Identify: Evaluate Eq.(30.29). Set Up: The angular frequency of the circuit is EMBED Equation.DSMT4 . Execute: (a) When EMBED Equation.DSMT4 (b) We want EMBED Equation.DSMT4 , so EMBED Equation.DSMT4 . This gives EMBED Equation.DSMT4 Evaluate: When R increases, the angular frequency decreases and approaches zero as EMBED Equation.DSMT4 . 30.42. Identify: L has units of H and C has units of F. Set Up: EMBED Equation.DSMT4 . EMBED Equation.DSMT4 says EMBED Equation.DSMT4 EMBED Equation.DSMT4 says EMBED Equation.DSMT4 . Execute: The units of EMBED Equation.DSMT4 are EMBED Equation.DSMT4 Therefore, the unit of EMBED Equation.DSMT4 is EMBED Equation.DSMT4 Evaluate: For Eq.(30.28) to be valid, EMBED Equation.DSMT4 and EMBED Equation.DSMT4 must have the same units, so R and EMBED Equation.DSMT4 must have the same units, and we have shown that this is indeed the case. 30.43. Identify: The emf EMBED Equation.DSMT4 in solenoid 2 produced by changing current EMBED Equation.DSMT4 in solenoid 1 is given by EMBED Equation.DSMT4 The mutual inductance of two solenoids is derived in Example 30.1. For the two solenoids in this problem EMBED Equation.DSMT4 where A is the cross-sectional area of the inner solenoid and l is the length of the outer solenoid. Set Up: EMBED Equation.DSMT4 . Let the outer solenoid be solenoid 1. Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 Evaluate: If current in the inner solenoid changed at 37.5 A/s, the emf induced in the outer solenoid would be EMBED Equation.DSMT4 30.44. Identify: Apply EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Set Up: EMBED Equation.DSMT4 is the flux through one turn. Execute: (a) EMBED Equation.DSMT4 EMBED Equation.DSMT4 . Therefore, EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 EMBED Equation.DSMT4 .Therefore, at EMBED Equation.DSMT4 , EMBED Equation.DSMT4 . The magnitude of the induced emf is 0.230 V. Evaluate: The maximum emf is when EMBED Equation.DSMT4 and at this instant EMBED Equation.DSMT4 30.45. Identify: EMBED Equation.DSMT4 Set Up: During an interval in which the graph of i versus t is a straight line, EMBED Equation.DSMT4 is constant and equal to the slope of that line. Execute: (a) The pattern on the oscilloscope is sketched in Figure 30.45. Evaluate: (b) Since the voltage is determined by the derivative of the current, the V versus t graph is indeed proportional to the derivative of the current graph. INCLUDEPICTURE "FIG30-045.tif" \* MERGEFORMAT Figure 30.45 30.46. Identify: Apply EMBED Equation.DSMT4 Set Up: EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 EMBED Equation.DSMT4 The graphs are given in Figure 30.46. (b) EMBED Equation.DSMT4 since the emf and current are EMBED Equation.DSMT4 out of phase. (c) EMBED Equation.DSMT4 since the emf and current are EMBED Equation.DSMT4 out of phase. Evaluate: The induced emf depends on the rate at which the current is changing. INCLUDEPICTURE "youn.ism.ch.30/FIG30-046.tif" \* MERGEFORMAT Figure 30.46 30.47. Identify: Apply EMBED Equation.DSMT4 to the series and parallel combinations. Set Up: In series, EMBED Equation.DSMT4 and the voltages add. In parallel the voltages are the same and the currents add. Execute: (a) Series: EMBED Equation.DSMT4 but EMBED Equation.DSMT4 for series components so EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . (b) Parallel: Now EMBED Equation.DSMT4 Therefore, EMBED Equation.DSMT4 . But EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Evaluate: Inductors in series and parallel combine in the same way as resistors. 30.48. Identify: Follow the steps outlined in the problem. Set Up: The energy stored is EMBED Equation.DSMT4 . Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 (d) EMBED Equation.DSMT4 (e) EMBED Equation.DSMT4 Evaluate: The magnetic field between the conductors is due only to the current in the inner conductor. 30.49. (a) Identify and Set Up: An end view is shown in Figure 30.49. INCLUDEPICTURE "FIG30-049.tif" \* MERGEFORMAT Apply Ampere s law to a circular path of radius r. EMBED Equation.DSMT4 Figure 30.49Execute: EMBED Equation.DSMT4 EMBED Equation.DSMT4 the current in the inner conductor Thus EMBED Equation.DSMT4 (b) Identify and Set Up: Follow the procedure specified in the problem. Execute: EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 EMBED Equation.DSMT4 (d) Eq.(30.9): EMBED Equation.DSMT4 Part (c): EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: The value of L we obtain from these energy considerations agrees with L calculated in part (d) of Problem 30.48 by considering flux and Eq.