60. (a) Our notation is as follows: his the height of the toroid, aits inner radius, and bits 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. 29-24 and the self-inductance using Eq. 30-33: ΦBababBdANirhdrNihba==FHGIKJ=FHGIKJzzµµ0022ππlnand 20ln2BNhNbLiaµΦ⎛⎞⎜⎟π⎝⎠. Now, since the inner circumference of the toroid is l= 2πa= 2π(10 cm) ≈62.8 cm, the number of turns of the toroid is roughly N≈62.8 cm/1.0 mm = 628. Thus
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