P31_052 - N 62 . 8 cm / 1 . 0 mm = 628. Thus L = N 2 h 2 ln...

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52. (a) Our notation is as follows: h is the height of the toroid, a its inner radius, and b its outer radius. Since it has a square cross section, h = b a =0 . 12 m 0 . 10 m = 0 . 02 m . . We derive the flux using Eq. 30-26 and the self-inductance using Eq. 31-35: Φ B = Z b a BdA = Z b a µ µ 0 Ni 2 πr hdr = µ 0 Nih 2 π ln µ b a and L = N Φ B /i =( µ 0 N 2 h/ 2 π )ln( b/a ) . We note that the formulas for Φ B and L can also be found in the Supplement for the chapter, in Sample Problem 31-11. Now, since the inner circumference of the toroid is l =2 πa =2 π (10 cm) 62 . 8 cm, the number of turns of the toroid is roughly
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Unformatted text preview: N 62 . 8 cm / 1 . 0 mm = 628. Thus L = N 2 h 2 ln b a ( 4 10 7 H / m ) (628) 2 (0 . 02 m) 2 ln 12 10 = 2 . 9 10 4 H . (b) Noting that the perimeter of a square is four times its sides, the total length ` of the wire is ` = (628)4(2 . 0 cm) = 50 m, the resistance of the wire is R = (50 m)(0 . 02 / m) = 1 . 0 . Thus L = L R = 2 . 9 10 4 H 1 . 0 = 2 . 9 10 4 s ....
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