# 28 - Kapoor(mk9499 – homework28 – Turner –(60230 1...

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Unformatted text preview: Kapoor (mk9499) – homework28 – Turner – (60230) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In an RL series circuit, an inductor of 4 . 05 H and a resistor of 7 . 29 Ω are connected to a 29 . 2 V battery. The switch of the circuit is initially open. Next close the switch and wait for a long time. Eventually the current reaches its equilibrium value. At this time, what is the corresponding energy stored in the inductor? Correct answer: 32 . 4889 J. Explanation: Let : L = 4 . 05 H , R = 7 . 29 Ω , and E = 29 . 2 V . The current in an RL circuit is I = E R parenleftBig 1- e- Rt/L parenrightBig . The final equilibrium value of the current, which occurs as t → ∞ , is I = E R = 29 . 2 V 7 . 29 Ω = 4 . 00549 A . The energy stored in the inductor carrying a current 4 . 00549 A is U = 1 2 LI 2 = 1 2 (4 . 05 H) (4 . 00549 A) 2 = 32 . 4889 J . 002 (part 1 of 3) 10.0 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. ℓ 2 ℓ 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 2 A 2 ℓ 2 I 2 2. U = μ 2 N 2 2 ℓ 2 A 2 I 2 2 3. U = μ 2 N 2 A 2 ℓ 2 I 2 2 4. U = μ 2 N 2 2 A 1 ℓ 2 I 2 2 5. U = 1 2 μ N 2 2 A 2 I 2 2 6. U = μ 2 N 2 2 A 2 I 2 2 7. U = μ 2 N 2 2 A 2 ℓ 2 I 2 8. U = μ 2 N 2 2 A 2 ℓ 2 I 2 2 correct Explanation: The magnetic energy density is given by B 2 2 μ . Inside the solenoid the magnetic field is B = μ N 2 I 2 ℓ 2 , and the volume enclosed by the solenoid is...
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28 - Kapoor(mk9499 – homework28 – Turner –(60230 1...

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