solution11_pdf

# That tries to counter the flow of the current which

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that tries to counter the flow of the current, which in this case would be in the clockwise direction. 005 10.0 points An 83 . 6 turns circular coil with radius 3 . 44 cm and resistance 1 . 67 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 . 094 T / s, a 2 = 0 . 0807 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 = 3 . 42 s. Correct answer: 0 . 200769 V. Explanation: Let : n = 83 . 6 turns , r = 3 . 44 cm = 0 . 0344 m , R = 1 . 67 Ω , a 1 = 0 . 094 T / s , a 2 = 0 . 0807 T / s 2 , and t = 3 . 42 s . The area of the circular coil is A = π r 2 = π (0 . 0344 m) 2 = 0 . 00371763 m 2 , so from Faraday’s law, E = - n d Φ B dt = - n d A B dt

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taherisefat (mt23852) – hw 11 – Opyrchal – (121012) 3 = - n A d B dt = - n A ( a 1 + 2 a 2 t ) = - (83 . 6 turns) (0 . 00371763 m 2 ) × [0 . 094 T / s + 2 (0 . 0807 T / s 2 ) (3 . 42 s)] = - 0 . 200769 V , which has a magnitude of 0 . 200769 V . 006 10.0 points Given: μ 0 = 1 . 25664 × 10 - 6 N / A 2 . A 2 m long large coil with a radius of 16 . 1 cm and 390 turns surrounds a 4 . 4 m long solenoid with a radius of 6 . 1 cm and 7000 turns, see figure below. The current in the solenoid changes as I = I 0 sin(2 π f t ) , where I 0 = 30 A and f = 60 Hz. 4 . 4 m 2 m 16 . 1 cm 6 . 1 cm 12 Ω E = E 0 sin ω t Find the maximum induced current I max in the large coil. Correct answer: 8 . 59014 A. Explanation: Let : R = 12 Ω , r 1 = 16 . 1 cm = 0 . 161 m , r 2 = 6 . 1 cm = 0 . 061 m , A 1 = π r 2 1 = 0 . 0814332 m 2 , A 2 = π r 2 2 = 0 . 0116899 m 2 , N 1 = 390 , N 2 = 7000 , 1 = 2 m , 2 = 4 . 4 m , n 1 = N 1 1 = 195 turns / meter , n 2 = N 2 2 = 1590 . 91 turns / meter , I 0 = 30 A , and ω = 2 π f = 376 . 991 rad / s . 2 1 A 1 A 2 R E = E 0 sin ω t Basic Concepts: Faraday’s law E = - N d Φ dt and Ohm’s law E = I R . Solution: The angular velocity is ω = 2 π f = 2 π (60 Hz) = 376 . 991 rad / s . The solenoid carries a current I = I 0 sin ω t . The maximum magnetic field in the solenoid is B max 2 = μ 0 N 2 I 0 2 = μ 0 (7000) (30 A) (4 . 4 m) = 0 . 0599759 T , where B 2 = B max 2 sin ω t . The flux Φ 12 through coil 1 due to coil 2 (the solenoid) is Φ 12 = B 2 A 2 , so the mutual
taherisefat (mt23852) – hw 11 – Opyrchal – (121012) 4 inductance is M 12 = N 1 Φ 12 I = N 1 B 2 A 2 I = μ 0 N 1 N 2 A 2 2 = μ 0 (390) (7000) (0 . 0116899 m 2 ) (4 . 4 m) = 0 . 00911443 H , where A 2 = π (0 . 061 m) 2 = 0 . 0116899 m 2 . The induced emf is E 1 = -M 12 d I dt = -M 12 I 0 d dt sin ω t = M 12 I 0 ω cos ω t = (0 . 00911443 H) (30 A) × (376 . 991 rad / s) cos ω t = (103 . 082 V) cos ω t , where the maximum emf is E max 1 = 103 . 082 V . We can then determine the maximum cur- rent in the resistor using Ohm’s law. I max = E max 1 R = 103 . 082 V 12 Ω = 8 . 59014 A 007 10.0 points A powerful electromagnet has a field of 1 . 45 T and a cross sectional area of 0 . 696 m 2 . Now we place a coil of 229 turns with a total resistance of 1 . 68 Ω around the electromagnet and turn off the power to the electromagnet in 0 . 0316 s. What will be the induced current in the coil? Correct answer: 4353 . 28 a. Explanation: Basic Concepts: Faraday’s Law of Induc- tion: E = - d Φ B dt Ohm’s law: I = V R For coil of many turns, the emf is the sum of induced emf in each turn. Besides, when calculating the induced emf in the coil, we actually are try to get a average emf. For one turn of coil, the induced emf is: E 1 = - d Φ B dt = ΔΦ B Δ t = Δ B · A Δ t = 31 . 9367 V So, the emf in the whole coil is: E =

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• Spring '08
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