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# solutions_33 - PHY2061 R D Field Chapter 33 Solutions I...

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PHY2061 R. D. Field Solutions Chapter 33 Page 1 of 11 Chapter 33 Solutions Problem 1: The bent wire carries current I , as shown the Figure . What is the magnetic field at point P , at distance d directly opposite the 90 o bend (k = µ 0 /4 π )? Answer: kI/d, out of page Solution: Wire A does not contribute at P since 0 ° = × r l d v and wire B (semi-infinite wire) contributes kI/d (one half of 2kI/d). Problem 2: Two semi-infinite straight pieces of wire are connected by a circular loop segment of radius R as shown in the Figure . If extrapolated to their intersection point straight wires would form a 30 o angle. If the wires carry a current I , what is the magnitude of the magnetic field at the center of the circular arc (k = µ 0 /4 π )? Answer: 4.62kI/R Solution: The magnetic field at the center of the circular loop segment is the superposition of three terms as follows: R kI R kI R kI R kI R kI B z 62 . 4 ) 62 . 2 2 ( 180 150 = + = + + = o o π . P I d B A I 30 o

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PHY2061 R. D. Field Solutions Chapter 33 Page 2 of 11 Problem 3: Three concentric current loops have radii R , 2R , and 3R , as shown in the Figure . The loop with radius 2R carries a current of 2I and the loop with radius 3R carries a current of 3I in the same direction ( counter-clockwise ). If the net magnetic field at the center of the three loops is zero, what is the magnitude and direction of the current in the loop with radius R ? Answer: 2I, clockwise Solution: If we define the z-axis to be out of the paper then the net magnetic field at the center of the three circles is given by 0 3 ) 3 ( 2 2 ) 2 ( 2 2 1 = + + = R I k R I k R kI B z π π π , where I 1 is the current ( clockwise ) in the smallest loop. Solving for I 1 gives I 1 = 2I . Problem 4: Two semi-infinite straight pieces of wire are connected by a circular loop segment of radius R as shown in the Figure . If the two straight pieces of wire are perpendicular to each other and if the circuit carries a current I , what is the magnitude of the magnetic field at the center of the circular arc (k = µ 0 /4 π )? Answer: 6.71kI/R Solution: If we define the z-axis to be out of the paper then the net magnetic field at the center of the circular arc is given by R kI R kI R kI R kI R kI B z 71 . 6 2 3 2 2 3 = + = + + = π π . 2I 3I 2R R 3R I R I
PHY2061 R. D. Field Solutions Chapter 33 Page 3 of 11 Problem 5: An infinitely long straight wire carries a current I 1 , and a wire loop of radius R carries a current I 2 as shown in the Figure . If I 1 = 4I 2 , and if the net magnetic field at the center of the circular loop is zero, how far is the center of the loop from the straight wire? Answer: 1.27R Solution: Setting magnitudes of the magnetic field from the loop and wire equal to each other yields R kI y kI 2 1 2 2 π = . Solving for y gives R R R I I y 27 . 1 4 2 1 = = = π π . Problem 6: Four infinitely long parallel wires each carrying a current I form the corners of square with sides of length L as shown in the Figure . If three of the wire carry the current I in the same direction ( out of the page ) and one of the wires carries the current I in the opposite direction (

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