(30.6) 30.50. Identify: Apply EMBED Equation.DSMT4 to each solenoid, as in Example 30.3. Use EMBED Equation.DSMT4 to calculate the mutual inductance M. Set Up: The magnetic field produced by solenoid 1 is confined to the space within its windings and is equal to EMBED Equation.DSMT4 . Execute: (a) EMBED Equation.DSMT4 EMBED Equation.DSMT4 . EMBED Equation.DSMT4 EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 . EMBED Equation.DSMT4 Evaluate: If the two solenoids are identical, so that EMBED Equation.DSMT4 , then EMBED Equation.DSMT4 . 30.51. Identify: EMBED Equation.DSMT4 . The self-inductance of a solenoid is found in Exercise 30.11 to be EMBED Equation.DSMT4 . Set Up: The length l of the solenoid is the number of turns divided by the turns per unit length. Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 . If EMBED Equation.DSMT4 is the number of turns per unit length, then EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . For this coil EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . This is not a practical length for laboratory use. Evaluate: The number of turns is EMBED Equation.DSMT4 . The length of wire in the solenoid is the circumference C of one turn times the number of turns. EMBED Equation.DSMT4 . The length of wire is EMBED Equation.DSMT4 . This length of wire will have a large resistance and EMBED Equation.DSMT4 electrical energy loses will be very large. 30.52. Identify: This is an R-L circuit and EMBED Equation.DSMT4 is given by Eq.(30.14). Set Up: When EMBED Equation.DSMT4 , EMBED Equation.DSMT4 . Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 so EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . Evaluate: The current after a long time depends only on R and is independent of L. The value of EMBED Equation.DSMT4 determines how rapidly the final value of i is reached. 30.53. Identify and Set Up: Follow the procedure specified in the problem. EMBED Equation.DSMT4 EMBED Equation.DSMT4 Execute: (a) Eq.(30.9): EMBED Equation.DSMT4 EMBED Equation.DSMT4 Then EMBED Equation.DSMT4 Exercise 30.27 (c): EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 which checks. (b) Exercise 30.27(a): The rate at which the battery supplies energy is EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 (d) Evaluate: EMBED Equation.DSMT4 The energy supplied by the battery equals the sum of the energy stored in the magnetic field of the inductor and the energy dissipated in the resistance of the inductor. 30.54. Identify: This is a decaying R-L circuit with EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . Set Up: EMBED Equation.DSMT4 , EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . The rate at which energy stored in the inductor is decreasing is EMBED Equation.DSMT4 . Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 . EMBED Equation.DSMT4 (c) In the resistor, EMBED Equation.DSMT4 . (d) EMBED Equation.DSMT4 . EMBED Equation.DSMT4 which is the same as part (a). Evaluate: During the decay of the current all the electrical energy originally stored in the inductor is dissipated in the resistor. 30.55. Identify and Set Up: Follow the procedure specified in the problem. 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h×E{5jh×E{5U hòc 5"hâZ)hâZ):OJPJQJ^Jo(hšO3hšO35:j¢Éh×E{h×E{EHúÿUj°YìI h×E{CJUVh×E{hòc jh×E{UjèÆh×E{h×E{EHúÿUj®YìI h×E{CJUVž ¢¤¦®ÐÒ B D F H L N | ~ € ‚ À Ò Ô È Ê òåÝÙÔÙÝÐÃ¶ÝÙÔ¬§š‹¬ÙÝÐ~qÝÙgUÙQhšO3"hâZ)hâZ):OJPJQJ^Jo(hšO3hšO35:jýâh×E{h~§EHæÿUj1›îI h~§CJUVjßh×E{hÏ)5EHìÿUjK¼íI hÏ)CJUV h×E{5jh×E{5UjÚh×E{hÏ)EHèÿUj?¼íI hÏ)CJUVh×E{ hòc 5hòc jh×E{UjŒÕh×E{h{zrEHèÿUj…lìI h{zrCJUVÊ Î Ö Ø ê ð ò `b’”–êìxxxx xx’xÀxÂxÄxÆxÈxØxÚxúxüx yy:y÷ïêàÜ×àÅÜ½¹¬Ÿ½Ü½¹¹ƒ½Ü½¹vi½ÜàÅÜdÜ½¹ hòc 6j˜íh×E{h×E{EHìÿUjÅYìI h×E{CJUVjðêh×E{h×E{EHöÿUjÂYìI h×E{CJUVUj#èh×E{h×E{EHöÿUjÀYìI h×E{CJUVh×E{jh×E{U"hâZ)hâZ):OJPJQJ^Jo( hòc 5hòc hšO3hšO35: h&u¢5homh&u¢5homhòc 5$ation.DSMT4 is the energy stored in the capacitor. The equation is EMBED Equation.DSMT4 Execute: Multiplying by i gives EMBED Equation.DSMT4 EMBED Equation.DSMT4 the second term. EMBED Equation.DSMT4 the third term. EMBED Equation.DSMT4 the rate at which electrical energy is dissipated in the resistance. EMBED Equation.DSMT4 the rate at which the amount of energy stored in the inductor is changing. EMBED Equation.DSMT4 the rate at which the amount of energy stored in the capacitor is changing. Evaluate: The equation says that EMBED Equation.DSMT4 the net rate of change of energy in the circuit is zero. Note that at any given time one of EMBED Equation.DSMT4 is negative. If the current and EMBED Equation.DSMT4 are increasing the charge on the capacitor and EMBED Equation.DSMT4 are decreasing, and vice versa. 