# hw13a - aldawsari(ma28683 HW13 Betancourt(16856 This...

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aldawsari (ma28683) – HW13 – Betancourt – (16856) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 29.1-29.3 001 10.0 points An 18 . 1 turns circular coil with radius 7 . 72 cm and resistance 5 . 85 Ω is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies in time according to the expression B = a 1 t + a 2 t 2 , where a 1 = 0 . 0537 T / s, a 2 = 0 . 0526 T / s 2 are constants, time t is in seconds and field B is in Tesla. Find the magnitude of the induced emf in the coil at t = 8 . 15 s. Correct answer: 0 . 308759 V. Explanation: Let : n = 18 . 1 turns , r = 7 . 72 cm = 0 . 0772 m , R = 5 . 85 Ω , a 1 = 0 . 0537 T / s , a 2 = 0 . 0526 T / s 2 , and t = 8 . 15 s . The area of the circular coil is A = π r 2 = π (0 . 0772 m) 2 = 0 . 0187234 m 2 , so from Faraday’s law, E = n d Φ B dt = n d A B dt = n A d B dt = n A ( a 1 + 2 a 2 t ) = (18 . 1 turns) (0 . 0187234 m 2 ) × [0 . 0537 T / s + 2 (0 . 0526 T / s 2 ) (8 . 15 s)] = 0 . 308759 V , which has a magnitude of 0 . 308759 V . 002 (part 1 of 2) 10.0 points In the arrangement shown in the figure, the resistor is 8 Ω and a 3 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 8 m / s . Assume the bar and rails have negligible resistance and friction. m 1 g 8 m / s 8 Ω 3 T 3 T I 4 m Calculate the applied force required to move the bar to the left at a constant speed of 8 m / s. Correct answer: 144 N. 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 = (3 T) (4 m) (8 m / s) = 96 V . From Ohm’s law, the current flowing through the resistor is I = E R = 96 V 8 Ω = 12 A .

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aldawsari (ma28683) – HW13 – Betancourt – (16856) 2 Thus, the magnitude of the force exerted on the bar due to the magnetic field is F B = I ℓ B = (12 A)(4 m)(3 T) = 144 N . To maintain the motion of the bar, a force must be applied on the bar to balance the magnetic force F = F B = 144 N . 003 (part 2 of 2) 10.0 points At what rate is energy dissipated in the resis- tor? Correct answer: 1152 W. Explanation: The power dissipated in the resistor is P = I 2 R = (12 A) 2 (8 Ω) = 1152 W .
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