spring_07_exam4_sol_phy2049

spring_07_exam4_sol_phy2049 - PHY2049 Spring 2007 Exam 4...

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PHY2049 Spring 2007 Exam 4 Solutions 1 Prof. Darin Acosta Prof. Greg Stewart April 28, 2007 Exam 4 Solutions 1. Which of the following statements is true? 1. In equilibrium all of any excess charge stored on a conductor is on the outer surface. 2. In equilibrium the electric field inside a conductor is zero. 3. In a spherical insulator with radius R on which charge q is uniformly distributed, the electric field at a distance of R/4 from the center of the sphere is ¼ of the field at the surface of the sphere. a. 1,2,and 3 b. 1 and 2, not 3 c. 1 and 3, not 2 d. 2 and 3, not 1 e. only 2 Solution: 1 and 2 are true, as shown by Gauss’ Law. For statement 3, use eq. 23-20 (derived from Gauss’ Law) E=(q/4 πε 0 )r/R 3 Since r=1/4 R, E at ¼ R is just ¼ of E at R =(q/4 πε 0 )(1/R 2 ) 2. A long straight wire is 2R long and carries current i upwards in the sketch. What is the magnetic field B at point P which is R away from the center of the wire as shown? | | R |______.P i | R | R | a. μ 0 i/2(2) 1/2 π R b. μ 0 i/2(3) 1/2 π R c. μ 0 i/4 π R d. μ 0 i(3) 1/2 /4 π R e. μ 0 i/2 π R Solution: the field at point P is just double the field produced by either half of the wire, so choose the upper half, solve for B from this segment, and double the result. R R B = 2 0 dB = μ 0 i/2 π 0 sin θ ds/r 2 where ds is some segment of the wire along its length between 0 and R (see Fig. 29-5, p 767 in the text), θ is the angle between the straight wire and the vector r connected ds and point P. The magnitude ‘r’ is (s 2 + R 2 ) 1/2 and sin θ = R/r = R/(s 2 + R 2 ) 1/2 R so B= μ 0 i/2 π 0 Rds/(s 2 + R 2 ) 3/2
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PHY2049 Spring 2007 Exam 4 Solutions 2 This integral gives s/R 2 (s 2 + R 2 ) 1/2 evaluated between s=0 and R This gives B= μ 0 i/2 π R R/(R 2 + R 2 ) 1/2 = μ 0 i/2(2) 1/2 π R which is just 1/(2) 1/2 smaller than the result for an infinitely long wire. 3. A current i goes through the half circle of radius 2R, straight segment of length R, half circle of radius R, and straight segment of length R shown above in a counter-clockwise direction. What is the value of the magnetic field B at the center of the two half- loops? a. 3 μ 0 i/8R b. μ 0 i/2R c. μ 0 i/4R d. 3 μ 0 i/16R e. 7 μ 0 i/16R solution: The contribution of a circular loop of wire of subtended angle Φ is μ 0 i Φ /4 π R where Φ is in radians. The contribution of the two straight segments of wire that go through the center point is zero. Both half circles produce B fields that add, thus B total = B from larger half loop + B from smaller half loop = μ 0 i π /4 π R + μ 0 i π /4 π *2R = μ 0 i(1/4R + 1/8R) = (3/8) μ 0 i/R 4. Two charges, +q and +3 q, are placed along the x-axis a distance ‘d’ apart (see picture). A third charge which is negative and has magnitude ‘Q’ is placed on the x- axis at x=0.37 d so that there is no net force in on the negative charge ‘Q’ from the sum of the Coulomb forces from +q and +3 q. x=0 is at +q.
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spring_07_exam4_sol_phy2049 - PHY2049 Spring 2007 Exam 4...

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