30.56. Identify: The energy stored in a capacitor is EMBED Equation.DSMT4 . The energy stored in an inductor is EMBED Equation.DSMT4 . Energy conservation requires that the total stored energy be constant. Set Up: The current is a maximum when the charge on the capacitor is zero and the energy stored in the capacitor is zero. Execute: (a) Initially EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . The total energy stored is EMBED Equation.DSMT4 . (b) The current is maximum when EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . EMBED Equation.DSMT4 so EMBED Equation.DSMT4 . EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . Evaluate: The maximum charge on the capacitor is EMBED Equation.DSMT4 . 30.57. Identify and Set Up: Use EMBED Equation.DSMT4 (energy stored in a capacitor) to solve for C. Then use Eq.(30.22) and EMBED Equation.DSMT4 to solve for the L that gives the desired current oscillation frequency. Execute: EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: f is in Hz and EMBED Equation.DSMT4 is in rad/s; we must be careful not to confuse the two. 30.58. Identify: Apply energy conservation to the circuit. Set Up: For a capacitor EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . For an inductor EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 , so EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 (d) If EMBED Equation.DSMT4 then EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . This gives EMBED Equation.DSMT4 Evaluate: EMBED Equation.DSMT4 for all times. 30.59. Identify: Set EMBED Equation.DSMT4 , where EMBED Equation.DSMT4 . Set Up: The energy density in the magnetic field is EMBED Equation.DSMT4 . Consider volume EMBED Equation.DSMT4 of sunspot material. Execute: The energy density in the sunspot is EMBED Equation.DSMT4 The total energy stored in volume V of the sunspot is EMBED Equation.DSMT4 The mass of the material in volume V of the sunspot is EMBED Equation.DSMT4 EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . The volume divides out, and EMBED Equation.DSMT4 . Evaluate: The speed we calculated is about 30 times smaller than the escape speed. 30.60. Identify: EMBED Equation.DSMT4 is given by Eq.(30.14). Set Up: The graph shows EMBED Equation.DSMT4 at EMBED Equation.DSMT4 and V approaches the constant value of 25 V at large times. Execute: (a) The voltage behaves the same as the current. Since EMBED Equation.DSMT4 is proportional to i, the scope must be across the EMBED Equation.DSMT4 resistor. (b) From the graph, as EMBED Equation.DSMT4 so there is no voltage drop across the inductor, so its internal resistance must be zero. EMBED Equation.DSMT4 . When EMBED Equation.DSMT4 , EMBED Equation.DSMT4 From the graph, EMBED Equation.DSMT4 at EMBED Equation.DSMT4 . Therefore EMBED Equation.DSMT4 . EMBED Equation.DSMT4 gives EMBED Equation.DSMT4 . (c) The graph if the scope is across the inductor is sketched in Figure 30.60. Evaluate: At all times EMBED Equation.DSMT4 . At EMBED Equation.DSMT4 all the battery voltage appears across the inductor since EMBED Equation.DSMT4 . At EMBED Equation.DSMT4 all the battery voltage is across the resistance, since EMBED Equation.DSMT4 . INCLUDEPICTURE "youn.ism.ch.30/FIG30-060.tif" \* MERGEFORMAT Figure 30.60 30.61. Identify and Set Up: The current grows in the circuit as given by Eq.(30.14). In an R-L circuit the full emf initially is across the inductance and after a long time is totally across the resistance. A solenoid in a circuit is represented as a resistance in series with an inductance. Apply the loop rule to the circuit; the voltage across a resistance is given by Ohm s law. Execute: (a) In the R-L circuit the voltage across the resistor starts at zero and increases to the battery voltage. The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the voltage drops, so the oscilloscope is across the solenoid. (b) At EMBED Equation.DSMT4 the current in the circuit approaches its final, constant value. The voltage doesn t go to zero because the solenoid has some resistance EMBED Equation.DSMT4 The final voltage across the solenoid is EMBED Equation.DSMT4 where I is the final current in the circuit. (c) The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current is zero and there is no voltage drop across any of the resistance in the circuit. (d) As EMBED Equation.DSMT4 EMBED Equation.DSMT4 and from the graph EMBED Equation.DSMT4 (the final voltage across the inductor), so EMBED Equation.DSMT4 (e) EMBED Equation.DSMT4 EMBED Equation.DSMT4 where EMBED Equation.DSMT4 includes the voltage across the resistance of the solenoid. EMBED Equation.DSMT4 EMBED Equation.DSMT4 (read approximately from the