OHW 24 - terry (ect328) oldhomework 24 Turner (59130) This...

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terry (ect328) – oldhomework 24 – Turner – (59130) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A toroid having a rectangular cross section ( a =0 . 96 cm by b =4 . 56 cm) and inner radius 2 . 3 cm consists oF N = 650 turns oF wire that carries a current I = I 0 sin ω t , with I 0 = 41 . 7 A and a Frequency f = 32 . 6 Hz. A loop that consists oF N ± = 15 turns oF wire links the toroid, as in the fgure. b a N R N l Determine the maximum E induced in the loop by the changing current I . Correct answer: 0 . 174735 V. Explanation: Basic Concept: ±araday’s Law E = - d Φ B dt . Magnetic feld in a toroid B = μ 0 NI 2 π r . Solution: In a toroid, all the ²ux is confned to the inside oF the toroid B = μ 0 2 . So, the ²ux through the loop oF wire is Φ B 1 = ± B dA = μ 0 0 2 π sin( ) ± b + R R adr r = μ 0 0 2 π a sin( ) ln ² b + R R ³ . Applying ±araday’s law, the induced emF can be calculated as Follows E = - N ± d Φ B 1 dt = - N ± μ 0 0 2 π ω a ln ² b + R R ³ cos( ) = -E 0 cos( ) where ω =2 πf was used. The maximum magnitude oF the induced emf , E 0 , is the coe³cient in Front oF cos( ). E 0 = - N ± d Φ B 1 dt = - N ± μ 0 0 ω 2 π a ln ´ b + R R µ = - (15 turns) μ 0 (650 turns) × (41 . 7 A) (32 . 6 Hz) (0 . 96 cm) × ln ´ (4 . 56 cm) + (2 . 3 cm) (2 . 3 cm) µ = - 0 . 174735 V |E| . 174735 V . 002 10.0 points A long straight wire carries a current 40 A. A rectangular loop with two sides parallel to the straight wire has sides 5 cm and 12 . 5 cm, with its near side a distance 3 cm From the straight wire, as shown in the fgure. 3 cm 5 cm 12 . 5 cm 40 A ±ind the magnetic ²ux through the rectan- gular loop.
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terry (ect328) – oldhomework 24 – Turner – (59130) 2 The permeability of free space is 4 π × 10 - 7 T · m / A. Correct answer: 9 . 80829 × 10 - 7 Wb. Explanation: Let : I = 40 A , a = 5 cm = 0 . 05 m , b = 12 . 5 cm = 0 . 125 m , d = 3 cm = 0 . 03 m , and μ 0 4 π =1 × 10 - 7 N / A 2 . d a b x dx I The magnetic Fux through the strip of area dA is d Φ = B dA = μ 0 2 π I x bdx = μ 0 4 π 2 bI dx x , so the total magnetic Fux through the rectan- gular loop is Φ total = ± d + a d d Φ = μ 0 4 π (2 bI ) ± d + a d dx x = μ 0 4 π (2 ) ln d + a d = (1 × 10 - 7 N / A 2 ) 2 (0 . 125 m) (40 A) × ln ² 0 . 03 m + 0 . 05 m 0 . 03 m ³ = 9 . 80829 × 10 - 7 Wb . 003 (part 1 of 4) 10.0 points A bar of negligible resistance and mass of 84 kg in the ±gure below is pulled horizon- tally across frictionless parallel rails, also of negligible resistance, by a massless string that passes over an ideal pulley and is attached to a suspended mass of 270 g. The uniform magnetic ±eld has a magnitude of 720 mT, and the distance between the rails is 20 cm. The rails are connected at one end by a load resistor of 90 mΩ. The acceleration of gravity is 9 . 8m / s 2 . 720 mT 720 mT 720 mT 20 cm 270 g a 90 m Ω 84 kg What is the magnitude of the terminal ve- locity ( i.e., the eventual steady-state speed v ) reached by the bar? Correct answer: 11 . 4844 m / s. Explanation: Let : m = 84 kg , M = 270 g , ' = 20 cm , B = 720 mT , and R = 90 mΩ .
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OHW 24 - terry (ect328) oldhomework 24 Turner (59130) This...

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