Physics Solution Manual for 1100 and 2101

16 b the resonant frequency f0 is given by f0 1 2 lc

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Unformatted text preview: of turns per unit length is the ratio of the total number N of turns to the length l : N n= (2) l Therefore, changing the length l of the solenoid also changes the number n of turns per unit length, since the total number N of turns is unchanged. Using Equations (1) and (2), we will derive an expression for the length of the solenoid in terms of its inductance L and the constant quantities N, µ0, and A. SOLUTION Substituting Equation (2) into Equation (1) and simplifying, we obtain 2 µ0 N 2 A N L = µ 0 n l A = µ 0 lA = l l 2 (3) Solving Equation (3) for the length of the solenoid yields l= µ0 N 2 A (4) L When the solenoid’s length is l1 , its inductance is L1 = 5.40×10−5 H. Reducing its length to l 2 increases its inductance to L2 = 8.60×10−5 H. The amount ∆l = l1 − l 2 by which the solenoid’s length decreases is, from Equation (4), ∆l = l1 − l 2 = µ0 N 2 A µ0 N 2 A L1 ( − ) L2 1 1 = µ0 N 2 A − L L 1 2 ( ) 1 1 2 = 4π ×10−7 T ⋅ m/A ( 65 ) 9.0 × 10−4 m 2 − = 0.033 m −5 −5 5.40 × 10 H 8.60 × 10 H 57. REASONING AND SOLUTION The energy density in an electric field is 2 Energy density = ε0E /2 (Equation 19.12 with κ = 1), and the energy density in a magnetic field is Energy density = B2/(2µ 0) (Equation 22.11, where ε0 = 8.85 × 10−12 C2/(N·m2) and µ 0 = 4π × 10−7 T·m/A. Equating these expressions and rearranging yields E= B ε 0 µ0 = ( 12 T )( ) = 3.6 × 109 N/C 8.85 × 10−12 C 2 / N ⋅ m 2 4π ×10−7 T ⋅ m/A ______________________________________________________________________________ Chapter 22 Problems 1227 58. REASONING The emf due to self-induction is given by ξ = − L ( ∆I / ∆t ) (Equation 22.9), where L is the self-inductance, and ∆I /∆t is the rate at which the current changes. Both ∆I and ∆t are known. We can find the self-inductance of a toroid by starting with that of a long solenoid, which we obtained in Example 13. SOLUTION The emf induced in the toroid is given by ∆I ∆t ξ = −L (22.9) As obtained in Example 13, the self-inductance of a long solenoid is L = µ0 n 2 Al , where l is the length of the solenoid, n is the number of turns per unit length, and A is the crosssectional area of the solenoid. A toroid is a solenoid that is bent to form a circle of radius R. The length l of the toroid is the circumference of the circle, l = 2π R. Substituting this expression for l into the equation for L, we obtain L = µ0 n 2 Al = µ0 n 2 A( 2π R ) Substituting this relation for L into Equation (22.9) yields ξ = −L ( ∆It ) = −µ n A( 2π R ) ( ∆It ) ∆ ∆ = − ( 4π × 10 2 0 −7 T ⋅ m/A )( 2400 m ) (1.0 ×10−6 m2 ) 2π ( 0.050 m ) −1 2 (1.1 A − 2.5 A ) 0.15 s = 2.1× 10−5 V ______________________________________________________________________________ 59. REASONING AND SOLUTION The mutual inductance is M= N2 Φ2 I1 The flux through loop 2 is µ N I 2 Φ 2 = B1 A2 = 0 1 1 π R2 2R 1 ( ) Then 2 µ0π N1 N 2 R2 N 2Φ 2 N 2 µ0 N1I1 2 = π R2 = I1 I1 2 R1 2 R1 ______________________________________________________________________________ M= ( ) 1228 ELECTROMAGNETIC INDUCTION 60. REASONING AND SOLUTION According to the transformer equation (Equation 22.12), we have V 4320 V Ns = s N p = (21) = 756 Vp 120.0 V ______________________________________________________________________________ 61. REASONING The air filter is connected to the secondary, so that the power used by the air filter is the power provided by the secondary. However, the power provided by the secondary comes from the primary, so the power used by the air filter is also the power delivered by the wall socket to the primary. This power is P = I PVP (Equation 20.15a), where IP is the current in the primary and VP is the voltage provided by the socket, which we know. Although we do not have a value for IP, we do have a value for IS, which is the current in the secondary. We will take advantage of the fact that IP and IS are related according to IS N P = (Equation 22.13). I P NS SOLUTION The power used by the filter is P = I PVP N Solving Equation 22.13 for IP shows that I P = IS S N P expression for the power gives N P = I PVP = IS S N P . Substituting this result into the 50 −3 1 VP = 1.7 × 10 A (120 V ) = 1.0 × 10 W 1 ( ) 62. REASONING Since the secondary voltage (the voltage to charge the batteries) is less than the primary voltage (the voltage at the wall socket), the transformer is a step-down transformer. In a step-down transformer, the voltage across the secondary coil is less than the voltage across the primary coil. However, the current in the secondary coil is greater than the current in the primary coil. Thus, the current that goes through the batteries is greater than the current from the wall socket. If the transformer has negligible resistance, the power delivered to the batteries is equal to the power coming from the wall socket. SOLUTION a. The turns ratio N s / N p is equal to the ratio of secondary voltage to the primary voltage: Chapter 22 Problems N s Vs 9.0 V = = = 1:13 N p Vp 120 V 1229 (22.12) b. The current from the wall socket is given by Equation 22.13: N 1 −3 −2 I p = I s s = 225...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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