Physics Solution Manual for 1100 and 2101

Equating the two expressions for itotal shows that b

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Unformatted text preview: 4 rad/s) sin 27° = 12 V ______________________________________________________________________________ 76. REASONING AND SOLUTION 1 2 The energy stored in a capacitor is given by 2 Equation 19.11b as CV . The energy stored in an inductor is given by Equation 22.10 as 1 2 L I 2 . Setting these two equations equal to each other and solving for the current I, we get C 3.0 × 10 −6 F V= (35 V ) = 0.86 A L 5.0 × 10 −3 H ______________________________________________________________________________ I= 77. REASONING The magnitude ξ of the average emf induced in the triangle is given by ∆Φ (see Equation 22.3), which is Faraday’s law. This expression can be used ∆t directly to calculate the magnitude of the average emf. Since the triangle is a single-turn coil, the number of turns is N = 1. According to Equation 22.2, the magnetic flux Ф is ξ = −N 1236 ELECTROMAGNETIC INDUCTION Φ = BA cos φ = BA cos 0° = BA (1) where B is the magnitude of the field, A is the area of the triangle, and φ = 0° is the angle between the field and the normal to the plane of the triangle (the magnetic field is perpendicular to the plane of the triangle). It is the change ∆Φ in the flux that appears in Faraday’s law, so that we use Equation (1) as follows: ∆Φ = BA − BA0 = BA where A0 = 0 m2 is the initial area of the triangle just as the bar passes point A, and A is the area after the time interval ∆t has elapsed. The area of a triangle is one-half the base (dAC) times the height (dCB) of the triangle. Thus, the change in flux is ∆Φ = BA = B ( 1 dACdCB ) 2 The base and the height of the triangle are related, according to dCB = dAC tan θ, where θ = 19º. Furthermore, the base of the triangle becomes longer as the rod moves. Since the rod moves with a speed v during the time interval ∆t, the base is dAC = v∆t. With these substitutions the change in flux becomes ∆Φ = B ( 1 d AC dCB ) = B 1 dAC ( d AC tan θ ) = 1 B ( v∆t )2 tan θ 2 2 2 (2) SOLUTION Substituting Equation (2) for the change in flux into Faraday’s law, we find that the magnitude of the induced emf is ∆Φ ξ = −N = −N ∆t 1 2 B ( v∆t ) tan θ 2 ∆t = N 1 Bv 2 ∆t tan θ 2 = (1) 1 ( 0.38 T )( 0.60 m/s ) ( 6.0 s ) tan19° = 0.14 V 2 2 Chapter 22 Problems 1237 78. REASONING AND SOLUTION S a. When the magnet is above the ring its N magnetic field points down through the B induced ring and is increasing as the magnet falls. The induced magnetic field attempts to reduce the increasing field I I and points up. The induced current in B induced the ring is as shown in the drawing at the S right, and then the ring looks like a magnet with its north pole at the top, N repelling the north pole of the falling magnet and retarding its motion. When the magnet is below the ring, its magnetic field still points down through the ring but is decreasing as the magnet falls. The induced magnetic field attempts to bolster the decreasing field and points down. The induced current in the ring is as shown in the drawing, and the ring then looks like a magnet with its north pole at the bottom, attracting the south pole of the falling magnet and retarding its motion. b. The motion of the magnet is unaffected, since no induced current can flow in the cut ring. No induced current means that no induced magnetic field can be produced to repel or attract the falling magnet. ______________________________________________________________________________ 79. SSM REASONING According to Equation 22.3, the average emf ξ induced in a single loop (N = 1) is ξ = −∆Φ / ∆t . Since the magnitude of the magnetic field is changing, the area of the loop remains constant, and the direction of the field is parallel to the normal to the loop, the change in flux through the loop is given by ∆Φ = ( ∆B )A . Thus the magnitude ξ of the induced emf in the loop is given by ξ = − ( ∆B ) A / ∆ t . Similarly, when the area of the loop is changed and the field B has a given value, we find the magnitude of the induced emf to be ξ = − B ( ∆A ) / ∆ t . SOLUTION a. The magnitude of the induced emf when the field changes in magnitude is ∆B 2 –3 = (0.018 m )(0.20 T/s) = 3.6 × 10 V ∆t ξ = −A b. At a particular value of B (when B is changing), the rate at which the area must change can be obtained from B∆A ξ=− ∆t or ∆A ξ 3.6 × 10 –3 V –3 2 = = = 2.0 ×10 m / s ∆t B 1.8 T 1238 ELECTROMAGNETIC INDUCTION In order for the induced emf to be zero, the magnitude of the magnetic field and the area of the loop must change in such a way that the flux remains constant. Since the magnitude of the magnetic field is increasing, the area of the loop must decrease, if the flux is to remain constant. Therefore, the area of the loop must be shrunk . ______________________________________________________________________________ 80. REASONING AND SOLUTION If the rectangle is made ∆I = If wide, then the top of the rectangle will intersect the line at LIf. The work is one-half the area of the rectangle, so W= 1 2 ( LI f ) I f = 1 LI f 2 2 ____________...
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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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