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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 − 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. 2.76858e-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 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