graph), so EMBED Equation.DSMT4 Then EMBED Equation.DSMT4 Evaluate: At t = 0 there is no current and the 50 V measured by the oscilloscope is the induced emf due to the inductance of the solenoid. As the current grows, there are voltage drops across the two resistances in the circuit. We derived an equation for EMBED Equation.DSMT4 the voltage across the solenoid. At t = 0 it gives EMBED Equation.DSMT4 and at EMBED Equation.DSMT4 it gives EMBED Equation.DSMT4 30.62. Identify: At EMBED Equation.DSMT4 , EMBED Equation.DSMT4 through each inductor. At EMBED Equation.DSMT4 , the voltage is zero across each inductor. Set Up: In each case redraw the circuit. At EMBED Equation.DSMT4 replace each inductor by a break in the circuit and at EMBED Equation.DSMT4 replace each inductor by a wire. Execute: (a) Initially the inductor blocks current through it, so the simplified equivalent circuit is shown in Figure 30.62a. EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . EMBED Equation.DSMT4 since no current flows through it. EMBED Equation.DSMT4 , since the inductor is in parallel with the EMBED Equation.DSMT4 resistor. EMBED Equation.DSMT4 . (b) Long after S is closed, steady state is reached, so the inductor has no potential drop across it. The simplified circuit is sketched in Figure 30.62b. EMBED Equation.DSMT4 . EMBED Equation.DSMT4 ; EMBED Equation.DSMT4 ; EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . Evaluate: Just after the switch is closed the current through the battery is 0.333 A. After a long time the current through the battery is 0.385 A. After a long time there is an additional current path, the equivalent resistance of the circuit is decreased and the current has increased. INCLUDEPICTURE "youn.ism.ch.30/FIG30-062a.tif" \* MERGEFORMAT INCLUDEPICTURE "youn.ism.ch.30/FIG30-062b.tif" \* MERGEFORMAT Figure 30.62 30.63. Identify and Set Up: Just after the switch is closed, the current in each branch containing an inductor is zero and the voltage across any capacitor is zero. The inductors can be treated as breaks in the circuit and the capacitors can be replaced by wires. After a long time there is no voltage across each inductor and no current in any branch containing a capacitor. The inductors can be replaced by wires and the capacitors by breaks in the circuit. Execute: (a) Just after the switch is closed the voltage EMBED Equation.DSMT4 across the capacitor is zero and there is also no current through the inductor, so EMBED Equation.DSMT4 and since EMBED Equation.DSMT4 are also zero. EMBED Equation.DSMT4 reads zero. EMBED Equation.DSMT4 then must equal 40.0 V, and this means the current read by EMBED Equation.DSMT4 is EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 all other ammeters read zero. EMBED Equation.DSMT4 and all other voltmeters read zero. (b) After a long time the capacitor is fully charged so EMBED Equation.DSMT4 The current through the inductor isn t changing, so EMBED Equation.DSMT4 The currents can be calculated from the equivalent circuit that replaces the inductor by a short circuit, as shown in Figure 30.63a. INCLUDEPICTURE "FIG30-063a.tif" \* MERGEFORMAT Figure 30.63a EMBED Equation.DSMT4 EMBED Equation.DSMT4 The voltage across each parallel branch is 40.0 V 24.0 V = 16.0 V EMBED Equation.DSMT4 EMBED Equation.DSMT4 means EMBED Equation.DSMT4 reads 0.160 A. EMBED Equation.DSMT4 means EMBED Equation.DSMT4 reads 0.320 A. EMBED Equation.DSMT4 reads zero. Note that EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 (d) At t = 0 and EMBED Equation.DSMT4 As the current in this branch increases from zero to 0.160 A the voltage EMBED Equation.DSMT4 reflects the rate of change of the current. The graph is sketched in Figure 30.63b. INCLUDEPICTURE "FIG30-063b.tif" \* MERGEFORMAT Figure 30.63b Evaluate: This reduction of the circuit to resistor networks only apply at t = 0 and EMBED Equation.DSMT4 At intermediate times the analysis is complicated. 