spring_07_exam2_sol_phy2049

spring_07_exam2_sol_phy2049 - PHY2049 Spring 2007 Exam 2...

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PHY2049 Spring 2007 Exam 2 Solutions 1 Prof. Darin Acosta Prof. Greg Stewart March 26, 2007 Exam 2 Solutions 1. A current i travels in the direction of the arrows shown in the figure in three wire segments. The straight segments (upper left and upper right in the figure) both are lined up with the center point, P , which is the center of the circular arc segment wire shown which subtends (takes up) an angle of 120 o . The straight wires are both of length R; the radius of the circular arc is also R. What is the correct expression for the magnitude of the total field produced by the current in all three wire segments at point P? a. μ 0 i/(6R) b. μ 0 i/(3R) c. μ 0 i/(2R) d. μ 0 i/(4R) e. μ 0 i/(12R) Solution: The current in the straight wires that would go thru center point P if extended contributes zero to the field (as discussed in lecture for chap. 29). The field at the center of a circular arc of wire (eq. 29-9) is B= μ 0 i φ /(4 π R), where φ is the angle taken up by the circular arc. 120 o is 1/3 of 360 o , where 360 o is 2 π radians, so 120 o =(1/3) 2 π radians, so B= μ 0 i φ /(4 π R)= = μ 0 i(2/3) π /(4 π R)= μ 0 i/(6R). 2. Consider the toroid shown in the figure. Wire is wrapped around the core in a doughnut geometry, current i=10 A in the wire. As shown, the wire carries current out of the page (this direction of the wire is shown as small 'o'), around the toroid, and back into
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PHY2049 Spring 2007 Exam 2 Solutions 2 the page (denoted as 'x'). There are 40 windings or turns on this toroid. Three circular Amperian loops – the smallest one (#1), the middle-sized one between the inner and outer
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spring_07_exam2_sol_phy2049 - PHY2049 Spring 2007 Exam 2...

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