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in this case would be in the counter-clockwise direction. 005 10.0 points A 39 . 2 turns circular coil with radius 6 . 92 cm and resistance 3 . 14 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 . 0141 T / s, a 2 = 0 . 0893 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 = 7 . 76 s. Correct answer: 0 . 825634 V. Explanation: Let : n = 39 . 2 turns , r = 6 . 92 cm = 0 . 0692 m , R = 3 . 14 Ω , a 1 = 0 . 0141 T / s , a 2 = 0 . 0893 T / s 2 , and t = 7 . 76 s . The area of the circular coil is A = π r 2 = π (0 . 0692 m) 2 = 0 . 015044 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 ) = - (39 . 2 turns) (0 . 015044 m 2 ) × [0 . 0141 T / s + 2 (0 . 0893 T / s 2 ) (7 . 76 s)] = - 0 . 825634 V ,
johnson (rj6247) – hw 11 – Opyrchal – (121014) 3 which has a magnitude of 0 . 825634 V . 006 10.0 points Given: μ 0 = 1 . 25664 × 10 - 6 N / A 2 . A 3 m long large coil with a radius of 15 . 1 cm and 280 turns surrounds a 7 . 1 m long solenoid with a radius of 5 . 1 cm and 6300 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. 7 . 1 m 3 m 15 . 1 cm 5 . 1 cm 16 Ω E = E 0 sin ω t 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 .
Find the maximum induced current I max in the large coil. The solenoid carries a current I =
johnson (rj6247) – hw 11 – Opyrchal – (121014) 4 where A 2 = π (0 . 051 m) 2 = 0 . 00817128 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 . 00255118 H) (30 A) × (376 . 991 rad / s) cos ω t = (28 . 8531 V) cos ω t , where the maximum emf is E max 1 = 28 . 8531 V . We can then determine the maximum cur- rent in the resistor using Ohm’s law. I max = E max 1 R = 28 . 8531 V 16 Ω = 1 . 80332 A 007 10.0 points A powerful electromagnet has a field of 1 . 39 T and a cross sectional area of 0 . 863 m 2 . Now we place a coil of 171 turns with a total resistance of 2 . 13 Ω around the electromagnet and turn off the power to the electromagnet in 0 . 0359 s. What will be the induced current in the coil? For coil of many turns, the emf is the sum of induced emf in each turn. Besides, when = ΔΦ B Δ t = Δ B · A Δ t = 33 . 4142 V So, the emf in the whole coil is: E = n · E 1 = 171 turns × 33 . 4142 V = 5713 . 83 V Apply Ohm’s law, we get the current as fol- lowing: I = V R = E R = 5713 . 83 V 2 . 13 Ω = 2682 . 55 a 008 (part 1 of 2) 10.0 points
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:
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