871_Physics ProblemsTechnical Physics

871_Physics ProblemsTechnical Physics - 211 Chapter 30...

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

View Full Document Right Arrow Icon
Chapter 30 211 P30.73 Note that the current I exists in the conductor with a current density J I A = , where Aa aa a =− L N M O Q P = π 2 22 2 44 2 . Therefore, J I a = 2 2 . To find the field at either point P 1 or P 2 , find B s which would exist if the conductor were solid, using Ampère’s law. Next, find B 1 and B 2 that would be due to the conductors of radius a 2 that could occupy the void where the holes exist. Then use the superposition principle and subtract the field that would be due to the part of the conductor where the holes exist from the field of the solid conductor. θθ θ B s - B 1 - B 2 P 1 r r r 2 + a 2 () 2 a /2 a /2 B s −′ B 1 B 2 P 2 FIG. P30.73 (a) At point P 1 , B Ja r s = µπ 0 2 2 ej , B ra 1 0 2 2 = bg di , and B 2 0 2 2 = + . BB B B r B I r rr a I r s =−−= + L N M M O Q P P = −− L N M M M O Q P P P = L N M O Q P 12 2 0 2 0 2 11 42 1 2 2 2 4 µ a f directed to the left (b) At point P 2 , B r s = 0 2 2 and ′= ′= + BB 0 2 2 2 2 . The horizontal components of B 1 and B 2 cancel while their vertical components add. BB B B r r B r
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online