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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 °
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 17

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online