Question

.200 500 Figure P30.19 Pour long, parallel conductors carry equal arrents of 1 . 4.00 A ng re P30 19·en end view of the conductor. The direction of the arre ts i t the page at pent-na in leat d ,/- crosses) and out of the page at C and D (indicated by the dots). Cakculate the magnitude and direction of the magnetic field at point P, located at the center of the square with an edge length 0.20o HT Direction -12 points 585 30 P 24 A cylindrical conductor of radius R-3.00 cm carries a current of r-2.75 A along its length. This current is uniformly distributed throughout the cross section of the conductor. (a) Calculate the magnitude of the magnetic field midway along the radius of the wire (that is, ar R/2). (b) Find the distance beyond the surface of the conductor at which the magnitude of the magnetic field has the same value as the magnitude of the field atrR/2.

Answer 1

First, let us pick a coordinate system by choosing unit vectors. Leti be down and to the lef. be down and to the right, and k be straight down. Using the right-hand rule, we determine the direction of the magnetic field at P generated by each wire to be (10) The magnitude of each B is given by JA ! 2tr And since the currents have the same magnitude, and each corner is equidistant from the square center, each magnetic field contribution will have the same magnitude. The distance r is given by (12) We still have to add our vector fields, which gives Ao! 2tr (13) Ttr