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hw9 solutions

# hw9 solutions - homework 09 SMITH TAYLOR Due 4:00 am...

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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. d B dt = I a R 2. d B dt = I a 2 R 3. d B dt = I R a 2 correct 4. d B dt = R a I 5. d B dt = I R a Explanation: The emf produced is given by E = d φ dt = a 2 d B dt . Using Ohm’s law we can solve for B I R = a 2 d B dt d B 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 0 sin ω t , with I 0 = 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: Faraday’s Law E = d Φ B dt . Magnetic field in a toroid B = μ 0 N I 2 π r . Solution: In a toroid, all the flux is confined to the inside of the toroid B = μ 0 N I 2 π r . So, the flux through the loop of wire is Φ B 1 = integraldisplay B dA = μ 0 N I 0 2 π sin( ω t ) integraldisplay b + R R a dr r = μ 0 N I 0 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 d Φ B 1 dt = N μ 0 N I 0 2 π ω a ln parenleftbigg b + R R parenrightbigg cos( ω t ) = −E 0 cos( ω t )

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homework 09 – SMITH, TAYLOR – Due: Mar 23 2008, 4:00 am 2 where ω = 2 πf was used. The maximum magnitude of the induced emf , E 0 , is the coefficient in front of cos( ω t ). E 0 = N d Φ B 1 dt = N μ 0 N I 0 ω 2 π a ln bracketleftbigg b + R R bracketrightbigg = (23 turns) μ 0 (630 turns) × (73 . 8 A) (95 . 6 Hz) (2 . 41 cm) × ln bracketleftbigg (1 . 76 cm) + (5 . 78 cm) (5 . 78 cm) bracketrightbigg = 0 . 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 Ω 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 1 g 7 m / s 4 Ω 4 T 4 T I 7 m 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 %). Explanation: Motional emf is E = B ℓ v . Magnetic force on current is vector F = I vector × vector B . Ohm’s Law is I = V R . The motional emf induced in the circuit is E = B ℓ v = (4 T) (7 m) (7 m / s) = 196 V . From Ohm’s law, the current flowing through the resistor is I = E R = 196 V 4 Ω = 49 A .
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