Physics 212 Spring 2009 Exam 3 Version C (857202)
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1 Page 1 of 6 Physics 212 Spring 2009 Exam 3 Version C (857202) Question Instructions Be sure to answer every question. Follow the rules shown on the first page for filling in the Scantron form. Each problem is worth 10% of the exam. When you are finished, check with Dr. Mike or his TA to be sure you have finished the scantron correctly. 1. Question DetailsMagnetic Flux (2) [ ] A small circular coil of 25 turns of wire lies in a uniform magnetic field of 0.5 T such that the normal to the plane of the coil makes an angle of 60 with the direction of B. The radius of the coil is 4 cm. What is the magnitude of the magnetic flux through the coil? nmlkj v*10*0.031 T m 2 nmlkj z*01*0.063 T m 2 nmlkj w*06*0.001 T m 2 nmlkj x*04*0.054 T m 2 nmlkj y*03*0.250 T m 2 The flux is = N B A cos. 2. Question DetailsMirror Equations and Magnification (3) [750390] Dr. Mike stands in front of a CONVEX mirror. He looks at the image of his eye. It is UPRIGHT and half the size of his real eye. If f is the focal point of the mirror, how far is he standing from the front of the mirror? nmlkj v*10*f nmlkj x*03*3f/8 nmlkj w*03*2f/3 nmlkj z*04*2f nmlkj y*04*f/2 For this situation, the linear magnification is +1/2. Thus, we have +1/2 = - d i / d o or d i = - d o / 2. Using this in the mirror equation, we find f = [1 / d o - 2 / d o ] -1 = - d o. Therefore, d o = - f. 3. Question DetailsFaraday's Law (2) [ ] A magnetic field changes from 2 T to 6 T over a time of 20 seconds. If it passes perpendicularly through a square loop of wire with a 400 cm 2 area and a total resistance of 30 ohms, what is the current in the wire induced by the magnetic field? nmlkj z*02*0.24 A nmlkj x*04*8.00 ma nmlkj v*10*0.27 ma nmlkj w*07*2.67 A
2 Page 2 of 6 nmlkj y*02*0.40 ma The change in flux is / t = N ( B / t) A cos. This is also the EMF. The current is then the EMF divided by the resistance. 4. Question DetailsMirror Equation or Ray Trace (3) [750394] A concave mirror (R = 64 cm) is used to project a transparent slide onto a wall. The slide is located at a distance of 37.0 cm from the mirror, and a small flashlight shines light through the slide and onto the mirror. The setup is similar to that in Figure 25.19a. Figure 25.19a. The height of the object on the slide is 1.10 cm. What is the height of the image? (Make sure you give the proper sign in this case!) nmlkj w*05*7.04 cm nmlkj v*10*-7.04 cm nmlkj z*02*0.17 cm nmlkj x*03*2.61 cm nmlkj y*04*-0.17 cm Since the mirror is concave and spherical, it has a focal length of +R/2 = cm. Also, since the image is in front of the mirror, it is positive as well cm is the object distance since the slide is the object and it is positive too. The image is the projection on the wall, so the distance to the wall from the mirror must be the image distance. Thus, we can use the mirror equation to find d i = (1/f - 1/d o ) -1 = cm. The height of the object on the slide is h o and may be positive or negative depending on whether the object is considered to be right side up or upside down. Normally, we assume the object to be positive in height regardless of its actual orientation. This is why we use a positive number here for the object height. Using the linear magnification equation, we have h i = - h o d i / d o = cm. Note that this is negative because it is inverted and the object height is positive. 5. Question DetailsWhat can we see? (3) [750393] A person 1.65 m tall wants to be able to see her full image in a plane mirror. How far above the floor should the bottom of the mirror be placed so that she can see her feet, assuming that the top of the person's head is 20 cm above her eye level? nmlkj x*03*1.55 m nmlkj w*05*1.025 m nmlkj v*10*0.725 m nmlkj y*06*0.825 m nmlkj z*01*0.2 m
3 Page 3 of 6 This problem is easy if you draw a diagram! The minimum height for the mirror must allow for light from her feet and the top of her head to reach her eyes. A ray diagram will easily show this to be 1/2 her height. If the top of her head is 20 cm above her eyes, then the light from her head as it goes to her eyes must hit the mirror 10 cm below her head (the position of the top of her head is her height.) Thus, the top of the mirror is 10 cm less than her height (1.55 m) and the bottom is half her height below that (0.825 m), or m. 6. Question DetailsRay Trace or Mirror Equation (1) [ ] Do a ray trace for the object and mirror shown here or use the number of blocks to do the mirror equation. Determine which of the following describes the image. nmlkj z*02*it is virtual, inverted and about 3 blocks in height. nmlkj y*02*it is real, upright, and about 3 blocks in height. nmlkj v*10*it is virtual, upright, and about 3 blocks in height. nmlkj x*04*it is virtual, upright, and about 10 blocks in height. nmlkj w*04*it is real, inverted, and about 10 blocks in height. The ray trace is shown below.
