chapter20-1

# chapter20-1 - Induced Voltages and Inductance 235 8 As...

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PROBLEM SOLUTIONS 20.1 The angle between the direction of the constant F eld and the normal to the plane of the loop is θ = 0°, so Φ B BA == ( ) × ( ) × ( ) −− cos . . 0 5 0 8 01 0 1 21 0 22 T m m ⋅= cos . . 04 8 1 0 4 8 3 ° T m m W b 2 20.2 The magnetic ﬂ ux through the loop is given by Φ B BA = cos , where B is the magnitude of the magnetic F eld, A is the area enclosed by the loop, and is the angle the magnetic F eld makes with the normal to the plane of the loop. Thus, Φ B BA × () cos . . 500 10 200 5 c m 10 m 1 cm 2 2 2 7 100 10 cos . cos θθ Tm 2 (a) When B ur is perpendicular to the plane of the loop, = 0° and Φ B 7 . T m 2 (b) If × ( ) 30 0 1 00 10 30 0 7 .. c o s . °, then T m 2 Φ B ° 866 10 8 T m 2 (c) If × ( ) 90 0 1 00 10 90 0 7 c o s . °, then T m 2 Φ B ° = 0 20.3 Φ B BA B r cos ( )cos θπ 2 where is the angle between the direction of the F eld and the normal to the plane of the loop. (a) If the F eld is perpendicular to the plane of the loop, = 0°, and B r B = ( ) = ×⋅ ( ) Φ πθ π 2 3 2 800 10 012 0 cos . .c o s m 2 ° = 0 177 T (b) If the F eld is directed parallel to the plane of the loop, 90 , and Φ B BA BA = cos cos 90 0 °

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236 Chapter 20 20.4 The magnetic ﬁ eld lines are tangent to the surface of the cylinder, so that no magnetic ﬁ eld lines penetrate the cylindrical surface. The total ﬂ ux through the cylinder is zero . 20.5 (a) Every ﬁ eld line that comes up through the area A on one side of the wire goes back down through area A on the other side of the wire. Thus, the net ﬂ ux through the coil is zero . (b) The magnetic ﬁ eld is parallel to the plane of the coil, so θ = 90 0 .°. Therefore, Φ B BA BA == = cos cos . 90 0 0 ° 20.6 (a) The magnitude of the ﬁ eld inside the solenoid is Bn I N I ( ) μμ π 00 7 41 0 400 0 360 l TmA m . ( ) = 5 00 6 98 10 6 98 3 .. . A T m T (b) The ﬁ eld inside a solenoid is directed perpendicular to the cross-sectional area, so 0 and the ﬂ ux through a loop of the solenoid is Φ B BA B ( ) ( ) × −− cos cos θπ r 2 3 6 98 10 3 00 10 T 2 2 5 0 1 97 10 mT m 2 ( ) cos . ° 20.7 (a) The magnetic ﬂ ux through an area A may be written as Φ B BA B = ( ) = cos component of perpendicular to AA ( ) ⋅ Thus, the ﬂ ux through the shaded side of the cube is Φ Bx =⋅= ( ) ⋅× ( ) 50 25 10 31 10 2 2 3 . T m T m 2 (b) Unlike electric ﬁ eld lines, magnetic ﬁ eld lines always form closed loops, without beginning or end. Therefore, no magnetic ﬁ eld lines originate or terminate within the cube and any line entering the cube at one point must emerge from the cube at some other point. The net ux through the cube, and indeed through any closed surface, is zero . 20.8 ε () = × ΔΦ Δ Δ Δ B t t cos 15 0 16 10 3 2 m ° × = cos 0 120 10 10 10 010 3 4 s V m V 20.9 With the constant ﬁ eld directed perpendicular to the plane of the coil, the ﬂ ux through the coil is Φ B BA BA cos 0° . As the enclosed area increases, the magnitude of the induced emf in the coil is = ( ) × ( ) = ΔΦ Δ Δ Δ B t B A t 030 50 10
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chapter20-1 - Induced Voltages and Inductance 235 8 As...

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