SOLUTION TO EXAM 2

SOLUTION TO EXAM 2 - Exam 2 solutions Name: Physics 204...

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Unformatted text preview: Exam 2 solutions Name: Physics 204 morning summer 2008 Instructions: Score • The responsibility is yours to show me that you understand what you are doing so please show all your work. If you feel you are stuck on a particular part explain as much of your thinking as you can so I can give you the maximum amount of partial credit (explain the strategy you are using, draw pictures/graphs, etc.) 1: • Use the backs of the exam pages if you need extra space. 4: • Each problem is worth 10 points. The problems are not all of the same difficulty though, so budget your time wisely. 5: • The last pages of the exam contain useful physical constants, conversion factors, and other data. Please check there first if you feel you need more information than is given in the problem. 2: 3: Total /50 Good luck! 1. A solenoid (1000 windings, radius 10cm, length 200cm) carries a current of 0.3A. Additionally, another wire runs along the axis of the solenoid and carries a current of 15A. a. Draw a diagram showing the magnetic field at a point halfway between the axis and the edge of the solenoid. Your diagram should illustrate how you decided what direction the field will point. Also, explain your reasoning in words. Hint: Be sure to show the magnetic field of the solenoid at that point. Be sure to show the magnetic field of the wire at that point. Then, determine the total magnetic field at that point and show that. b. Determine the magnitude of the magnetic field at that point. c. What magnetic force, if any, will the wire feel? Explain your reasoning. The point being considered is a point halfway between the wire and the surface of the solenoid that is facing you (the reader). Using the right hand rule, the magnetic field created by the straight wire points down, and the magnetic field of the solenoid points to the right. The total magnetic field points down and to the right with a magnitude and direction given by the following analysis. Is Is r Bs Iw r Bw r B Bs = μ0 N Is L ⇒ Bs = ( 4π × 10−7 T ⋅ m A ) 1000 ( 0.3A ) 2m ⇒ Bs = 1.88 × 10 −4 T μI Bw = 0 w 2π r ( 4π × 10−7 T ⋅ m A ) (15A ) ⇒ Bw = 2π ( 0.05m ) ⇒ Bw = 6.0 × 10 −5 T B = Bs2 + Bw 2 = (1.88 × 10 −4 T )2 + (−6.0 × 10 −5 T )2 = 1.98 × 10 −4 T The wire will not feel a magnetic force because the current is flowing in the same direction as the magnetic field of the solenoid. 2. It is the distant future, and a new power source called a “Tachyon Reversal” engine has been developed. The core of the engine takes the shape of a cylinder (radius 2m, length 10m). Unfortunately this engine emits potentially harmful radiation in the form of exotic −19 negatively charged particles (charge −1.6 × 10 C , mass 2.4 × 10 −23 kg ). These particles are emitted radially outward from the edge of the cylinder at 15% of light speed. The engine core is surrounded with a layer of shielding that takes the shape of a thin cylindrical shell (radius 6m, length 10m, with the engine core at the center). You have been asked to design a magnetic field for the region between the surface of the engine core and the shielding that will cause the particles to be deflected back into the core. You have found a way to make the field have the same strength throughout the region, and have the field lines encircle the engine core. You are convinced this configuration will work. a. Draw a diagram showing the magnetic field and the trajectory one of the particles will take. Hint: Visualize the region between the engine core and the shielding. Now, visualize a rectangular “slice” of that region extending from the outer edge of the engine core to the inner edge of the shielding. b. What should the strength of the magnetic field in this region be? The uniform magnetic field will cause the negatively charged exotic particles to move in a semicircular path before returning to the engine core. In order for this to happen the magnetic field must have a strength of −23 8 mv ( 2.4 × 10 kg )( 0.15 × 3.00 × 10 m s ) B= = qR (1.6 × 10−19 C ) ( 6m − 2m ) ⇒ B = 1.69 × 103 T r B This is a very strong field to sustain, but in the distant future it might be possible. Engine core Shielding 3. A coil of wire (diameter 20cm) with 100 windings is suspended between the poles of a strong horseshoe magnet by an axle that can rotate. The average strength of the magnetic field between the poles of the magnet is 3.0T. The axle is connected to a waterwheel which causes the coil to spin in between the poles of the magnet at a rate of 10 revolutions per second. This causes a current to flow in the