P31_022 - 22(a First we observe that a large portion of the figure contributes flux which “cancels out.” The field(due to the current in the

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Unformatted text preview: 22. (a) First, we observe that a large portion of the figure contributes flux which “cancels out.” The field (due to the current in the long straight wire) through the part of the rectangle above the wire is out of the page (by the right-hand rule) and below the wire it is into the page. Thus, since the height of the part above the wire is b − a, then a strip below the wire (where the strip borders the long wire, and extends a distance b − a away from it) has exactly the equal-but-opposite flux which cancels the contribution from the part above the wire. Thus, we obtain the non-zero contributions to the flux: a µ0 ib a µ0 i ΦB = B dA = (b dr ) = ln . 2πr 2π b−a b −a Faraday’s law, then, (with SI units and 3 significant figures understood) leads to E = = = dΦB d µ0 ib a =− ln dt dt 2π b−a µ0 b a di µ0 b − ln =− ln 2π b − a dt 2π −µ0 b(9t − 10) a ln . 2π b−a − a b−a d dt 92 t − 10t 2 With a = 0.120 m and b = 0.160 m, then, at t = 3.00 s, the magnitude of the emf induced in the rectangular loop is |E| = 4π × 10−7 (0.16)(9(3) − 10) 0.12 ln 2π 0.16 − 0.12 = 5.98 × 10−7 V . di (b) We note that dt > 0 at t = 3 s. The situation is roughly analogous to that shown in Fig. 31-5(c). From Lenz’s law, then, the induced emf (hence, the induced current) in the loop is counterclockwise. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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