This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: C HAPTER 20 Quick Quizzes 1. (b), (c), (a). At each instant, the magnitude of the induced emf is proportional to the rate of change of the magnetic field (hence, proportional to the slope of the curve shown on the graph). 2. The left wingtip on the west side of the airplane. The magnetic field of the Earth has a downward component in the northern hemisphere. As the airplane flies northward, the right-hand rule indicates that positive charge experiences a force to the left side of the airplane. Thus, the left wingtip becomes positively charged and the right wingtip negatively charged. 3. (b). According to Equation 20.3, because B and v are constant, the emf depends only on the length of the wire moving in the magnetic field. Thus, you want the long dimension moving through the magnetic field lines so that it is perpendicular to the velocity vector. In this case, the short dimension is parallel to the velocity vector. From a more conceptual point of view, you want the rate of change of area in the magnetic field to be the largest, which you do by thrusting the long dimension into the field. 4. (c). In order to oppose the approach of the north pole, the magnetic field generated by the induced current must be directed upward. An induced current directed counterclockwise around the loop will produce a field with this orientation along the axis of the loop. 5. (b). When the iron rod is inserted into the solenoid, the inductance of the coil increases. As a result, more potential difference appears across the coil than before. Consequently, less potential difference appears across the bulb and its brightness decreases. 155 C H A P T E R 2 0 Problem Solutions 20.1 The magnetic flux through the area enclosed by the loop is ( 29 ( 29 ( 29 2 2 cos cos0 0.30 T 0.25 m B BA B r = = &#192; = = 2 2 5 .9 1 0 T m- &#0; 20.2 ( 29 ( 29 cos cos B BA B A = = ( 29 ( 29 2 0.950 T 0.850 T 0.200 m cos 30.0 =- &#0; = 2 2 1 .0 9 1 0 T m- &#0; 20.3 When the loop lies on a horizontal table, the angle between the field and the normal to the plane of the loop is 90.0 62.0 28.0 = &#151;- &#192; = * . The flux through the area enclosed by the loop is then cos B BA = ( 29 ( 29 ( 29 [ ] 4 0.520 10 T 0.150 m 0.25 m cos 28.0- = &#0; &#0; = 6 2 1 .7 2 1 0 T m- &#0; When the loop is mounted vertically on a north wall, 62.0 = &#0; and ( 29 ( 29 ( 29 [ ] 4 0.520 10 T 0.150 m 0.25 m cos62.0 B- = @ &#192; = 7 2 9 .1 5 1 0 T m- &#0; When the loop is mounted vertically on an east wall, 90.0 = &#0; , so cos cos90.0 B BA BA = = &#192; = 20.4 The magnetic field lines are tangent to the surface of the cylinder, so that no magnetic field lines...
View Full Document
- Spring '08