workout 30

# workout 30 - HPHY 253 Workout 30 The Magnetic Field Dr A E...

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HPHY 253 Workout 30 The Magnetic Field Dr. A. E. Bak Name 1. Serway & Jewett Problem 30 : 2 . Calculate the magnitude of the magnetic °eld at a point 100 cm perpendicularly from an in°nitely long, thin straight wire carrying a current of 1 : 00 A . It is convenient to use a system of cylindrical coordinates f s; °; z g . Take the z axis along the length of the wire with the current in the positive direction. Take s as the perpendicular distance from the wire to the observation point. For the in°nitesimal current element lying between z and z + dz , we have I dl = I dz e z : The displacement vector from this in°nitesimal current element to the observation point is r = s e s ° z e z ; having magnitude r = q s 2 + ( ° z ) 2 = p s 2 + z 2 : Therefore, the magnetic °eld at the observation point, due to the in°nitesimal current element I dl , is given by dB = ° ± 0 4 ² ± I dl ± r r 3 = ° ± 0 4 ² ± ( I dz e z ) ± ( s e s ° z e z ) ² p s 2 + z 2 ³ 3 = ´ ± 0 Is e ° 4 ² µ dz ( s 2 + z 2 ) 3 = 2 : On making use of the principle of superposition, we °nd that the total magnetic °eld at the observation point, due to the entire wire current, is B = Z dB = ± 0 Is e ° 4 ² Z + 1 °1 dz ( s 2 + z 2 ) 3 = 2 = ´ ± 0 Is e ° 4 ² µ z s 2 p s 2 + z 2 z =+ 1 z = °1 = ± 0 I e ° 2 ²s : Note that we can treat the unit vector e ° here as a constant because the wire and observation point determine a plane of constant ° . For the values I = 1 : 00 A ; s = 1 : 00 m ; we have B = ± 0 I 2 ²s = ² 4 ² ± 10 ° 7 ³ (1 : 00) 2 ² (1 : 00) = 2 : 00 ± 10 ° 7 T = 0 : 200 ±T as the magnetic-°eld magnitude. 2. Serway & Jewett Problem 30 : 5 . Determine the magnetic °eld at a point P located a distance x from the corner of an in°nitely long wire bent at a right angle as shown in Figure P 30 : 5 . The wire carries a steady current I . Relative to the °gure, take the origin at the wire-bend, with the + x direction as straight right and the + y direction as straight down. Let us refer to the left-to-right piece of wire as Piece #1 and to the top-to-bottom piece as Piece #2 . For any in°nitesimal current element of Piece #1 , we °nd that dl and r point in opposite directions, and so dl ± r = 0 for it. Therefore, we have B 1 = 0 as the magnetic °eld at P produced by the current in Piece #1 . For the in°nitesimal current element of Piece #2 lying between y and y + dy , we have I dl = I ( ° dy e y ) = ° I dy e y : The displacement vector from this in°nitesimal current element to point P is r = x e x ° y e y ; having magnitude r = q x 2 + ( ° y ) 2 = p x 2 + y 2 : 1

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