Ch32-p006 - 6 The integral of the field along the indicated path is by Eq 32-18 and Eq 32-19 equal to(4.0 cm(2.0 cm enclosed area 0id = 52 nT m =

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6. The integral of the field along the indicated path is, by Eq. 32-18 and Eq. 32-19, equal to 00 2 enclosed area
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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