30.64. Identify: At all times EMBED Equation.DSMT4 . The voltage across the resistor depends on the current through it and the voltage across the inductor depends on the rate at which the current through it is changing. Set Up: Immediately after closing the switch the current thorough the inductor is zero. After a long time the current is no longer changing. Execute: (a) EMBED Equation.DSMT4 so EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . The ammeter reading is EMBED Equation.DSMT4 . (b) After a long time, EMBED Equation.DSMT4 and EMBED Equation.DSMT4 EMBED Equation.DSMT4 and EMBED Equation.DSMT4 The ammeter reading is EMBED Equation.DSMT4 (c) None of the answers in (a) and (b) depend on L so none of them would change. Evaluate: The inductance L of the circuit affects the rate at which current reaches its final value. But after a long time the inductor doesn t affect the circuit and the final current does not depend on L. 30.65. Identify: At EMBED Equation.DSMT4 , EMBED Equation.DSMT4 through each inductor. At EMBED Equation.DSMT4 , the voltage is zero across each inductor. Set Up: In each case redraw the circuit. At EMBED Equation.DSMT4 replace each inductor by a break in the circuit and at EMBED Equation.DSMT4 replace each inductor by a wire. Execute: (a) Just after the switch is closed there is no current through either inductor and they act like breaks in the circuit. The current is the same through the EMBED Equation.DSMT4 resistors and is equal to EMBED Equation.DSMT4 EMBED Equation.DSMT4 (b) After a long time the currents are constant, there is no voltage across either inductor, and each inductor can be treated as a short-circuit . The circuit is equivalent to the circuit sketched in Figure 30.65. EMBED Equation.DSMT4 . EMBED Equation.DSMT4 The voltage across each parallel branch is EMBED Equation.DSMT4 . EMBED Equation.DSMT4 reads EMBED Equation.DSMT4 . EMBED Equation.DSMT4 reads EMBED Equation.DSMT4 . EMBED Equation.DSMT4 reads EMBED Equation.DSMT4 Evaluate: Just after the switch is closed the current through the battery is 0.455 A. After a long time the current through the battery is 0.585 A. After a long time there are additional current paths, the equivalent resistance of the circuit is decreased and the current has increased. INCLUDEPICTURE "youn.ism.ch.30/FIG30-065.tif" \* MERGEFORMAT Figure 30.65 30.66. Identify: Closing EMBED Equation.DSMT4 and simultaneously opening EMBED Equation.DSMT4 produces an L-C circuit with initial current through the inductor of 3.50 A. When the current is a maximum the charge q on the capacitor is zero and when the charge q is a maximum the current is zero. Conservation of energy says that the maximum energy EMBED Equation.DSMT4 stored in the inductor equals the maximum energy EMBED Equation.DSMT4 stored in the capacitor. Set Up: EMBED Equation.DSMT4 , the current in the inductor just after the switch is closed. Execute: (a) EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . (b) When EMBED Equation.DSMT4 is maximum, EMBED Equation.DSMT4 . Evaluate: In the final circuit the current will oscillate. 30.67. Identify: Apply the loop rule to each parallel branch. The voltage across a resistor is given by iR and the voltage across an inductor is given by EMBED Equation.DSMT4 The rate of change of current through the inductor is limited. Set Up: With S closed the circuit is sketched in Figure 30.67a. INCLUDEPICTURE "FIG30-067a.tif" \* MERGEFORMAT The rate of change of the current through the inductor is limited by the induced emf. Just after the switch is closed the current in the inductor has not had time to increase from zero, so EMBED Equation.DSMT4 Figure 30.67aExecute: (a) EMBED Equation.DSMT4 (b) The voltage drops across R, as we travel through the resistor in the direction of the current, so point a is at higher potential. (c) EMBED Equation.DSMT4 EMBED Equation.DSMT4 (d) The voltage rises when we go from b to a through the emf, so it must drop when we go from a to b through the inductor. Point c must be at higher potential than point d. (e) After the switch has been closed a long time, EMBED Equation.DSMT4 Then EMBED Equation.DSMT4 EMBED Equation.DSMT4 Set Up: The rate of change of the current through the inductor is limited by the induced emf. Just after the switch is opened again the current through the inductor hasn t had time to change and is still EMBED Equation.DSMT4 The circuit is sketched in Figure 30.67b. INCLUDEPICTURE "FIG30-067b.tif" \* MERGEFORMAT Execute: The current through EMBED Equation.DSMT4 in the direction b to a. Thus EMBED Equation.DSMT4 EMBED Equation.DSMT4 Figure 30.67b(f) Point where current enters resistor is at higher potential; point b is at higher potential. (g) EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 Then EMBED Equation.DSMT4 As you travel counterclockwise around the circuit in the direction of the current, the voltage drops across each resistor, so it must rise across the inductor and point d is at higher potential than point c. The current is decreasing, so the induced emf in the inductor is directed in the direction of the current. Thus, EMBED Equation.DSMT4 (h) Point d is at higher potential. Evaluate: The voltage across EMBED Equation.DSMT4 is constant once the switch is closed. In the branch containing EMBED Equation.DSMT4 just after S is closed the voltage drop is all across L and after a long time it is all across EMBED Equation.DSMT4 Just after S is opened the same current flows in the single loop as had been flowing through the inductor and the sum of the voltage across the resistors equals the voltage across the inductor. This voltage dies away, as the energy stored in the inductor is dissipated in the resistors. 