P32_058

# 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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