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# 25 - Shie Gary Homework 25 Due Nov 1 2004 4:00 am Inst...

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Shie, Gary – Homework 25 – Due: Nov 1 2004, 4:00 am – Inst: Turner 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 3) 10 points A long solenoid carries a current I 2 . Another coil (of larger diameter than the solenoid) is coaxial with the center of the solenoid, as in the figure below. 2 1 Outside solenoid has N 1 turns Inside solenoid has N 2 turns A 1 A 2 The current I 2 is held constant. The energy stored in the solenoid is given by 1. U = μ 0 2 N 2 2 A 2 2 I 2 2. U = μ 0 2 N 2 2 2 A 2 I 2 2 3. U = μ 0 2 N 2 2 A 2 I 2 2 4. U = 1 2 μ 0 N 2 2 A 2 I 2 2 5. U = μ 0 2 A 2 2 I 2 6. U = μ 0 2 N 2 2 A 1 2 I 2 2 7. U = μ 0 2 N 2 2 A 2 2 I 2 2 correct 8. U = μ 0 2 N 2 A 2 2 I 2 2 Explanation: Basic concept: Magnetic flux magnetic field of a solenoid at the center B = μ 0 N I , magnetic energy, mutual induction. Solution: The magnetic energy density is given by B 2 2 μ 0 . Inside the solenoid the mag- netic field is B = μ 0 N 2 I 2 2 , and the volume enclosed by the solenoid is A 2 2 , so U = 1 2 μ 0 μ 0 N 2 2 I 2 2 A 2 2 = μ 0 2 N 2 2 A 2 2 I 2 2 . 002 (part 2 of 3) 10 points The mutual inductance M 12 between the coil and the solenoid is given by 1. M 12 = μ 0 N 1 N 2 A 1 2 2. M 12 = μ 0 N 1 N 1 N 2 3. M 12 = μ 0 N 1 N 2 A 2 2 correct 4. M 12 = μ 0 N 1 N 2 A 1 5. M 12 = μ 0 N 2 A 2 2 6. M 12 = μ 0 2 N 1 N 2 A 2 7. M 12 = μ 0 N 1 N 2 2 Explanation: The mutual inductance M 12 of loop 1 with respect to loop 2 is defined as M 12 N 1 Φ 12 I 2 , where Φ 12 is the flux through a single loop 1 due to loop 2. Since the magnetic field inside the coil is restricted to the part inside the solenoid, we have Φ 12 = μ 0 N 2 2 I 2 A 2 , so M 12 = μ 0 N 2 N 1 A 2 2 . 003 (part 3 of 3) 10 points Let the current I 2 be dependent on time, I = I 0 exp( - a t ), where I 0 = 1 . 6 A, a = 5 s - 1 , and t is measured in seconds.

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Shie, Gary – Homework 25 – Due: Nov 1 2004, 4:00 am – Inst: Turner 2 If the mutual inductance between the coil and the solenoid is 6 mH, what is the mag- nitude of the emf induced in the coil at time t 1 = 3 . 7 s ? Correct answer: 4 . 43398 × 10 - 7 mV. Explanation: The magnitude of the induced emf on the circular coil is E ind = M 12 fl fl fl fl dI 2 dt fl fl fl fl = M 12 I 0 a e - a t 1 = 4 . 43398 × 10 - 7 mV . 004 (part 1 of 2) 10 points A small-diameter inductor having a self- inductance of 120 mH and a large-diameter inductor having a self-inductance of 76 mH are connected in series as shown in the figure. Consider the two inductors to be far apart. Thus, the mutual inductance between the two inductors is M = 0 mH. a b Determine the equivalent self-inductance L eq for the system. Correct answer: 196 mH. Explanation: Let : L s = 120 mH and L = 76 mH . Let E be the emf across the larger-diameter inductor, E s be the emf across the smaller- diameter inductor, and E be the total emf across both inductors.
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25 - Shie Gary Homework 25 Due Nov 1 2004 4:00 am Inst...

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