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Physics1C_Chapter30_part1

# Physics1C_Chapter30_part1 - CHAPTER 30 INDUCTANCE Vahé...

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Unformatted text preview: 4/11/10 CHAPTER 30 INDUCTANCE Vahé Peroomian INDUCTANCE   Is there a difference between a straight wire and a a coil made of the same wire placed in a circuit? MUTUAL INDUCTANCE       In a system with two coils, current flowing in first coil produces a B at coil 2 and ΦB through coil 2. If current through coil 1 is changing, flux through coil 2 changes, inducing an emf in coil 2, creating a current in the second coil. The reverse of this is also true. Mutual inductance M is defined by M= N 2 d Φ B 2 N1d Φ B1 = i1 i2 1 4/11/10 MUTUAL INDUCTANCE: UNITS   SI unit of mutual inductance is the henry (1H). 1 H = 1 Wb/A = 1 V . s/A = 1 Ω . S = 1 J/A2 CHAPTER | Inductance In mutual induction, the emf induced in one coil is always proportional to the rate at which the current in the other coil is changing. Although the proportionality constants M12 and M 21 have been treated separately, it can be shown that they are equal. Therefore, with M12 5 M 21 5 M, Equations 32.16 and 32.17 become I e2 5 2M ddt1 and e1 5 2M dI2 dt These two equations are similar in form to Equation 32.1 for the self-induced emf e 5 2L (dI/dt). The unit of mutual inductance is the henry. EXAMPLE 1. Quick Quiz In Figure 32.8, coil 1 is moved closer to coil 2, with the orientation of both coils remaining fixed. Because of this movement, the . b y Braun GmbH, Kronberg An electricutual inductionhasthebasecoils (a) increases, (b) the m toothbrush of a two designed to hold decreases, or (c) is unaffected. toothbrush handle when not in use. When the handle is placed on the base, a changing current in a solenoid inside the base cylinder induces a current in a coil inside the handle. This induced current charges the battery in the E x a m pl e . “Wireless” Battery the base handle. We can modelCharger as a solenoid of length , with NB turns, carrying a current I, and having a crossAn electric toothbrush has a base designed to hold the sectional area A. The toothbrush handle when not in use. As shown in Figure Coil 1 (base) handle coil contains NH 32.9a, the handle has a cylindrical hole that fits loosely NB over a matching cylinder on urns andWhen the handle t the base. completely is placed on the base, a changing current in a solenoid Coil 2 surrounds the abaseinside coil. (handle) inside the base cylinder induces a current in coil Find charges the battery t he handle. This induced current the mutual inducin NH t he handle. tance of the system. We can model the base as a solenoid of length , w ith | Inductance CHAPTER N B turns (Fig. 32.9b), carrying a current I, and having a cross-sectional area A. The handle coil contains N H turns and completely surrounds the base coil. Find the mutual emf a In mutual induction, the induced in one coil is always proportional to the rate b inductance of the system. at which the current in the other coil is changing. Although the proportionality constants M12 and M 21 haveFigure treated separately,(a) This electric toothbrushthey are been . (Example 32.5) it can be shown that uses the mutual induction of solenoids as part of its batterySOLUTION equal. Therefore, with M 5 M 5 M, Equations 32.16 and 32.17 become Conceptualize Be sure you can identify the two coils in the situation and understand that a changing current in one coil induces a current in the second coil. 12 charging system. (b) A coil of N t urns wrapped around the 21 H center of a solenoid of N B t urns. I e2 5 2M ddt1 and e1 5 2M dI2 dt Categorize We will determine the Theseusingequationsdiscussed in this section,Equation 32.1 for the self-induced emf result two concepts are similar in form to so we categorize this example as a substitution problem. e 5 2L (dI/dt). The unit of mutual inductance is the henry. NB Use Equation 30.17 to express the magnetic field in the I B 5 m0 , interior of the base solenoid: Quick Quiz . In Figure 32.8, coil 1 is moved closer to coil 2, with the orientation of both coils 5 NHFBH 5fixed. Because BNH A movement, the remaining NH BA 5 m N of this Find the mutual inductance, noting that the magnetic M I S coil caused by current IB of the two 106 (a) i ) It, NH =04 , NB the the magand f lux F BH t hrough the handle’s uppose mutual induction= (3.00 × coils A/sncreases, (b),decreases, or (c) is netic field of the base coil is BA: 0, l = unaffected. and A = 1.00 × 10-4 m2. =2 0.025 m, EXAMPLE 2 Wireless charging is used in(a number of other .5 µs, what averagesignificant example is the inductive charging