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Unformatted text preview: terry (ect328) – oldhomework 24 – Turner – (59130) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find 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 sin ω t , with I = 41 . 7 A and a frequency f = 32 . 6 Hz. A loop that consists of N ℓ = 15 turns of wire links the toroid, as in the figure. 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: Faraday’s Law E = d Φ B dt . Magnetic field in a toroid B = μ N I 2 π r . Solution: In a toroid, all the flux is confined to the inside of the toroid B = μ N I 2 π r . So, the flux through the loop of wire is Φ B 1 = integraldisplay B dA = μ N I 2 π sin( ω t ) integraldisplay b + R R adr r = μ N I 2 π a sin( ω t ) ln parenleftbigg b + R R parenrightbigg . Applying Faraday’s law, the induced emf can be calculated as follows E = N ℓ d Φ B 1 dt = N ℓ μ N I 2 π ω a ln parenleftbigg b + R R parenrightbigg cos( ω t ) =E cos( ω t ) where ω = 2 πf was used. The maximum magnitude of the induced emf , E , is the coefficient in front of cos( ω t ). E = N ℓ d Φ B 1 dt = N ℓ μ N I ω 2 π a ln bracketleftbigg b + R R bracketrightbigg = (15 turns) μ (650 turns) × (41 . 7 A) (32 . 6 Hz) (0 . 96 cm) × ln bracketleftbigg (4 . 56 cm) + (2 . 3 cm) (2 . 3 cm) bracketrightbigg = . 174735 V E = 0 . 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 figure. 3cm 5 cm 12 . 5cm 40A Find the magnetic flux through the rectan gular loop. 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 4 π = 1 × 10 7 N / A 2 . d a b x dx I The magnetic flux through the strip of area dA is d Φ = B dA = μ 2 π I x bdx = μ 4 π 2 bI dx x , so the total magnetic flux through the rectan gular loop is Φ total = integraldisplay d + a d d Φ = μ 4 π (2 bI ) integraldisplay d + a d dx x = μ 4 π (2 bI ) ln d + a d = (1 × 10 7 N / A 2 ) 2 (0 . 125 m) (40 A) × ln parenleftbigg . 03 m + 0 . 05 m . 03 m parenrightbigg = 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 figure 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 field has a magnitude of 720 mT,...
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 Fall '09
 Turner
 Magnetic Field, Terry

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