Physics 1 Problem Solutions 243

Physics 1 Problem Solutions 243 - Chapt. 31 Faradays Law of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapt. 31 Faradays Law of Induction and Inductors 239 evaluate it in terms of B , , , the cylindrical radial coordinate r , and the differential dz . Note that there are two cases. The differential bit of contour on the side of the contour counterclockwise from n has velocity r n and that on the side of the contour clockwise from n has velocity in the direction r n . HINT: Note that z dvectors = dz where dz could be implicitly positive for negative. c) To absorb the two cases for the two sides of n in a single formula lets introduce coordinate y which is in the plane of the contour and perpendicular to the z axis. We let y > 0 for the counterclockwise side and y < 0 for the clockwise side. We can now right r as y . Give the formula for the differential motional emf now. d) We want to integrate the differential emf over the contour by summing up the contributions to each dz interval first about each height z first and then summing these up for the total.first and then summing these up for the total....
View Full Document

Ask a homework question - tutors are online