P32_058 - 58. (a) The magnitude of the toroidal field is...

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Unformatted text preview: 58. (a) The magnitude of the toroidal field is given by B0 = µ0 nip , where n is the number of turns per unit length of toroid and ip is the current required to produce the field (in the absence of the ferromagnetic material). We use the average radius (ravg = 5.5cm) to calculate n: n= Thus, ip = 400 turns N = = 1.16 × 103 turns/m . 2πravg 2π (5.5 × 10−2 m) B0 0.20 × 10−3 T = = 0.14 A . −7 T · m/A)(1.16 × 103 /m) µ0 n (4π × 10 (b) If Φ is the magnetic flux through the secondary coil, then the magnitude of the emf induced in that coil is E = N (dΦ/dt) and the current in the secondary is is = E /R, where R is the resistance of the coil. Thus N dΦ is = . R dt The charge that passes through the secondary when the primary current is turned on is q= is dt = N R dΦ N dt = dt R Φ dΦ = 0 NΦ . R The magnetic field through the secondary coil has magnitude B = B0 + BM = 801B0 , where BM is 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 A is the area of the Rowland ring (the field is inside the ring, not in the region between the ring and coil). If r is the radius of the ring’s cross section, then A = πr2 . Thus Φ = 801πr2 B0 . The radius r is (6.0 cm − 5.0 cm)/2 = 0.50 cm and Φ = 801π (0.50 × 10−2 m)2 (0.20 × 10−3 T) = 1.26 × 10−5 Wb . Consequently, q= 50(1.26 × 10−5 Wb) = 7.9 × 10−5 C . 8.0 Ω ...
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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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