Physics Solution Manual for 1100 and 2101

# Physics Solution Manual for 1100 and 2101

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Unformatted text preview: = 0.40 N ⋅ m 2 π f π f π ( 60.0 Hz ) Chapter 22 Problems 49. 1223 SSM REASONING The energy density is given by Equation 22.11 as Energy density= Energy 12 = B Volume 2µ0 The energy stored is the energy density times the volume. SOLUTION The volume is the area A times the height h. Therefore, the energy stored is B 2 Ah (7.0 × 10 –5 T)2 (5.0 ×108 m 2 )(1500 m) = = 1.5 ×109 J –7 2 µ0 2(4π ×10 T ⋅ m/A) ______________________________________________________________________________ Energy = 50. REASONING When the current through an inductor changes, the induced emf ξ is given by Equation 22.9 as ∆I ξ =−L ∆t where L is the inductance, ∆I is the change in the current, and ∆t is the time interval during which the current changes. For each interval, we can determine ∆I and ∆t from the graph. SOLUTION a. ξ =−L 4.0 A − 0 A ∆I = − 3.2 × 10−3 H = −6.4 V −3 ∆t 2.0 × 10 s − 0 s b. ξ =−L ∆I 4.0 A − 4.0 A −3 = − 3.2 × 10 H = 0V −3 −3 ∆t 5.0 × 10 s − 2.0 × 10 s ( ( ) ) ∆I 0 A − 4.0 A = − 3.2 × 10−3 H = +3.2 V −3 −3 ∆t 9.0 × 10 s − 5.0 × 10 s ______________________________________________________________________________ c. ξ =−L ( ) 51. REASONING We will designate the coil containing the current as the primary coil, and the other as the secondary coil. The emf ξs induced in the secondary coil due to the changing current in the primary coil is given by ξs = − M ( ∆I p / ∆t ) (Equation 22.7), where M is the mutual inductance of the two coils, ∆Ip is the change in the current in the primary coil, and ∆t is the change in time. This equation can be used to find the mutual inductance. 1224 ELECTROMAGNETIC INDUCTION SOLUTION Solving Equation 22.7 for the mutual inductance gives M =− ξs ∆t ∆I p =− (1.7 V ) ( 3.7 ×10−2 s ) = 2.5 ×10−2 H ( 0 A − 2.5 A ) ______________________________________________________________________________ 52. REASONING According to Faraday’s law of electromagnetic induction, expressed as ξ = − L ( ∆I /∆t ) (Equation 22.9), an emf is induced in the solenoid as long as the current is changing in time. The amount of electrical energy E stored by an inductor is E = 1 LI (Equation 22.10), 2 where L is the inductance and I is the current. 2 The power P is equal to the energy removed divided by the time t or P = E 1 LI =2 t t (Equation 6.10b). SOLUTION a. The emf induced in the solenoid is ∆I Emf = − L ∆t 0 A − 15 A = +620 V = − ( 3.1 H ) −3 75 × 10 s (22.9) b. The energy stored in the solenoid is E = 1 LI = 2 2 1 2 ( 3.1 H ) (15 A ) 2 = 350 J (22.10) c. The rate (or power, P) at which the energy is removed is 1 ( 3.1 H )(15A ) = 4700 W E 1 LI (6.10b) P= = 2 =2 t t 75 ×10-3 s ______________________________________________________________________________ 2 2 53. REASONING AND SOLUTION The induced emf in the secondary coil is proportional to the mutual inductance. If the primary coil is assumed to be unaffected by the metal, that is ∆I1/∆t is the same for both cases, then New emf = 3(0.46 V) = 1.4 V ______________________________________________________________________________ 2 Chapter 22 Problems 54. REASONING According to ξs = − M 1225 ∆I p (Equation 22.7), a change ∆Ip in the current in ∆t the primary coil induces an emf ξs in the secondary, where M is the mutual inductance of the two coils and ∆t is the time interval of the current change. We are interested only in the magnitude of the current change ∆Ip, so we will omit the minus sign in Equation 22.7. The induced emf ξs in the secondary coil will drive a current Is, as we see from Ohm’s law: ξs = Is R (Equation 20.2), where R is the resistance of the circuit that includes the secondary coil. SOLUTION Omitting the minus sign in ξs = − M yields ∆I p = ∆I p ∆t (Equation 22.7) and solving for ∆Ip ξs ∆t (1) M Substituting ξs = Is R (Equation 20.2) into Equation (1), we obtain ∆I p = 55. SSM ξs ∆t M = Is R∆t M = ( 6.0 ×10−3 A ) (12 Ω ) ( 72 ×10−3 s ) 3.2 ×10−3 H REASONING AND SOLUTION = 1.6 A From the results of Example 13, the self- inductance L of a long solenoid is given by L = µ0 n 2 Al . Solving for the number of turns n per unit length gives n= L 1.4 × 10−3 H = = 4.2 × 103 turns/m −7 −3 2 µ0 Al (4π × 10 T ⋅ m/A)(1.2 × 10 m )(0.052 m) Therefore, the total number of turns N is the product of n and the length l of the solenoid: N = nl = (4.2 × 10 3 turns/m)(0.052 m) = 220 turns ______________________________________________________________________________ 56. REASONING As demonstrated in Example 13, the inductance L of a solenoid of length ℓ and cross-sectional area A is given by L = µ 0 n 2 lA (1) where µ 0 = 4π × 10 −7 T ⋅ m/A is the permeability of free space and n is the number of turns per unit length. It may appear that solving Equation (1) for the length l of the solenoid 1226 ELECTROMAGNETIC INDUCTION would determine l as a function of the inductance L, and therefore solve the problem. However, the number n...
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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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