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

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Unformatted text preview: 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 = & ¡ 4 ¢ ¡ I dl ¡ r r 3 = & ¡ 4 ¢ ¡ ( I dz e z ) ¡ ( s e s & z e z ) ¢ p s 2 + z 2 £ 3 = ¤ ¡ Is e & 4 ¢ ¥ 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 = ¡ Is e & 4 ¢ Z + 1 &1 dz ( s 2 + z 2 ) 3 = 2 = ¤ ¡ Is e & 4 ¢ ¥ z s 2 p s 2 + z 2 ¦ ¦ ¦ ¦ z =+ 1 z = &1 = ¡ I e & 2 ¢s : 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 = ¡ I 2 ¢s = ¢ 4 ¢ ¡ 10 & 7 £ (1 : 00) 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 = for it. Therefore, we have B 1 = 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 ( &...
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This note was uploaded on 05/01/2011 for the course HPHY 253 taught by Professor Bak during the Spring '09 term at Morehouse.

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

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