hw9 solutions - homework 09 SMITH, TAYLOR Due: Mar 23 2008,...

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Unformatted text preview: homework 09 SMITH, TAYLOR 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 a 2 R 3. dB dt = I R a 2 correct 4. dB dt = Ra I 5. dB dt = I Ra Explanation: The emf produced is given by E = d dt = a 2 dB dt . Using Ohms 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 . 41 cm by b = 1 . 76 cm) and inner ra- dius 5 . 78 cm consists of N = 630 turns of wire that carries a current I = I sin t , with I = 73 . 8 A and a frequency f = 95 . 6 Hz. A loop that consists of N = 23 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 . 82299 V (tolerance 1 %). Explanation: Basic Concept: Faradays Law E = d B dt . Magnetic field in a toroid B = N I 2 r . Solution: In a toroid, all the flux is confined to the inside of the toroid B = N I 2 r . So, the flux through the loop of wire is B 1 = integraldisplay B dA = N I 2 sin( t ) integraldisplay b + R R adr r = N I 2 a sin( t ) ln parenleftbigg b + R R parenrightbigg . Applying Faradays law, the induced emf can be calculated as follows E = N d B 1 dt = N N I 2 a ln parenleftbigg b + R R parenrightbigg cos( t ) = E cos( t ) homework 09 SMITH, TAYLOR Due: Mar 23 2008, 4:00 am 2 where = 2 f was used. The maximum magnitude of the induced emf , E , is the coefficient in front of cos( t ). E = N d B 1 dt = N N I 2 a ln bracketleftbigg b + R R bracketrightbigg = (23 turns) (630 turns) (73 . 8 A) (95 . 6 Hz) (2 . 41 cm) ln bracketleftbigg (1 . 76 cm) + (5 . 78 cm) (5 . 78 cm) bracketrightbigg = . 82299 V |E| = 0 . 82299 V . Question 3, chap 31, sect 1. part 1 of 2 10 points In the arrangement shown in the figure, the resistor is 4 and a 4 T magnetic field is directed into the paper. The separation between the rails is 7 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 7 m / s . Assume the bar and rails have negligible resistance and friction. m 1g 7 m / s 4 4 T 4 T I 7m Calculate the applied force required to move the bar to the left at a constant speed of 7 m / s. Correct answer: 1372 N (tolerance 1 %)....
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This homework help was uploaded on 04/10/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw9 solutions - homework 09 SMITH, TAYLOR Due: Mar 23 2008,...

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