This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: of length, dl . Fortunately, dl and are perpendicular in this case, greatly simplifying our equation for dB : dB = However, this vector is at an angle θ to the z axis. Thus the component of the field produced by dl in the zaxis is given by: dB z = cos θ = The geometry used to get this equation can be seen from the . Now we integrate this expression over the entire circle. Notice, however, that dl = 2 Πb , or simply the circumference of the circle. Thus: B z = = This equation applies to any point on the axis of the ring. To find the field at the center of the ring, we simply plug in z = 0 : B z = Thus we have a set of equations for the field of a ring. Though the derivation required calculus, and may not be useful, it allowed us to get some experience using our complex equation from the last section. Next we stack a number of rings on top of each other, and analyze the resultant field....
View
Full
Document
This note was uploaded on 02/09/2012 for the course PHY PHY2053 taught by Professor Davidjudd during the Fall '10 term at Broward College.
 Fall '10
 DavidJudd
 Physics, Current, Power

Click to edit the document details