(b) If Φis the magnetic flux through the secondary coil, then the magnitude of the emf induced in that coil is ε= N(dΦ/dt) and the current in the secondary is is= /R, where Ris the resistance of the coil. Thus, iNRddts=FHGIKJΦ. The charge that passes through the secondary when the primary current is turned on is 0.sNdNNqi dtdtdRdtRRΦΦΦ===Φ=∫∫∫The magnetic field through the secondary coil has magnitude B = B0+ BM= 801B0, where BMis the field of the magnetic dipoles in the magnetic material. The total field is perpendicular to the plane of the secondary coil, so the magnetic flux is Φ= AB, where Ais the area of the Rowland ring (the field is inside the ring, not in the region between the ring and coil). If ris the radius of the ring’s cross section, then A =
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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.