31. (a) Lettingxbe the distance from the right end of the rails to the rod, we Fnd an expression for themagnetic ﬂux through the area enclosed by the rod and rails. By Eq. 30-19, the Feld isB=µ0i/2πr,whereris the distance from the long straight wire. We consider an inFnitesimal horizontal strip oflengthxand widthdr, parallel to the wire and a distancerfrom it; it has areaA=xdrand theﬂuxdΦB=(µ0ix/2)dr. By Eq. 31-3, the total ﬂux through the area enclosed by the rod andrails isΦB=µ0ix2πZa+Ladrr=µ0ix2πlnµa+La¶.According to ±araday’s law the emf induced in the loop isE=dΦBdt=µ0i2πdxdtlnµa+La¶=µ0iv2πlnµa+La¶=(4π×10−7T·m/A)(100A)(5.00 m/s)2πlnµ1.00 cm + 10.0cm1.00 cm¶=2.40×10−4V.(b) By Ohm’s law, the induced current isi`=E/R=(2.40×10−4V)/(0.400 Ω) = 6.00×10−4A. Sincethe ﬂux is increasing the magnetic Feld produced by the induced current must be into the page inthe region enclosed by the rod and rails. This means the current is clockwise.
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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.