983BhW44M7q9_1210016733_jwk572

# 983BhW44M7q9_1210016733_jwk572 - homework 09 KIM JI Due...

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homework 09 – KIM, JI – 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 2 R 2. d B dt = R a I 3. d B dt = I R a 4. d B dt = I R a 2 correct 5. d B dt = I a R 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 . 62 cm by b = 4 . 11 cm) and inner ra- dius 2 . 34 cm consists of N = 210 turns of wire that carries a current I = I 0 sin ω t , with I 0 = 31 . 6 A and a frequency f = 42 . 9 Hz. A loop that consists of N = 29 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 . 275601 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 – KIM, JI – 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 = (29 turns) μ 0 (210 turns) × (31 . 6 A) (42 . 9 Hz) (2 . 62 cm) × ln bracketleftbigg (4 . 11 cm) + (2 . 34 cm) (2 . 34 cm) bracketrightbigg = 0 . 275601 V |E| = 0 . 275601 V . Question 3, chap 31, sect 1. part 1 of 2 10 points In the arrangement shown in the figure, the resistor is 2 Ω and a 2 T magnetic field is directed out of the paper. The separation between the rails is 4 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 4 m / s . Assume the bar and rails have negligible resistance and friction. m 1 g 4 m / s 2 Ω 2 T 2 T I 4 m Calculate the applied force required to move the bar to the left at a constant speed of 4 m / s. Correct answer: 128 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 = (2 T) (4 m) (4 m / s) = 32 V . From Ohm’s law, the current flowing through the resistor is I = E R = 32 V 2 Ω = 16 A .
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