Physics Solution Manual for 1100 and 2101

13 for 0 and equation 106 for into equation 224 gives

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Unformatted text preview: n, so that Φoutside = 0 Wb for both cases. Moreover, the field is perpendicular to the plane of the coil and has the same magnitude B over the entire area A of the coil, once it has completely entered the field region. Thus, Φinside = BA in both cases, according to Equation 22.2. The time interval required for the coil to enter the field region completely can be expressed as the distance the coil travels divided by the speed at which it is pushed. In part a of the drawing the distance traveled is W, while in part b it is L. Thus, we have ( ∆ t )a = W / v ( 4) ( ∆ t )b = L / v (5) Substituting Equations (3), (4), and (5) into Equations (1) and (2), we find ∆Φ BA = ∆t a W / v ξa = − (6) ∆Φ BA = ∆t b L / v ξb = − (7) Dividing Equation (6) by Equation (7) gives BA ξ L 0.15 V ξb = a = = W /v = = 3.0 or = 0.050 V BA 3.0 3.0 W ξb L/v ______________________________________________________________________________ ξa 26. REASONING An emf is induced in the coil because the magnetic flux through the coil is changing in time. The flux is changing because the angle φ between the normal to the coil and the magnetic field is changing. The amount of induced current is equal to the induced emf divided by the resistance of the coil (see Equation 20.2). According to Equation 20.1, the amount of charge ∆q that flows is equal to the induced current I multiplied by the time interval ∆t = t − t0 during which the coil rotates, or ∆q = I( t − t0). Chapter 22 Problems 1207 SOLUTION According to Equation 20.1, the amount of charge that flows is ∆q = I∆t. The current is related to the emf ξ in the coil and the resistance R by Equation 20.2 as I = ξ /R. The amount of charge that flows can, therefore, be written as ξ ∆q = I ∆t = ∆t R The emf is given by Faraday’s law of electromagnetic induction as BA cos φ − BA cos φ0 ∆Φ =−N ∆t ∆t ξ =−N where we have also used Equation 22.2, which gives the definition of magnetic flux as Φ = BA cos φ . With this emf, the expression for the amount of charge becomes BA cos φ − BA cos φ0 −N ∆t ∆q = I ∆t = R ∆t = − NBA ( cos φ − cos φ0 ) R Solving for the magnitude of the magnetic field yields ( ) − (140 Ω ) 8.5 × 10 C − R ∆q B= = = 0.16 T NA ( cos φ − cos φ0 ) ( 50 ) 1.5 × 10−3 m 2 ( cos 90° − cos 0° ) ( −5 ) ______________________________________________________________________________ 27. SSM WWW REASONING The energy dissipated in the resistance is given by Equation 6.10b as the power P dissipated multiplied by the time t, Energy = Pt. The power, according to Equation 20.6c, is the square of the induced emf ξ divided by the resistance R, 2 P = ξ /R. The induced emf can be determined from Faraday’s law of electromagnetic induction, Equation 22.3. SOLUTION Expressing the energy consumed as Energy = Pt, and substituting in P = ξ 2/R, we find ξ 2t Energy = P t = R The induced emf is given by Faraday’s law as ξ = − N ( ∆Φ/∆t ) , and the resistance R is –2 equal to the resistance per unit length (3.3 × 10 Ω/m) times the length of the circumference of the loop, 2π r. Thus, the energy dissipated is 1208 ELECTROMAGNETIC INDUCTION 2 Φ − Φ0 ∆Φ − N t − t t −N t ∆t 0 = Energy = −2 −2 3.3 × 10 Ω / m 2π r 3.3 × 10 Ω / m 2π r 2 ( ) ( ) 2 2 BA cos φ − B0 A cos φ B − B0 − N − NA cos φ t t − t t t − t0 0 = = −2 −2 3.3 × 10 Ω / m 2π r 3.3 × 10 Ω / m 2π r ( ) ( ) 2 0.60 T − 0 T 2 − (1) π ( 0.12 m ) ( cos 0° ) ( 0.45 s ) 0.45 s − 0 s −2 = = 6.6 × 10 J −2 3.3 × 10 Ω / m 2π ( 0.12 m ) ( ) ______________________________________________________________________________ 28. REASONING The magnitude BI of the magnetic field at the center of a circular coil with N turns and a radius r, carrying a current I is given by BI = N µ0 I 2r (21.6) where µ 0 = 4π × 10 −7 T ⋅ m/A is the permeability of free space. We use the symbol r to denote the radius of the coil in order to distinguish it from the coil’s resistance R. The coil’s induced current I is caused by the induced emf ξ. We will determine the induced current with Ohm’s law (Equation 20.2), the magnitude ξ of the induced emf, and the resistance R of the coil: ξ (20.2) I= R ∆Φ ∆t (Equation 22.3), where ∆Φ/∆t is the rate of change of the flux caused by the external magnetic field B. At any instant, the flux through each turn of the coil due to the external magnetic field is given by The magnitude of the induced emf is found from Faraday’s law: Φ = BA cos φ = BA cos 0o = BA ξ = −N (22.2) where A is the cross-sectional area of the coil. The angle φ between the direction of the external magnetic field and the normal to the plane of the coil is zero, because the external magnetic field is perpendicular to the plane of the coil. Chapter 22 Problems 1209 SOLUTION In this situation, the cross-sectional area A of the coil is constant, but the magnitude B of the external magnetic field is changing. Therefore, substituting ∆Φ Equation 22.2 into ξ = − N (Equation 22.3) yields ∆t ξ = −N ∆ ( BA ) ∆...
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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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