# hw9 - homework 09 – LIN KEVIN – Due 4:00 am 1 Question...

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Unformatted text preview: homework 09 – LIN, KEVIN – Due: Mar 23 2008, 4:00 am 1 Question 1, chap 31, sect 2. part 1 of 1 10 points A square loop of wire of resistance R and side a is oriented with its plane perpendicular to a magnetic field vector B , as shown in the figure. B B a I What must be the rate of change of the magnetic field in order to produce a current I in the loop? 1. dB dt = I a R 2. dB dt = I R a 2 correct 3. dB dt = I Ra 4. dB dt = I a 2 R 5. dB dt = Ra I Explanation: The emf produced is given by E = dφ dt = a 2 dB dt . Using Ohm’s law we can solve for B I R = a 2 dB dt dB dt = I R a 2 . Question 2, chap 31, sect 2. part 1 of 1 10 points A toroid having a rectangular cross section ( a = 2 . 5 cm by b = 2 . 88 cm) and inner ra- dius 5 . 52 cm consists of N = 370 turns of wire that carries a current I = I sin ω t , with I = 72 . 4 A and a frequency f = 48 . 3 Hz. A loop that consists of N ℓ = 31 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 . 52905 V (tolerance ± 1 %). 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 ) homework 09 – LIN, KEVIN – Due: Mar 23 2008, 4:00 am 2 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 = − (31 turns) μ (370 turns) × (72 . 4 A) (48 . 3 Hz) (2 . 5 cm) × ln bracketleftbigg (2 . 88 cm) + (5 . 52 cm) (5 . 52 cm) bracketrightbigg = − . 52905 V |E| = 0 . 52905 V . Question 3, chap 31, sect 1. part 1 of 2 10 points In the arrangement shown in the figure, the resistor is 3 Ω and a 1 T magnetic field is directed out of the paper. The separation between the rails is 6 m . Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 4 m / s . Assume the bar and rails have negligible resistance and friction. m ≪ 1g 4 m / s 3Ω 1 T 1 T I 6m Calculate the applied force required to move the bar to the right at a constant speed of 4 m / s. Correct answer: 48 N (tolerance ± 1 %)....
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hw9 - homework 09 – LIN KEVIN – Due 4:00 am 1 Question...

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