This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 wirebend, with the + x direction as straight right and the + y direction as straight down. Let us refer to the lefttoright piece of wire as Piece #1 and to the toptobottom 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 ( &...
View
Full
Document
This note was uploaded on 05/01/2011 for the course HPHY 253 taught by Professor Bak during the Spring '09 term at Morehouse.
 Spring '09
 bak

Click to edit the document details