a) At time t = 2 “cordless” devices. One magnetic flux through used by some manufacturers of electric cars that avoids direct metal-to-metal contact between the car and the charging each turn of the solenoid in the handle is caused by apparatus. E x a m pl e . current in the base coil? “Wireless” Battery Charger (b) What is Oscillations in in the handle’s coil? . the induced emf an LC Circuit . b y Braun GmbH, Kronberg An electric toothbrush has a base designed to hold the When a capacitor is connected to an inductor as illustrated in Figure 32.10, the toothbrush handle when not in use. As shown in Figure combination is an LC circuit. If the capacitor is initially charged and the switch is Coil 1 (base) 32.9a, the handle has a cylindrical hole that fits loosely NB over a matching cylinder on the base. When the handle is placed on the base, a changing current in a solenoid Coil 2 (handle) inside the base cylinder induces a current in a coil inside t he handle. This induced current charges the battery in NH t he handle. 27819_32_c32_p927-952.indd 936 27819_32_c32_p927-952.indd 936 We can model the base as a solenoid of length , w ith N B turns (Fig. 32.9b), carrying a current I, and having a cross-sectional area A. The handle coil contains N H turns and completely surrounds the base coil. Find the mutual a b inductance of the system. Figure . (Example 32.5) (a) This electric toothbrush uses the mutual induction of solenoids as part of its batterySOLUTION Conceptualize Be sure you can identify the two coils in the situation and understand that a changing current in one coil induces a current in the second coil. 10/6/09 8:50:05 AM 10/6/09 8:50:05 AM charging system. (b) A coil of N H t urns wrapped around the center of a solenoid of N B t urns. Categorize We will determine the result using concepts discussed in this section, so we categorize this example as a substitution problem. NB Use Equation 30.17 to express the magnetic field in the I B 5 m0 , interior of the base solenoid: NBNH NHFBH NH BA Find the mutual inductance, noting that the magnetic 5 5 m0 A M5 I I , f lux F BH t hrough the handle’s coil caused by the magnetic field of the base coil is BA: Wireless charging is used in a number of other “cordless” devices. One significant example is the inductive charging 2 4/11/10 SELF-INDUCTANCE   Self-induced emfs can occur in any circuit with a changing current: self-inductance. SELF-INDUCTANCE     Self-inductance is greatly enhanced if a coil is present in the circuit. Self-inductance or inductance of a coil with N turns is L= N ΦB i INDUCTORS AS CIRCUIT ELEMENTS   Inductor is shown with   Purpose of inductor in a circuit is to oppose variations in the current through the circuit. 3 4/11/10 A QUICK REVIEW OF KIRCHHOFF’S LOOP RULES   Kirchhoff’s junction rule: the algebraic sum of the currents into any junction is zero ∑I = 0 A QUICK REVIEW OF KIRCHHOFF’S LOOP RULES   Kirchhoff’s loop rule: the algebraic sum of the potential differences in any loop, including those associated with emfs and those of resistive elements, must equal zero. ∑V = 0 SIGN CONVENTION FOR KIRCHHOFF’S LOOP RULE     Minus to plus across emf is positive Through resistor in same direction as current is negative. 4 4/11/10 EXAMPLE 26.4 In the circuit shown, a 12-V power supply with unknown internal resistance r is connected to a run-down rechargeable battery with unknown emf ε and internal resistance 1 Ω and to an indicator light bulb of resistance 3 Ω carrying a current of 2 A. The current through the rundown battery is 1 A in the direction shown. Find the unknown current I, the internal resistance r, and the emf ε. INDUCTORS AS CIRCUIT ELEMENTS   Potential across inductor is Vab = Va − Vb = L di dt POTENTIAL DIFFERENCE ACROSS INDUCTOR   Potential difference across inductor depends on the rate of change of the current. 5 4/11/10 EXAMPLE 3 Consider a uniformly wound solenoid having N turns and length l. Assume l is much longer than the radius of the windings and the core of the solenoid is air. (a) Find the inductance of the solenoid. (b) Calculate the inductance of the solenoid if it contains 300 turns, its length is 25.0 cm, and its cross-sectional area is 4.00 cm2. (c) Calculate the self-induced emf in the solenoid if the current it carries decreases at the rate of 50.0 A/s. ENERGY STORED IN AN INDUCTOR   Energy input U needed to establish a final current I (initial current zero) in and inductor of inductance L is given by U=L 1 I ∫0 i di = 2 LI 2 MAGNETIC ENERGY DENSITY   For a toroidal solenoid, magnetic energy density is given by u= B2 2 µ0 (vacuum) u= B2 2µ (material) 6 ...
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