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Unformatted text preview: Version 068 – Final 2010 – Hoffmann – (56745) 1 This printout should have 30 questions. Multiplechoice 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 − ω 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. 2.76858e07 2. 3.20222e08 3. 1.26754e07 4. 1.46768e07 5. 1.56775e07 6. 1.20083e07 7. 1.43433e07 8. 2.20152e07 9. 2.73523e07 10. 2.13481e07 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 xaxis to positive infinity. If the linear charge den sity is given by λ = λ x /x , where λ is a constant, determine the electric field at the origin. (Here ˆ ı denotes the unit vector in the positive x direction.) 1. k λ 2 x ( − ˆ ı ) correct 2. k λ 2 x 2 (ˆ ı ) 3. k λ x ( − ˆ ı ) 4. k λ 2 x (ˆ ı ) 5. k λ x (ˆ ı ) 6. k λ 2 2 x (ˆ ı ) 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 Coulomb’s law, the electric field cre ated by each small element with charge dq is dE = k dq x 2 where dq = λdx = λ 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 λ x dx x 3 = − k λ x 2 1 x 2 vextendsingle vextendsingle vextendsingle ∞ x = k λ 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 λ is positive. That is, a positive charge at the origin would experience a force in the direction of − ˆ ı 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 λ 2 x ( − ˆ ı ) . 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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This note was uploaded on 01/27/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner

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