30.68. Identify: Apply the loop rule to the two loops. The current through the inductor doesn't change abruptly. Set Up: For the inductor EMBED Equation.DSMT4 and EMBED Equation.DSMT4 is directed to oppose the change in current. Execute: (a) Switch is closed, then at some later time EMBED Equation.DSMT4 The top circuit loop: 60.0 EMBED Equation.DSMT4 The bottom loop: EMBED Equation.DSMT4 (b) After a long time: EMBED Equation.DSMT4 and immediately when the switch is opened, the inductor maintains this current, so EMBED Equation.DSMT4 Evaluate: The current through EMBED Equation.DSMT4 changes abruptly when the switch is closed. 30.69. Identify and Set Up: The circuit is sketched in Figure 30.69a. Apply the loop rule. Just after EMBED Equation.DSMT4 is closed, i = 0. After a long time i has reached its final value and di/dt = 0. The voltage across a resistor depends on i and the voltage across an inductor depends on di/dt. INCLUDEPICTURE "youn.ism.ch.30/FIG30-069a.tif" \* MERGEFORMAT Figure 30.69a Execute: (a) At time EMBED Equation.DSMT4 By the loop rule EMBED Equation.DSMT4 so EMBED Equation.DSMT4 ( EMBED Equation.DSMT4 so this potential difference of 36.0 V is across the inductor and is an induced emf produced by the changing current.) (b) After a long time EMBED Equation.DSMT4 so the potential EMBED Equation.DSMT4 across the inductor becomes zero. The loop rule gives EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 Thus EMBED Equation.DSMT4 (Note that EMBED Equation.DSMT4 ) (c) EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 Integrate from t = 0, when i = 0, to t, when EMBED Equation.DSMT4 EMBED Equation.DSMT4 so EMBED Equation.DSMT4 EMBED Equation.DSMT4 Taking exponentials of both sides gives EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Substituting in the numerical values gives EMBED Equation.DSMT4 At EMBED Equation.DSMT4 (agrees with part (a)). At EMBED Equation.DSMT4 (agrees with part (b)). EMBED Equation.DSMT4 EMBED Equation.DSMT4 At EMBED Equation.DSMT4 (agrees with part (a)). At EMBED Equation.DSMT4 (agrees with part (b)). The graphs are given in Figure 30.69b. INCLUDEPICTURE FIG30-069b.tif \* MERGEFORMAT \* MERGEFORMAT Figure 30.69b Evaluate: The expression for i(t) we derived becomes Eq.(30.14) if the two resistors EMBED Equation.DSMT4 and R in series are replaced by a single equivalent resistance EMBED Equation.DSMT4 30.70. Identify: Apply the loop rule. The current through the inductor doesn't change abruptly. Set Up: With EMBED Equation.DSMT4 closed, EMBED Equation.DSMT4 must be zero. Execute: (a) Immediately after EMBED Equation.DSMT4 is closed, the inductor maintains the current EMBED Equation.DSMT4 through EMBED Equation.DSMT4 The loop rule around the outside of the circuit yields EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . (b) After a long time, EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Thus EMBED Equation.DSMT4 , EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . (c) EMBED Equation.DSMT4 , EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . EMBED Equation.DSMT4 The graphs of the currents are given in Figure 30.70. Evaluate: EMBED Equation.DSMT4 is in a loop that contains just EMBED Equation.DSMT4 and EMBED Equation.DSMT4 , so the current through EMBED Equation.DSMT4 is constant. After a long time the current through the inductor isn't changing and the voltage across the inductor is zero. Since EMBED Equation.DSMT4 is zero, the voltage across R must be zero and EMBED Equation.DSMT4 becomes zero. INCLUDEPICTURE FIG30-070.tif \* MERGEFORMAT \* MERGEFORMAT Figure 30.70 30.71. Identify: The current through an inductor doesn't change abruptly. After a long time the current isn't changing and the voltage across each inductor is zero. Set Up: Problem 30.47 shows how to find the equivalent inductance of inductors in series and parallel. Execute: (a) Just after the switch is closed there is no current in the inductors. There is no current in the resistors so there is no voltage drop across either resistor. A reads zero and EMBED Equation.DSMT4 reads 20.0 V. (b) After a long time the currents are no longer changing, there is no voltage across the inductors, and the inductors can be replaced by short-circuits. The circuit becomes equivalent to the circuit shown in Figure 30.71a. EMBED Equation.DSMT4 . The voltage between points EMBED Equation.DSMT4 and EMBED Equation.DSMT4 is zero, so the voltmeter reads zero. (c) Use the results of Problem 30.49 to combine the inductor network into its equivalent, as shown in Figure 30.71b. EMBED Equation.DSMT4 is the equivalent resistance. Eq.