P30_073 - 73. (a) The field in this region is entirely due...

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: 73. (a) The field in this region is entirely due to the long wire (with, presumably, negligible thickness). Using Eq. 30-19, µ0 iw |B | = = 4.8 × 10−3 T 2πr where iw = 24 A and r = 0.0010 m. (b) Now the field consists of two contributions (which are antiparallel) – from the wire (Eq. 30-19) and from a portion of the conductor (Eq. 30-22 modified for annular area): |B | = = µ0 ienc µ0 iw − 2πr 2πr 2 µ0 ic π r2 − πRi µ0 iw − 2 − πR2 2πr 2πr πRo i where r = 0.0030 m, Ri = 0.0020 m, Ro = 0.0040 m and ic = 24 A. Thus, we find |B | = 9.3 × 10−4 T. (c) Now, in the external region, the individual fields from the two conductors cancel completely (since ic = iw ): B = 0. ...
View Full Document

Ask a homework question - tutors are online