4 Page 4 of 6 7. Question DetailsReflection and Mirrors (1) [750389] Bailey-the-Wonderdog sees a reflection of herself in a mirror. When she does so she notices that the dog she sees in the image is smaller than she is. What is the shape of the mirror? nmlkj z*01*there is no mirror that would give this result. nmlkj x*01*flat nmlkj v*10*convex or Concave would both give this result. nmlkj y*05*convex nmlkj w*05*concave There are cases where this is true for both types of mirror. For instance, if we combine the mirror equation with the magnification equation, we get m = 1 / (d o / f -1). Then, when f = 1 and d o = 3, we have m = -1/2. Also, when f = -3 and and d o = 3, we get m = 1/2. In both cases, the absolute value of the magnification is less than 1 and the image is smaller. There really is no way to tell without drawing the ray trace or doing the equations. Although both mirrors can give smaller images, the two types of mirrors do NOT both give larger images. Only one does this. Can you figure out which one?
5 Page 5 of 6 8. Question DetailsLenz's Law (2) [ ] A conducting loop lies in the plane of this page. A decreasing external magnetic field points into the page. In what direction does the induced magnetic field point? nmlkj v*10*into the page. nmlkj z*00*there is no way to know. nmlkj y*03*counterclockwise. nmlkj w*05*out of the page. nmlkj x*02*clockwise. The induced magnetic field will oppose the change in the external magnetic field. In this case, it must point in the same direction as the external field in order to keep it from decreasing. 9. Question DetailsMirror Equation (4) [750392] Plane mirrors and convex mirrors form virtual images. With a plane mirror, the image may be infinitely far behind the mirror, depending on where the object is located in front of the mirror. For an object in front of a single convex mirror, what is the greatest distance behind the mirror at which the image can be found? Assume that R is the radius of the mirror nmlkj z*03*4r/3 nmlkj y*03*3r/4 nmlkj v*10*r/2 nmlkj w*05*r nmlkj x*05*2r Look at the mirror equation, and solve for the image distance. d i = (1/f - 1/d o ) -1. If f is a negative number (as it is for a convex mirror) and d o is positive (as it always is for a single mirror or lens), then the largest possible distance of the image behind the mirror is when the object is infinitely far away. Thus, the image is at d i = -f = -R/2, which is R/2 behind the mirror. 10. Question DetailsLenz's Law (3) [747007] A conducting loop lies in the plane of this page and carries a clockwise induced current. Which of the following statements could be true? nmlkj v*10*a decreasing magnetic field is directed into the page. nmlkj x*01*a constant magnetic field is directed out of the page. nmlkj y*05*an increasing magnetic field is directed into the page. nmlkj z*05*a decreasing magnetic field is directed out of the page. nmlkj w*01*a constant magnetic field is directed into the page. We are trying to understand where the INDUCED current came from. This is not the current that produces the field! The field came from somewhere else. In other words, the field caused the current. The current did not cause the field. We need to use Lenz's law for this one. The induced current went from zero to some value of current going clockwise. The current does cause a new field that increases into the page. This new field, which is caused by the current, opposes the CHANGE in the original field. Therefore, the original field was either decreasing into the page or increasing out of the page. Only one of these is given in the answers to be selected.
6 Page 6 of 6 Assignment Details Name (AID): Physics 212 Spring 2009 Exam 3 Version C (857202) Submissions Allowed: 100 Category: Exam Code: Locked: No Author: DeAntonio, Michael ( mdeanton@nmsu.edu ) Last Saved: Apr 14, :59 AM MDT Permission: Protected Randomization: Person Which graded: Last Feedback Settings Before due date Nothing After due date Nothing
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