(30.14) says EMBED Equation.DSMT4 with EMBED Equation.DSMT4 . EMBED Equation.DSMT4 , EMBED Equation.DSMT4 , EMBED Equation.DSMT4 so EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . The ammeter reads 0.147 A and the voltmeter reads 9.0 V. Evaluate: The current through the battery increases from zero to a final value of 0.267 A. The voltage across the inductor network drops from 20.0 V to zero. INCLUDEPICTURE "youn.ism.ch.30/FIG30-071a.tif" \* MERGEFORMAT INCLUDEPICTURE "FIG30-071b.tif" \* MERGEFORMAT Figure 30.71 30.72. Identify: At steady state with the switch in position 1, no current flows to the capacitors and the inductors can be replaced by wires. Apply conservation of energy to the circuit with the switch in position 2. Set Up: Replace the series combinations of inductors and capacitors by their equivalents. For the inductors use the results of Problem 30.47. Execute: (a) At steady state EMBED Equation.DSMT4 . (b) The equivalent circuit capacitance of the two capacitors is given by EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . The equivalent circuit is sketched in Figure 30.72a. Energy conservation: EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . As shown in Figure 30.72b, the capacitors have their maximum charge at EMBED Equation.DSMT4 . EMBED Equation.DSMT4 Evaluate: With the switch closed the battery stores energy in the inductors. This then is the energy in the L-C circuit when the switch is in position 2. INCLUDEPICTURE "youn.ism.ch.30/FIG30-072a.tif" \* MERGEFORMAT INCLUDEPICTURE "youn.ism.ch.30/FIG30-072b.tif" \* MERGEFORMAT Figure 30.72 30.73. Identify: Follow the steps specified in the problem. Set Up: Find the flux through a ring of height h, radius r and thickness dr. Example 28.19 shows that EMBED Equation.DSMT4 inside the toroid. Execute: (a) EMBED Equation.DSMT4 (b) EMBED Equation.DSMT4 (c) EMBED Equation.DSMT4 Evaluate: EMBED Equation.DSMT4 is the cross-sectional area A of the toroid and a is approximately the radius r, so this result is approximately the same as the result derived in Example 30.3. 30.74. Identify: The direction of the current induced in circuit A is given by Lenz s law. Set Up: When the switch is closed current flows counterclockwise in the circuit on the left, from the positive plate of the capacitor. The current decreases as a function of time, as the charge and voltage of the capacitor decrease. Execute: At loop A the magnetic field from the wire of the other circuit adjacent to A is into the page. The magnetic field of this current is decreasing, as the current decreases. Therefore, the magnetic field of the induced current in A is directed into the page inside A and to produce a magnetic field in this direction the induced current is clockwise. Evaluate: The magnitude of the emf induced in circuit A decreases with time after the switch is closed, because the rate of change of the current in the other circuit decreases. 30.75. (a) Identify and Set Up: With switch S closed the circuit is shown in Figure 30.75a. INCLUDEPICTURE "FIG30-075a.tif" \* MERGEFORMAT Apply the loop rule to loops 1 and 2. Execute: loop 1 EMBED Equation.DSMT4 EMBED Equation.DSMT4 (independent of t)Figure 30.75aloop (2) EMBED Equation.DSMT4 This is in the form of equation (30.12), so the solution is analogous to Eq.(30.14): EMBED Equation.DSMT4 (b) Evaluate: The expressions derived in part (a) give that as EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Since EMBED Equation.DSMT4 at steady-state, the inductance then has no effect on the circuit. The current in EMBED Equation.DSMT4 is constant; the current in EMBED Equation.DSMT4 starts at zero and rises to EMBED Equation.DSMT4 (c) Identify and Set Up: The circuit now is as shown in Figure 30.75b. INCLUDEPICTURE "FIG30-075b.tif" \* MERGEFORMAT Let t = 0 now be when S is opened. At t = 0, EMBED Equation.DSMT4 Figure 30.75bApply the loop rule to the single current loop. Execute: EMBED Equation.DSMT4 (Now EMBED Equation.DSMT4 is negative.) EMBED Equation.DSMT4 Integrate from t = 0, when EMBED Equation.DSMT4 EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Taking exponentials of both sides of this equation gives EMBED Equation.DSMT4 (d) Identify and Set Up: Use the equation derived