coil. The coil is connected through a “floating contact” to the power grid which acts as a single resistor of 10Ω . a. Explain in detail why this causes a current to flow in the coil. Will this current always have the same value? Will it always flow in the same direction? b. Determine the average current that flows in the coil as it rotates by 180 degrees. This is a generator! It will generate a sinusoidally varying EMF, or alternating current. This occurs because the magnetic flux through the coil is changing with time according to Φ B = BA cosθ = BA cos (ωt ) where B is the average strength of the magnetic field and ω is the angular frequency of the coil’s rotation. The average EMF can be determined using Faraday’s law, and the average current with some help from Kirchhoff’s loop rule and Ohm’s law. When the coil rotates by 180 degrees the magnetic flux through it changes sign. Since each rotation (360 degrees) takes 0.1s, it will take 0.05s to rotate 180 degrees Iav = ε av R = 1 ΔΦ B 1 Φ B , f − Φ B , f 1 NBA cos0° − ( −NBA cos180° ) = = R Δt R R Δt Δt 2NBA 2NBπ ( d 2 ) 2 (100 )( 3.0T ) π ( 0.2m 2 ) ⇒ Iav = = = = 37.7 A RΔt RΔt (10Ω )( 0.05s ) 2 2 4. Two plane mirrors are arranged so that they are at right angles to one another, one oriented vertically along a wall, one horizontally along the floor. They meet at a common edge to form an ‘L’ shape. A cube (length 20cm) made of a transparent material (index of refraction 1.25 for red light, 1.26 for violet light) is placed in the corner formed by the two mirrors. A beam of white light is incident on the top face of the cube 5cm from the vertical mirror, headed towards the vertical mirror, with an angle of incidence of 60 degrees. a. Draw a ray diagram for this situation. Make it large since you’ll be labeling it in part b. Your diagram should show everything that happens to the beam of light. Hint: This might involve doing some calculations. b. Label all angles of incidence, reflection, and refraction with their numerical values. The first diagram shows what will happen to the beam of light. Because of the wavelength dependence of the index of refraction short wavelengths will refract slightly more than long wavelengths. This will separate the white light into its component colors. The diagram shows the rays representing the limits of the visible spectrum. The uncertainty is when the light is incident on the left face of the cube. Total internal reflection might occur. To decide this I’ll need to determine the critical angle for both colors. ⎛ nair ⎝ ncube ,red ⎞ −1 ⎛ 1 ⎞ ⎟ = sin ⎜ ⎟ = 53.1° ⎟ ⎝ 1.25 ⎠ ⎠ ⎛ nair ⎞ −1 −1 ⎛ 1 ⎞ = sin ⎜ ⎟ = sin ⎜ ⎟ = 52.5° ⎜n ⎟ ⎝ 1.26 ⎠ ⎝ cube ,violet ⎠ θ c ,red = sin−1 ⎜ ⎜ θ c ,violet θi θr Next I’ll need to determine what the incident angle will be. I’ll draw a second diagram and keep it general enough so that it can represent either the red or the violet light. Using the law of reflection and some geometry a few times I’ve determined that the angle of incidence will be 90° − θ r . To find θ r I’ll use θr 2 90° − θ r 90° − θ r 90° − θ r 90° − θ r θr θr the law of refraction where the white light is incident on the top face of the cube. θ r will have two values, one for each color. ⎛ nair ⎞ ⎛ 1 ⎞ sinθ i ⎟ = sin−1 ⎜ sin60° ⎟ = 43.9° ⎟ ⎝ 1.25 ⎠ ⎝ ncube ,red ⎠ θ r ,red = sin−1 ⎜ ⎜ θ r ,violet ⎛ nair ⎞ ⎛ 1 ⎞ sinθ i ⎟ = sin−1 ⎜ sin60° ⎟ = 43.4° = sin−1 ⎜ ⎜n ⎟ ⎝ 1.26 ⎠ ⎝ cube ,violet ⎠ So the incident angles on the left face of the cube will be: θ i ,red = 90° − 43.9° = 46.1° θ i ,violet = 90° − 43.4° = 46.6° In neither case is the incident angle greater than the critical angle, so total internal reflection will not occur. The angles of refraction as the light leaves the left face of the cube are: ⎛ ncube ,red ⎞ ⎛ 1.25 ⎞ sinθ i ,red ⎟ = sin−1 ⎜ sin46.1° ⎟ = 64.2° ⎝ 1 ⎠ ⎝ nair ⎠ ⎛ ncube ,violet ⎞ ⎛ 1.26 ⎞ sinθ i ,violet ⎟ = sin−1 ⎜ sin46.6° ⎟ = 66.3° = sin−1 ⎜ nair ⎝ 1 ⎠ ⎝ ⎠ θ r 2,red = sin−1 ⎜ θ r 2,violet 5. You have been asked to design a mirror for a distortion room in a haunted house at an amusement park. People will be standing 2m front of it. The idea of the mirror will be to create an upright image of the person that is 3 times larger. a. Draw a ray diagram for the mirror that will accomplish this. Is the image real or virtual? b. What radius of curvature must the mirror have? The mirror needs to be converging and the object must be between the focal point and the mirror to create an upright, magnified image. The image will be virtual. −1 −1 F di 1 ⎛1 1 ⎞ 1 ⎛2 1 ⎞ =− ⎜ − ⎟ =− ⎜ − ⎟ do do ⎝ f do ⎠ d o ⎝ R do ⎠ 2 1 1 ⇒ − =− R do mdo m=− ⇒R = 2mdo 2 ( 3)( 2m ) = = 6m m −1 ( 3) − 1 k= ...
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