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Unformatted text preview: Version 068 Final 2010 Hoffmann (56745) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points The electric field in an electromagnetic wave (in vacuum) is described by E = E max sin( k x t ) , where E max = 44 N / C and k = 6 . 8 10 8 m 1 . The speed of light is 2 . 99792 10 8 m / s. Find the amplitude of the corresponding magnetic wave. 1. 2.76858e-07 2. 3.20222e-08 3. 1.26754e-07 4. 1.46768e-07 5. 1.56775e-07 6. 1.20083e-07 7. 1.43433e-07 8. 2.20152e-07 9. 2.73523e-07 10. 2.13481e-07 Correct answer: 1 . 46768 10 7 T. Explanation: Let : c = 2 . 99792 10 8 m / s and E max = 44 N / C . The relation between the magnitude of the electric field and that of the magnetic field for an electromagnetic wave is given by B = E c . Thus, the magnitude of the magnetic wave is B max = E max c = 44 N / C 2 . 99792 10 8 m / s = 1 . 46768 10 7 T . 002 10.0 points A line of charge starts at x = x , where x is positive, and extends along the x-axis to positive infinity. If the linear charge den- sity is given by = x /x , where is a constant, determine the electric field at the origin. (Here denotes the unit vector in the positive x direction.) 1. k 2 x ( ) correct 2. k 2 x 2 ( ) 3. k x ( ) 4. k 2 x ( ) 5. k x ( ) 6. k 2 2 x ( ) Explanation: First we realize that we are dealing with a continuous distribution of charge (as opposed to point charges). We must divide the dis- tribution into small elements and integrate. Using Coulombs law, the electric field cre- ated by each small element with charge dq is dE = k dq x 2 where dq = dx = x x dx. Now we integrate over the entire distribution (i.e. from x = x to x = + ) and insert our dq : | vector E | = integraldisplay k dq x 2 = integraldisplay x k x dx x 3 = k x 2 1 x 2 vextendsingle vextendsingle vextendsingle x = k 2 x . Since the distribution is to the right of the point of interest, the electric field is directed Version 068 Final 2010 Hoffmann (56745) 2 along the x axis if is positive. That is, a positive charge at the origin would experience a force in the direction of from this charge distribution. In fact, the direction of an elec- tric field at a point P in space is defined as the direction in which the electric force acting on a positive particle at that point P would point. So vector E = k 2 x ( ) . 003 10.0 points The electric field between the plates of an air capacitor of plate area 0 . 47 m 2 is changing at a rate of 41000 V / m / s....
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- Spring '08