in part (c) and solve for EMBED Equation.DSMT4 Execute: EMBED Equation.DSMT4 EMBED Equation.DSMT4 We are asked to find EMBED Equation.DSMT4 Use the expression derived in part (c). EMBED Equation.DSMT4 EMBED Equation.DSMT4 gives EMBED Equation.DSMT4 EMBED Equation.DSMT4 EMBED Equation.DSMT4 Then EMBED Equation.DSMT4 (e) Identify and Set Up: Use the expressions derived in part (a). Execute: The current through the light bulb before the switch is opened is EMBED Equation.DSMT4 Evaluate: When the switch is opened the current through the light bulb jumps from 0.0353 A to 0.600 A. Since the electrical power dissipated in the bulb (brightness) depend on EMBED Equation.DSMT4 the bulb suddenly becomes much brighter. 30.76. Identify: Follow the steps specified in the problem. Set Up: The current in an inductor does not change abruptly. Execute: (a) Using Kirchhoff s loop rule on the left and right branches: Left: EMBED Equation.DSMT4 Right: EMBED Equation.DSMT4 (b) Initially, with the switch just closed, EMBED Equation.DSMT4 (c) The substitution of the solutions into the circuit equations to show that they satisfy the equations is a somewhat tedious exercise but straightforward exercise. We will show that the initial conditions are satisfied: EMBED Equation.DSMT4 . EMBED Equation.DSMT4 (d) When does EMBED Equation.DSMT4 first equal zero? EMBED Equation.DSMT4 . EMBED Equation.DSMT4 and EMBED Equation.DSMT4 EMBED Equation.DSMT4 Evaluate: As EMBED Equation.DSMT4 , EMBED Equation.DSMT4 , EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . 30.77. Identify: Apply EMBED Equation.DSMT4 to calculate L. Set Up: In the air the magnetic field is EMBED Equation.DSMT4 . In the liquid, EMBED Equation.DSMT4 Execute: (a) EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . EMBED Equation.DSMT4 (b) Using EMBED Equation.DSMT4 we can find the inductance for any height EMBED Equation.DSMT4 . _______________________________________________________________________________ Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury EMBED Equation.DSMT4 0.63024 H 0.63000 H EMBED Equation.DSMT4 0.63048 H 0.62999 H EMBED Equation.DSMT4 0.63072 H 0.62999 H EMBED Equation.DSMT4 0.63096 H 0.62998 H ________________________________________________________________________________ The values EMBED Equation.DSMT4 have been used. Evaluate: (d) The volume gauge is much better for the liquid oxygen than the mercury because there is an easily detectable spread of values for the liquid oxygen, but not for the mercury. 30.78. Identify: The induced emf across the two coils is due to both the self-inductance of each and the mutual inductance of the pair of coils. Set Up: The equivalent inductance is defined by EMBED Equation.DSMT4 , where EMBED Equation.DSMT4 and i are the total emf and current across the combination. Execute: Series: EMBED Equation.DSMT4 But EMBED Equation.DSMT4 , so EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . Parallel: We have EMBED Equation.DSMT4 and EMBED Equation.DSMT4 , with EMBED Equation.DSMT4 and EMBED Equation.DSMT4 . To simplify the algebra let EMBED Equation.DSMT4 So EMBED Equation.DSMT4 Now solve for EMBED Equation.DSMT4 EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . EMBED Equation.DSMT4 and EMBED Equation.DSMT4 But EMBED Equation.DSMT4 or EMBED Equation.DSMT4 . Substitute EMBED Equation.DSMT4 in EMBED Equation.DSMT4 back into original equation: EMBED Equation.DSMT4 and EMBED Equation.DSMT4 Finally, EMBED Equation.DSMT4 . Evaluate: If the flux of one coil doesn't pass through the other coil, so EMBED Equation.DSMT4 , then the results reduce to those of problem 30.47. 30.79. Identify: Apply Kirchhoff s loop rule to the top and bottom branches of the circuit. Set Up: Just after the switch is closed the current through the inductor is zero and the charge on the capacitor is zero. Execute: EMBED Equation.DSMT4 EMBED Equation.DSMT4 . EMBED Equation.DSMT4 . (b) EMBED Equation.DSMT4 (c) As EMBED Equation.DSMT4 A good definition of a long time is many time constants later. (d) EMBED Equation.DSMT4 Expanding the exponentials like EMBED Equation.DSMT4 EMBED Equation.DSMT4 and EMBED Equation.DSMT4 if we have assumed that EMBED Equation.DSMT4 Therefore: EMBED Equation.DSMT4 (e) At EMBED Equation.DSMT4 (f) We want to know when the current is half its final value. We note that the current EMBED Equation.DSMT4 is very small to begin with, and just gets smaller, so we ignore it and find: EMBED Equation.DSMT4 EMBED Equation.DSMT4 . Evaluate: EMBED Equation.DSMT4 is initially zero and rises to a final value of 1.92 A. 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